MYSQL :: Count User Comments In Each Category (was "help On A Query")

Count Query Causing "excessive Processor Usage"
For the second time in the past 6 months my hosts have shut down my site because it was apparently causing a shared mySQL 5 server to max out at 100% usage. I'm not much a programmer, know even less about mySQL and obviously don't have access to the server from the back end but I'd appreciate any advice as to whether it is my query that is causing the problem:

Set rsMail = objConnC.Execute("SELECT * ,(SELECT COUNT(*) FROM comments WHERE comments.submissionID = submissions.submissionID AND comments.commentInclude = 1) as CommentCount FROM submissions WHERE PigeonHole = 'mailbag' AND Status <> 'hold' ORDER BY submissionID DESC LIMIT 20")

Apart from two instances, this query has been working fine on a page that receives between 10,000 and 30,000 visits a day.

The site is running on Windows and coded in ASP but is calling a separate mySQL server.

Count Of *unique* Visits By Page For Each Day (was "Need Help With Query")
Using MySQL 4.1

I have a table 'orders' with the following fields:
ipaddress
pagenum
xtime (which is just the yearmonthday)

this is the query:
SELECT distinct pagenum,ipaddress,xtime,count(pagenum) AS pagenumtotal FROM `orders` group by pagenum,xtime ORDER by xtime, pagenum asc

What I am trying to do is return a count of *unique* visits by page for each day. This query returns a count of *all* visits by page for each day, including repeats. So if a visitors hits a certain page on the site 10 times and adds a record 10 times, the count result will include all 10 records in the count. I want this to only count as 1.

List All Clients Not Assigned To Current User (was "Query Help Please")
I have two tables like this:

*clients
-- client_id
-- client_name

*manage
-- user_id (multiple entries possible)
-- client_id

To assign clients to users there is a entry in the manage table of the
user_id and client id.

I want to list all the client_ids and client_names that are not assigned to the current user_id say 1.

Find All Games A User Has A Top 3 Score In (was "Need Help With A Query")
I am trying to create a query that will go through a scores table for me and find all the games that a user has a top 3 score in. Here are the table that are relevant I think...

CREATE TABLE files (
fileid int(11) NOT NULL auto_increment,
file text NOT NULL,
icon text NOT NULL,
filelocation int(1) NOT NULL default Ƈ',
iconlocation int(1) NOT NULL default Ƈ',
customcode text NOT NULL,
title text NOT NULL,
description text NOT NULL,
keywords text NOT NULL,
width int(11) NOT NULL default Ɔ',
height int(11) NOT NULL default Ɔ',
category int(11) NOT NULL default Ɔ',
timesplayed int(11) NOT NULL default Ɔ',
status int(1) NOT NULL default Ɔ',
filetype int(2) NOT NULL default Ɔ',
dateadded date NOT NULL default &#55612;&#57200;-00-00',
rating int(11) NOT NULL default Ɔ',
totalvotes int(11) NOT NULL default Ɔ',
totalvotepoints int(11) NOT NULL default Ɔ',
scores enum(Ɔ',Ƈ') NOT NULL default Ɔ',
PRIMARY KEY (fileid)
) TYPE=MyISAM AUTO_INCREMENT=2285 ;

CREATE TABLE scores (
scoreid int(11) NOT NULL auto_increment,
fileid int(11) NOT NULL default Ɔ',
score int(11) NOT NULL default Ɔ',
username text NOT NULL,
userid int(11) NOT NULL default Ɔ',
usercomment text NOT NULL,
dateadded date NOT NULL default &#55612;&#57200;-00-00',
PRIMARY KEY (scoreid)
) TYPE=MyISAM AUTO_INCREMENT=16 ;

CREATE TABLE users (
userid int(11) NOT NULL auto_increment,
username text NOT NULL,
password text NOT NULL,
email text NOT NULL,
status enum(Ƈ',Ɔ') NOT NULL default Ɔ',
usergroup int(11) NOT NULL default Ƈ',
joined date NOT NULL default &#55612;&#57200;-00-00',
played int(11) NOT NULL default Ɔ',
comments int(11) NOT NULL default Ɔ',
avatar text NOT NULL,
avatar_uploaded enum(Ƈ',Ɔ') NOT NULL default Ɔ',
location text NOT NULL,
website text NOT NULL,
gender int(1) NOT NULL default Ɔ',
favourite int(11) NOT NULL default Ɔ',
receiveemails enum(Ƈ',Ɔ') NOT NULL default Ƈ',
newpm enum(Ƈ',Ɔ') NOT NULL default Ɔ',
confirmation text NOT NULL,
PRIMARY KEY (userid)
) TYPE=MyISAM AUTO_INCREMENT=8 ;
(
Basically the query would need to search through the scores table to find if the specified user $userid has the top 3 (by top 3 i mean 1st, 2nd, or 3rd highest score) scores of any game in the scores table. The end resukt would be a list that contained the fileid of the game the user has a top 3 score in, the actual score, and the position (1, 2, or 3) Im not sure if I am even explaining this right cause im kind of tired. I am a novice MYSQL user but this just has me stumped

Add COUNT(*) To Already Existing Query (was "How To Do This")
I have this query:
PHP

SELECT entries.*, categories.*, uploads.*, entries.blog_id FROM entriesINNER JOIN categories ON (categories.category_id = entries.category_id) LEFT JOIN uploads ON(uploads.blog_id = entries.blog_id) LEFT JOIN images ON (images.blog_id = entries.blog_id) ORDER BY entries.date_submitted DESC LIMIT 3

Now i want to add this
PHP

SELECT COUNT(*) FROM comments

but i dont know how to add it into the already existing query. The messages table has a column blog_id column in it so i could do a join like thisd LEFT JOIN comments ON
PHP

(comments.blog_id = entries.blog_id)

but i dont know how to get the first bit in there

User Problems... Only Root User "works".
i have been trying to figure this strange problem out for a week or so. here it is.

i have a set of users with various permissions for different reasons. about 2 weeks ago, my hardware suffered a fairly nasty failure. i recovered everything. things seemed to work out. i didnt think to check my phpBB forums or my blog (or other mysql based stuff). never had any problems with the DB... so, maybe i am missing something simple here. i hope so.

what my users can do:

they can log in via command line, a gui i use on mac os x and it does not matter if it is local, or remote. they work just fine.

what all but one users cannot do:

none of my users can run phpbb or the blog stuff.. except for the root user. any other user returns the error message written in the code of which ever script i try to access telling me that "could not connect to DB" etc..

root works fine.

now, i thought maybe it was a problem with the forums and other stuff. but i cant be. same thing happens with all things... even a game server i occasionally run that keeps account info in a DB.

all these users are set up properly and have worked just fine in the past. nothing new has been installed, no conf changes..im at a loss...

Unable To Connect To Server (was "Novice User Problem")
I've been reading Kevin Yank's Build Your Own Database Driven Website and am unable to connect to the MySQL server as he describes in Chapter 4 (p. 74-75).

After modifying his code to produce a mysql_error, I get:

Unable to connect to server at this time.Client does not support authentication protocol requested by server; consider upgrading MySQL client

Even though I am using the latest MySQL server (5.0).
My
?php

$dbcnx = @mysql_connect('localhost', 'root', 'my_actual_password');
if (!$dbcnx) {
exit('Unable to connect to server at this time.' . mysql_error() . '');
}

if (!@mysql_select_db('ijdb')) {
exit('Unable to locate the joke database at this time.');
}

?>
Here are all the jokes in our MySQL database:</P.
<blockquote>
<?php
$result = @mysql_query('SELECT joketext FROM joke');
if (!$result) {
exit('Error performing query: ' . mysql_error() . '');
}
while ($row = mysql_fetch_array($result)) {
echo '' . $row['joketext'] . ''
}
?>

Access Denied For User 'root'@'localhost' (was "help!!! Just A Beginner..")
Hello i recently installed WAMP5 to handle my phpmysqlapache localhost sever. but when i go to the control panel and clicked the localhost and phpmyadmin everything seems fine because i can see the phpinfo window and also the phpmyadmin.. but not when i clicked the SQLite Manager.Unable to connect to database manager.
Could not connect: Access denied for user 'root'@'localhost' (using password: YES).

ERROR 1045: Access Denied For User: '"root"@localhost' (Using Password: YES)
I have been working on this for three days (about 16 hours) to no avail. I have root access on a dedicated server running whm/cpanel on Redhat 9. I have followed every instruction from mysql.com as well as every suggestion I've been able to find.

These are the two errors I bounce between:

ERROR 1045: Access denied for user: '"root"@localhost' (Using password: YES)

and

ERROR 2002: Can't connect to local MySQL server through socket '/var/lib/mysql/mysql.sock' (111)

I'm out of ideas.

How To Change "#1226 - User 'root' Has Exceeded The 'max_questions' Resource" Error
I seem to have set a max_questions variable on my system and my admin console and command prompt don't work with this response:

"#1226 - User 'root' has exceeded the 'max_questions' resource"

How to fix?

And how do I not make thisw mistake again?

Displaying User Last Login Like "3 Days, 2 Hours Ago"
I have a table "user" with a field "lastlogin" that stores a basic datetime of the last time the user logged in
Do you know of a fast function or class that can convert that field into a more readable "User last logged in: 3 days, 2 hours ago" or something similar?
I've seen this everywhere and I don't like reinventing the wheel

What Privileges Need A User To Run "SHOW SLAVE STATUS" ?
i use mysql 4.1.22
i need to create an user who must have rights to run only "SHOW SLAVE STATUS" query.
from online documentation i wasn't able to find what privilege i must grant to this user to be able to do only this query.

How To Fix "Column Count Doesn't Match Value Count At Row 1" ?
I've been trying to install a portal to my installation of Invision Power Board, and I keep coming up with this problem. I post here for two reasons: firstly is that no one is responding over at the portal creator's thread and I need help, second is that this error looks like a mySQL error that could be solved without the need of the creator if possible.

Here is the error returned:

mySQL query error: INSERT INTO `ibf_portal_box` VALUES (1, 'lang.last_topics', '<box><menu>10,20,30,50</menu>
<topics_show>10</topics_show>
<order>last_post</order>
<exforums></exforums>
</box>', 0, 'middle', '*', 0, 4, 'last_topics', '1')

mySQL error: Column count doesn't match value count at row 1
mySQL error code:
Date: Tuesday 21st of December 2004 08:43:10 PM

I am very new to mySQL and don't really understand it. What is wrong with the above that it is trying to insert, and how is it fixed?

User Names For Both Players (was "I'm In A Bit Over My Head")
I have two tables:
* users, with field user_id and user_nick
* games, with field player1_id and player2_id
games.player1_id and games.player2_id are foreign keys for users.user_id

If I only have one player in games I would do

SELECT games.player1_id, users.user_nick FROM users, games WHERE users.user_id = games.player1_id
However, now that i have two players, and i want the nicks of them both, how would I go about it? Is it possible in one query only?
Of course this the following doesn't work, but something like this?:

SELECT games.player1_id, games.player2_id, users.user_nick AS player1_nick, users.user_nick AS player2_nick FROM users, games WHERE users.user_id = games.player1_id

"Access Denied For User" Mystery
I'm having an odd problem. I have a MySQL database, and am trying to connect to it through PHP. I've done this before. However, when I try to connect a particular user, I get:

Could not select database because Access denied for user 'myuser'@'localhost' (using password: YES)

If I change my PHP script to login as root, everything works fine, so it doesn't seem to be a PHP&MySQL configuration issue. Also, I CAN login as "myuser" through the MySQL command prompt AND access the database I'm trying to use in the PHP script.

Any ideas why a user would be able to access a DB through the command line, but not through a PHP script? I granted them all privileges on the database. I've retyped the password several times.

What Is The Sql Way To Restrict "superadmin" Priv To One User?
Is there some way to restrict in mysql that certan field has only ONE ROW
with CERTAIN VALUE X while other rows can have any values but not this
one(so "unique" is not the answer..)?

This can be of course done in application level, but it would just be nice
to know. I would build my application so that only one user can be
superadmin, while others are normal admins or something else.

Of course this restriction would not be enough anyway, because I still need
to figure out what happens in different cases. So actual benefit would be
close to zero, but nerd idealist inside of me requires to know - "we have a
right to know".

"Warnings" Count When Using Mysqlimport
When using mysqlimport, the final statistics include a count of
"Warnings", for example:

Connecting to foo.whatever.com
Selecting database foo
Loading data from LOCAL file: foo_userInfo.imp into foo_userInfo
foo.foo_userInfo: Records: 248 Deleted: 0 Skipped: 6 Warnings: 248
Disconnecting from foo.whatever.com

I'd like to know what the warnings were, however there doesn't appear
to be a command line option that might be used to expand the statistics
or otherwise output the text of the warnings (--verbose does not do
it).

Is there any way to view the warnings?

Count(*) Returns "Resource Id#3"
I have table:

CREATE TABLE users(
id INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
nick varchar(32),
passw varchar(32),
email varchar(128),
email_poskytovatel varchar(64));

and I want to read variabile id . I am using command SELECT Count(id) FROM users. But it returns "Resource id#3". Where is my problem?

About "Select Count(*)" Always Returns 0.
my mysql version is:

Ver 14.7 Distrib 4.1.14, for pc-linux-gnu (i686) using EditLine wrapper

OS: uname -a
Linux backup_srv1 2.4.29 #4 SMP Thu Mar 24 10:31:26 CST 2005 i686 unknown unknown GNU/Linux

Desc:

My statement is
"SELECT COUNT(1) FROM t_card_info_035 WHERE Fcommodity_id = 0x1111 AND Fseller_uin =111111 AND Fdeal_id = 12345;"

my client is a daemon, when it started, it can return the correct result, (such as 4 ) but after two or three days later, it always returns 0. I had gdb into it and find that the result in the result row is 0. but when I restart my daemon program, it will be ok again. This made me crazy... it is a mysql c API bug or something else?

My daemon client is programmed by c++ , and use the c API to connect mysql. I had check that there is nothing wrong with my code, and it has no memory leak or somthing else.

Searching For Numbers - "close To" Or "approximate"
I would like to know how to search a table for records where a particular field is "close to" a particular number.

For instance, let's say I have a table containing students and their most recent test scores. I'd like to see students who scored 7/10. That's easy enough (select * from table where score = 7). I'd also like to run another query to identify other students whose scores are not 7 but "close to" 7. Ie students who scored 6 or 8.

As an added bonus, I'd like to be able to do a related search that shows students who scored 7, then shows the other students ordered by how "close" they are to 7. Ie students who scored 6 or 8 are "closest" matches, and students who scored 1 are "furthest". I hope this makes sense!

Is there any pre-existing MySQL command that will do this? Or will I have to run separate queries for each value?

REQ: MySQL 4.0 Equivilent Of "DateDiff("m", Date, Now())=1"
I have a query that works in the rest of the SQL world

"SELECT invoice.*, invoice.Date
FROM invoice
WHERE (DateDiff("m", Date, Now())=1);"

which will give me all of last month's invoices.

However it doesn't work with MySQL 4.0.

While

"SELECT
Invoice.*
FROM
invoice
WHERE
(Month(Invoice.Date) = ((Month(Now()))-1))"

Is a close substituent, it will blowup in January.

Anyone have an equivalent expression for MySQL 4.0?

Get Average Of A "count"
How to get average of a count(user);
Following syntax run very well
$records = mysql_query( "SELECT user,action,count(user) FROM log where date='$datetr2' and action='add' group by action,user")

But now I also want avg(count(user)), but it didn't work

$records = mysql_query( "SELECT user,action,count(user) ,avg(count(user)) FROM log where date='$datetr2' and action='add' group by action,user")

count(user) only get hot many times a user log into the webside a day. Right now I want to know the average times of user logging into the web within month. I don't know to get average number. avg() function doesn't work here.

Create Single Query From Queries On Two Tables (was "Help With Query...")
I read from other thread that query inside loop is not good idea. May I ask some help how can I create a single query to the following code which I use loop.

$sql = "SELECT * FROM mytable order by points desc limit 10";
$rec = mysql_query($sql) or die(mysql_error());
$datas = mysql_fetch_array($rec);

do{
$sq = "Select * from secondtable where linkid = '$datas[id]'";
$rst = mysql_query($sq) or die(mysql_error());
$rows = mysql_fetch_array($rst);
echo "$rows[somefield]";
}while($datas= mysql_fetch_array($rec));
This works perfectly but I want the second query to be out of the loop if there is a way and how.

Grab 'title' From The Table 'forum' Within The Same Query (was "Help With Query")
I have the following query for my vBulletin database:
PHP

$get_stats_newthreads = $db->query_read(" SELECT thread.forumid, thread.postuserid, thread.postusername, thread.threadid, thread.title, thread.lastpost, thread.forumid, thread.replycount, thread.lastposter, thread.dateline, thread.iconid, thread.views, IF(views<=replycount, replycount+1, views) AS views, thread.visible FROM " . TABLE_PREFIX . "thread AS thread WHERE NOT ISNULL(thread.threadid) AND $weekold<lastpost AND thread.visible!=0 AND (forumid=34 OR forumid=7 OR forumid=8 OR forumid=11 OR forumid=10) ORDER BY rand() DESC LIMIT 5");

and would like to grab 'title' from the table 'forum' where forum.forumid=thread.forumid

Count(*) In A Select Returns "1". It Should Behave Like Select Count(*)
i'm trying to make a query work properly but I got lost:

SELECT *,count(*)
FROM cancons c, musics m, discos d, r_discos_cancons rdc
WHERE c.c_id_music = m.m_id
AND rdc.rdc_id_disc = d.d_id
AND d.d_id_music = m.m_id
AND m.m_id = 24
GROUP BY c_id

note:
cancons (ca) = songs (en)
discos (ca) = cd's (en)
music (ca) = musician (en)

don't worry for the WHERE part. i need it because of the foreign keys.
this query returns a table with the title of the song and some more data. on the right side, I get another column called "count(c_id)" with the number "1" in it for each row. That's supposed to be due to the "group by" clause, I think.

I'd like to get the table with the songs, as usual, and, with the same query, I'd like to get the total number of rows selected.

Design For Stock Items And Checkin/checkout Process (was "Query Help")
I'm currently in the process of trying to bodge together a simple(ish) stock system to be used on a company intranet but I've ground to a halt with a particular problem.

Using a MySQL database I've got tables with 11 fields, including 'Quantity' (self explanatory) and BookOut. Now this BookOut field contains information on the user who has booked out an item including the users name, the date/time etc

The problem I have is this. If an item has a quantity of 7 and a user books out 1, then his/her info is stored in BookOut but when another user books out another 1 of said item at a later stage then the previous BookOut info is overwritten.

Is there a way in that everytime another item is booked out that a new line is created within a specific field?

}

if(!sizeof($error)){

$update_SQL = "Update joystick Set BookOut = '" . mysql_real_escape_string($BookOut) . "', PCserial = '" . mysql_real_escape_string($PCserial) . "', Quantity = Quantity - " . $Quantity . ", MachineNo = '" . mysql_real_escape_string($MachineNo) . "', Date = now() WHERE ID = " . $ID;
if(! @mysql_query($update_SQL)){

$error[] = "Update SQL failed";

}
The above is the query I am currently using. I'm not totally proficient with .php and have already had a lot of help getting this far.

Query To Populate A Calendar (was "Wanted The Experts Opinions (that Means YOU!)")
I am building my first calendar for a client to list All their events on a month to month basis.

There are several tables of data that I need to query to populat the calendar from: Events, Committee Meetings, Lectures, and Group Actions.

I was wondering:

#1) Is it more effecient to query all the tables once and to store the results into an array that I can check as I loop through and output the days for the calendar?

#2) OR, is it better to query all the tables for each day as I loop through and construct the calendar ouput?

After formulating my question it seems like a no-brainer: That option #1 would be the most effecient, but I still would like to know your opinions. Especially if you have made a calendar before.

What Is Execution Time Of A Query Based On? (was "a Mysql Question")
when selecting data from the database, does the time taken to retrieve it vary from 56k connections to T3 connections? or does it all depends ont he general server speed/amount of connections on the db?

im not sure if ive explained that in the best way for people to understand
but im sure someone will get what i mean

Unique Combinations Of Model And Make (was "Query Question")
I'm trying to write a query and was wondering if someone could help me formulate it. Here is an example table I'm working with:

Table: vehicles
Model Make
Ford Taurus
Ford Taurus
Ford F150
Chevy Tahoe
Chevy Blazer
Chevy Tahoe

I want a query that returns all unique combinations of model and make. So the result set should be:
Model Make
Ford Taurus
Ford F150
Chevy Blazer
Chevy Tahoe

I know how to return distinct values for one variable, but not for two like that. Any help is greatly appreciated.

Adjacent Theater Seats (was "mySQL Query Issue")
Having not been working with more than a simple SELECT * from table query, I am having a few issues when it comes to using more complex queries.

I am currently using a query to select seats from a database using the following query:

$query1 = mysql_query("SELECT struct.seatid, struct.row, struct.seatnum FROM seats as seat LEFT JOIN structure as struct USING (seatid) WHERE seat.showtimeid=".(int) $_GET['id']." && seat.available = Ƈ' && seat.timelimit < $minutes2 ORDER BY seat.seatid ASC");
For each record that is selected I want to check if either of the seats either side of it are also selected, if not then it is not to be shown.

I have added a while loop below the above query and run another query following it (not sure if that makes sense, so I will paste it below!)

while($rows = mysql_fetch_assoc($query1)){

$minusseat = $rows[seatid] - 1;
$plusseat = $rows[seatid] + 1;

$queryminusseat = mysql_query("SELECT * from seats LEFT JOIN structure WHERE seat.showtimeid=".(int) $_GET['id']." && seat.available = Ƈ' && seat.seatid == $minusseat && seat.timelimit < $minutes2");

$num_rows = mysql_num_rows($queryminusseat)or die(mysql_error());

echo "$num_rows";

//if($minusnum > 0){
//echo "$minusseat " . $rows[seatid] ." $plusseat ";
//} else {
//echo "$minusseat " . $rows[seatid] ." $plusseat ";
//}

}
The code above is used below the initial query and the following error is being displayed:

Quote:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:Program Filesxampphtdocsprojectprototype1 ickets.php on line 154
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE seat.showtimeid=1 && seat.available = Ƈ' && seat.seatid == 1 && seat.time' at line 1

Invalid Use Of Group Function (was "Baffled By Query Error")
trying to figure out why I keep getting this error with the following query:

SELECT c.account_id,a.name,a.company,SUM(c.agent_product_time) AS mins
FROM account a LEFT JOIN calls c ON c.account_id = a.id
WHERE c.calldate >= DATE_SUB(NOW(),INTERVAL 14 DAY) AND c.agent_id = 2
AND SUM(c.agent_product_time) >= '500' GROUP BY c.account_id ORDER BY mins

ERROR: #1111 - Invalid use of group function

Most Products Flagged In The Past 24 Hours (was "Query Help")
I need a little help setting up the logic behind a query. Here is the situation:

I have 2 tables. Product table. Flag table (flagID, productID, date_flagged). Users can flag products. *date_flagged is a INTEGER unix timestamp.

I would like to be able to setup a query where I can pull up products that have been flagged the most in the past 24 hours.

At first I figured I would COUNT() the flags and GROUP BY productID, but this of course will only give me the most flagged all time, and because of the grouping by productID will not allow me to filter it by date.

User_id And Password Error? (was "mysql Query")
Following Query right or not

$sql = "SELECT * FROM login WHERE user_id = ".$username." AND password = ".$passwd."";

if error then point out error

Delete Except Latest 2500 Entries (was "help With Query")
I have a table that holds information on activities.
For each activity I store the following:
ID, When, Description
The ID is the key for the table
The When column holds the datetime when the activity was added

The table gets filled up pretty quickly.
I would like to run a query that will delete ALL the entries except for the 2500 latest ones. The When column should be the one to use when deciding which are the 2500 latest activities

Can someone show me the most efficient way of doing this?

Next And Prev DisplayOrder Values (was "Help With MySQL Query")
I have a table 'Album' containing columns 'PageID' and 'DisplayOrder' (both INT).
On each page that is displayed I want to have 'previous' and 'next' links to other pages, which are arranged via the DisplayOrder column.

At the moment I'm doing this in 2 separate queries:

SELECT DisplayOrder FROM Album WHERE PageID = $currentPage
$currentNo = [result of last query]
SELECT PageID FROM Album WHERE DisplayOrder IN ($currentNo-1, $currentNo+1)
Highly simplified but you get the idea. Would it be possible to combine this into one query?

Script To Change All Columns (was "mysql Query")
I have a table. I need to make some changes like change column name, change int() to var() etc. So I want to write a query that alter all the columns of the table then the data which were in the database before can be reinserted.

Models Where The Years Overlap (was "Query Help Needed")
I'm trying to formulate a query for the following example:

Table: model_by_year
model start_year end_year
model_1 2001 2004
model_1 2003 2004
model_2 2000 2005
model_2 1998 1999
model_3 1998 2003
model_3 2002 2006

I want a query that will return all like models where the years overlap. In other words, I'd want to return model_1 and model_3, but not model_2.

Dates In The Past 3 Weeks (was "SELECT Query Help")
I'm trying to select data from a table that has been submitted in the past 3 weeks. I am not having any success.

"Lost Connection To MySQL Server During Query"
Bog standard SuSE 8.2 Linux install.

I have installed mySQL (off the SuSE 8.2 distro CDs). I can connect locally,
e.g:

bealzebub> mysql -u root -p mysql -h localhost

....logs me in just fine. However, this:

bealzebub> mysql -u root -p mysql -h bealzebub

....causes the connection to fail with: "ERROR 2013: Lost connection to MySQL
server during query". Its the same if I try to log in from another machine.

"Lost Connection To MySQL Server During Query"
Hi, I wonder if anyone can help me with this.

Bog standard SuSE 8.2 Linux install.

I have installed mySQL (off the SuSE 8.2 distro CDs). I can connect locally,
e.g:

bealzebub> mysql -u root -p mysql -h localhost

....logs me in just fine. However, this:

bealzebub> mysql -u root -p mysql -h bealzebub

....causes the connection to fail with: "ERROR 2013: Lost connection to MySQL
server during query". Its the same if I try to log in from another machine.

Anyone?

How To Get The Total Number Of Rows With A Query "limit" ?
I would like to paginate the results of a query on several pages. So I
use a query with a limit X offset Y to display X results on a page,
ok.

But for the first page, I need to run the same query with a count(*)
to know how many pages I will get (number total of rows/ X).

The problem is my query is very slow (maybe 5s) because there is much
worch to do, and on the first page, I need to run this query twice
(not exactly, but ...) so the page is very very slow to load.

My question is : is there a function to get the total number of rows
even on a query with "limit" ? Or what could I do else?

Value "000" Stored As "0" In Text Field?
I've managed to import my data from a CSV using SQLyog Enterprise, however, I have a problem with the table storing the text value "000", MySQL truncates it to "0", unfortunately, I need the value to be stored as "000", does anyone know of any way that I can force the field to store all the characters, rather than treating it as a number? Surely a text field should treat any numbers stored within it as text?

What Is The Meaning Of The Letters "AB" In "MySQL AB"?
What does "AB" stand for in "MySQL AB".

Return A "true" Or "false"
i wanna do a "select", where if the script finds any matches, i dont wanna know about them. Instead, i just want it to return a a value: true (1) ou false (0).

ie.:

$sth = $dbh->prepare("SELECT * FROM maps WHERE name='algo' LIMIT 1");
$sth->execute();

if (match is true) {
print "True Match. A match was found";
}
else {
print "False Match. No matches were found.";
}

"Group By" Forgets "Order By"
I'm trying to build a simple RSS reader for a client and am running into some serious problems with the GROUP BY in MySQL. Basically I have two tables: one that holds all the entries and one that holds all the feeds. Then what I can do is say:
[MYSQL]SELECT * FROM entries WHERE feed_id='13' ORDER BY date DESC[/MYSQL]
That will select all the entries that belong to feed #13. This is fine, but what I'm doing now is building a library that will display the feed_id with the latest entry with that id. So I would think it would be something like the following:
[MYSQL]SELECT * FROM entries GROUP BY feed_id ORDER BY date DESC[/MYSQL]
That should get me the single latest entry from each feed_id. However it's forgetting the ORDER BY clause all together and seemingly ordering it by the auto-incrememnt value in the entries field.