PHP :: Display The Latest Database Entry?

REQ How Would I Compare Multiple Date Fields In One Table To Find The Latest Entry Opps
i didn't explain it correctly before
my table is like this

example fields:
ID name username outcome date1 date2 date3 (etc..) - date15 price1 price2 price3 (etc..)

I know that Mysql query order by will compare records on a specific date, but how do i compare
multiple fields within the same record. Want to find the latest date within the record..

Display Latest Input Without Refresh Page
I have a mysql database that store the results of a game. This database will frequently update (new result will insert after game finished). There is any script ( i think javascript) that will display the latest result without refresh the page.

Developing A Page That Displays Latest 5 Images In A Database
I am currently changing my web site so that it is PHP and mySQL powered. On the front page I have an area that will display the latest images in my portfolio and the latest text description.

Therefore, what I need is an admin page which allows me to upload images to a database which has a timestamp/date field, a thumbnail field, an image field, a text field for the title and a text field for the description. I also need a HTML form with inputs for the thumbnail upload, image upload, title and text description (in a textarea).

It would also be useful to have the page display the database contents and allow me to edit the details and also, delete rows. I then need code for that will display the most recent 5 thumbnails and the title on my index.php page.

Mulitple Database Entry
If I want to enter some info into 2 database's with some fields being the same eg name and pass and the key 'uid' to be the same i got this code and it give me a error, PHP Code:

How To Change A Number Entry In A Database
I need to change an entry in a database and I don't know how...they said..i should use update..but still ..it's all too vague for me... See, I need to modify a function which deletes ads. The ads are grouped in one table aptly named "ads".

Now, in a seperate table "category" is where I have saved the total number of ads in a record aptly titled = "total". These two tables have two things in common, they are joined by the category number=catid

Now, my problem is that whenever somebody deletes an ad. It just does that. It deletes the ad in table "ads". Now I need to change the total number of ads ("total") in table "category". Please do tell me how do i do this?

Link Entry In MySQL Database
i have a database with rateid, bandid and a rating,

how can i like each bandid to different pages, so bandid 1 will link to bandid 1's profile page and bandid 2 will link to bandid 2's profile page. any ideas?!

PHP Devloping A Post Variable Name With A Database Entry
I am developing a regisration page that will present a set of check
boxes to the viewer. This list of checkboxes is developed from a list
in a database so the amount and names of the boxes will change. This
portion works just fine.

The page content has a user name and password along with several sets
of these checkboxes. All the form content of this page is presented in
one form.

The problem:

I want to store the values presented in these checkboxes back into the
database based on what the user selects. The data will be stored in a
table that has the id for the user and the id for the checkbox item.
As anyone knows, checkbox's only send a boolean value through POST so
once the data is sent it is nearly impossible to test for its value
and work on it since these are dynamically presented.

What I mean by dynamically presented is that the list is pulled from a
database. The checkboxes are build as such: <p><input
type="checkbox" name="{$row['name']}">{$row['name']}</p>. $row is
the reference to the database result set that has been affected by the
row_assoc function.

Since I have no idea what the contents of $row['name'] will be. I
cannot use isset($_POST['checkboxname']) to test its value as you
would normally do. As I see it 'checkboxname' is just a string and so
it seems logical to me that this would work:
isset($_POST['$row['name']']), but it doesn't.

My request:

I need to be able to check to see if the checkbox is check like such:
isset($_POST['$row['name']']) so I can check the value of
$_POST['$row['name']']. What is the correct way to run this check if
it is at all possible? Or is there another way to do this or a more
effective way to do this?

Pre-selecting Options In Html-form According To Database Entry
I've got a html-form consisting - among other fields - of several drop-down-boxes and radio button groups. With this form you should be able to either create a new dataset or to edit an existing one.

If the latter is the case, my application should propose values (the ones which are already in the database) and write them into the input-fields. My problem is, it should have the according options pre-selected and the checkboxes checked, and I've no idea how to do this.

(Example: I've a dropdown box for languages, and my dataset says the language is English, then in my dropdown box the option "English" should be preselected when this dataset is to be edited).

Database Display
I am calling all fields from a table, just to many to call the ones i need. I would like to drop the first 2 from being shown, how can I do that by the code below or do I need to rewrite completely?

while ($line = mysql_fetch_array($store_data, MYSQL_ASSOC))
{
foreach ($line as $data)
{
echo $data . ', ';
}
}

Display Database Record
i have a php script that read from mysql and display then in textbox.
I have problem with the way textbox display the data.

<input type=text name="package_name" size=20 maxlength=10
value='.$package['package_name'].'>

when the package_name is 'china tour' it will only show china in the
text box. How can i show all the content of the record?

Display A Div Only If A Value In Database Is Equal To 1
Im trying to get a div to to only display if the the row in the database called "stock" is equal to "1" if it is equal to "0 " then do not display the specific div.
Here is the code:

Display Then Update Sql Database
I have an html drop down list that I have populated from a sql database, a selection is made and the information on the selection retrieved from the database and output in text forms on another page. I want to be able to edit or add data and then update the database. Code for the second page is below.

I have echo'd the post variable and it does make it through. From what I understand the post variable is being removed when I submit the second form and so it all fails. How can i sustain the variable from the first page? Code:

Image Display From Database
i have one problem in my programming.there is a table containing some details of the product with its image.. in view section in my php page ,i need to show this image and the rest of the fields in the table in a row...

How To Connect To Database And Display It In Flash?
how can i connect to mysql database and display the data in flash?

Don't Display Duplicate Database Entries?
I have a restaurants database that has a lot of restaurants listed with the same name in the same city with only the category changed. This is the code that I have that displays the list of restaurants. Can someone tell me what to add to it so that it will only display one of each restaurant of a certain name? PHP Code:

How To Display Data From Database In Different Web Pages?
When we get data from database by use mssql_query(),suppose the data is very large ,it has many rows, and we want display the data in different web pages like many search engines do, my question is how to do that using the same result without call mssql_query() again?

Trying To Display Image From Mysql Database
I have a mysql db, that holds images.
Images are encoded using base64_decode/encode, etc. Image data seems
fine.
I have a view.php page that is supposed to show the image, and an
image.php page that accesses
the database, retrives the image data, and then (theoretically) prints
the decoded data to the page.
below is the view.php page code: problem area the img tag src no
worky.

What am i missing (having stolen the idea for this from the web!!!!
thanks,
eric
/************************************************** *****************/
<html>
<head>
<title>Get The Image</title>
<link rel="stylesheet" type="text/css" href="css/csssmall.css" />
</head>
<body>
<center><img src="testimage/Sunset.jpg" width="300" border="1"
alt="Image of Sunset"></center>
<p class="para-color"><?php echo 'Beautiful Image' ?></p>
<br>
<center><img src="image.php?img=3" width="200" border="1" alt="Image of
Thelma Todd"></center>
<p class="para-color"><?php echo ''Nother Beautiful Image' ?></p>
</body>
</html>
/************************************************** ****************/

below is the image.php code:
/************************************************** ********************
<?php
$dbhost = 'localhost'
$dbuser = 'auser'
$dbpass = 'apassword'
$dbcnx = @mysql_connect($dbhost,$dbuser,$dbpass);

if (!$dbcnx)
{
echo( "connection to database server failed!");
exit();
}
if (! @mysql_select_db("portfolio") )
{
echo( "Image Database Not Available!" );
exit();
}
?>
<?php
$img = $_REQUEST["img"];
?>

<?php
$result = @mysql_query("SELECT * FROM images WHERE id=" . $img . "");

if (!$result)
{
echo("Error performing query: " . mysql_error() . "");
exit();
}
while ( $row = @mysql_fetch_array($result) )
{
$imgid = $row["id"];
$encodeddata = $row["sixfourdata"];
$title = $row['title'];
}
?>
<?php
//echo $encodeddata;print??
echo base64_decode($encodeddata);
//echo $prefix . $encodeddata;
?>
/************************************************** ***********************

Display Randomized Questions From A Database?
I would like to store 300 questions in a MySQL database, and configure a webpage to display them in groups of 50 at random, then store the responses. Could anyone recommend a script or strategy to do this please?

How Do I Display All Database Entries Between Two Dates
I have two dates in a form " startdate" and enddate"

when i display the results i would like to show all results between, and including the two dates. how do i do this... this is my current query:

My Variables:

$userid = $_GET['idr'];
$fromdate = $_GET['fromdate'];
$todate = $_GET['todate'];

The query:

$result = mysql_query("SELECT date, status, time, notes FROM timeclock WHERE userid='".$userid."' ORDER BY time");

Function To Display Form From Database!
I'm trying to write a function which gets all the client names from a table and displays them in a drop down list. My function is displaying a form with nothing in! Code:

How To Connect To Mysql Database And Display It In Flash?
i made a code for news that get the data in php and store it in database, then display it using flash.. how to do this..

#news database fields:
newstitle
newsbody
newsdate

#variables:
$newstitle=$_POST['title'];
$newsbody=$_POST['body'];
$newsdate=date("Y-m-d");

Storing Some Html Text In Database & Display Them
i have some text in HTML format ..i just want to store HTML text in database and retrive and display them as a HTML ...

Display Results From A Database On Multiple Pages
I am new to PHP and Mysql but am really determined to learn these techniques. I have some script that reads and displays results from the databse onto a page. However, I really wish to limit the number of results that are displayed to 10 per page and then have the next/privious links under the results. With these results, the results under the email field should be hyperlinked so that the person viewing the results should double click on the email link and send the person an email. How can i do that? I have to learn this technique.

The code is:-

How To Make Display Links From Mysql Database?
i have the following problem and i am breaking my mind on it:
I have a mysql database with (id,title,topic,body) in the database i have
stored some documents.
How can i display a link with only the title of the document stored?(easy
one) but...
I want to view the whole document in the same page!
Let me explain in code:

<?php
$ip = "localhost";
$user = "kostas";
$password = "kostas";
$basename = "test";
$table ="docs";
$db = mysql_connect($ip, $user, $password);
mysql_select_db($basename,$db);
$result1= mysql_query("SELECT * FROM $table ORDER BY id DESC LIMIT 5 ");
$myrow = mysql_fetch_array($result1);
$title = $myrow["title"];
$intro = $myrow["intro"];
$body = $myrow["body"];
?>

PHP Function To Display A Database / Table Structure?
My webhosts cpanel is down for maintenence, including phpMyAdmin, and I need to view my database structure... Is there any function in PHP to display the database structure (like the "Export" function in PHPMyAdmin)?

Display Items From A Mysql Database In A Table
I am trying to display items from a mysql database in a table but I need it to start a new table row each time multiples of 3 rows of the database table have been displayed like the following example:

item 1 item 2 item3
item 4 item 5 item 6

Does anybody know how I can make this work using php.

Submit/Approve/Display Content In Database In PHP?
I want a page where users can submit an image, text, and links and then what is submitted goes to an administration page where such submitted can be approved or declined. And then another page where all the things that have been approved go into a table where the image, text, and links are displayed in a numbered fashion. Here are some picture examples of what I mean: Code:

Display Links From Database, Constructs And Expressions
My users need to enter links into a MySQL text field (along with their text) to a file on the server.

For example; the file <a href="somefile.pdf"> somefilea is really big.However, I also want to calculate the file size for them.

I've used some of your code to do this, <?php getFileSize('somefile.pdf') ?>, so that the user would enter their text as; the file size <?php getFileSize('cover.pdf') ?> is really big.However, I've been told that putting PHP code in the database field is both problematic and requires the use of eval().

But, I've also been told that I can use something like:
the file {link:somefile.pdf} is really big.Then, when the text appears on the page a script looks through the text and constructs an html link from the text {link:somefile.pdf}.

But ... how? What would that script look like, any clues?

How To Display Mysql Database Information In Multiple Pages?
how to display mysql database information in multiple pages? i would like to display all my member with consist of 400 people and i wan it list in mulitiple pages, and contain a "delete tick" for me,and when i click the member, can redirect me to member detail imformation. it just similiar to email pages of yahoo inbox.

Limit A Database Field To Displaying Only One Time And Still Display All?
I have a data table with a field that contains the exact same text (example "some text") on several different rows. How can I display all my rows without getting duplicates of this field; I only want to display this field once. I tried to use a LIMIT, but this is not what I needed. Code:

Display Database Output In Mutli Column Html Table
I am building an HTML form using data from a mysql query. The data to be displayed on the form is 1 column from the db - but multiple rows. Because there are many rows (well, only about 100+) in the table I want to be able to put 2 or 3 (space permitting) rows from the result on each row of the HTML table with a checkbox beside each entry from the db table. The idea is that the user will scroll down the list and check up to 4 or 5 boxes before submitting the form and I didn't want them to have to scroll ALLLL the way to the bottom (100+ rows). I understand how to get data from a db table and put it into an html table, but I cannot figur out how to put more than 1 row from the result on to the same row of the html table.

I hope I haven't confused anyone reading this -- If I am way off base with what I am trying to do please give me alternatives. I am trying to avoid paginating the data for a couple reasons a) clicking a next and/or prev button is more hassle than just scolling down a list and b) the data in the column being displayed is only about 35 - 40 characters and it seems to be if i only placed one result row on each html table row, there would be a lot of wasted space.

No Array Entry Vs Null Array Entry
Consider this statement related to Fields sent to PHP via an HTTP POST.

$A = $_POST["FieldName"];

If the page did not have a form field with FiledName, the result is
that $A is empty.

If the page DID have a form field with FieldName, but the field was
left empty, then $A is empty also.

How should I test $_POST to distinguish between these two conditions.

Migrating Php 4.0.5 To The Latest Php
will there be any problem in migrating from php 4.0.5 to the latest php version 4.2.3?

i been reading the feature of 4.2.3 where they require session/post/get to be addressed as $_session['var'],$_post['var'] and $_get['var'].

will this cause any problem to my existing codes in which i uses the plain $form_var and $session_var and not $HTTP_POST_VARS['var'] or similar.

Getting Latest Binaries
I've only used PHP under Windows OS. Now I'm trying to set it up in Linux. My problem is that my linux distro has an old version of PHP. So where can I get the latest binary? PHP.net doesn't distribute linux binaries anymore.

CW - My Latest Php Website
Its also the first shopping cart i've wrote .

Personally i find it a bit strange that given all the php scripters on
this newsgroup nobody ever posts a url to their work .

I suspect thats sometimes [1] because a lot of people never really
finish a website and they are more interested in php code snippets - not
saying thats a bad thing of course - it just seems a bit pointless if
you never really do anything with the gained knowledge .

[1] and non disclosure in some cases .

Anyway , As you will see from the website below i also have a interest
in web design and these days try to blend the 2 together .

Comments (good or bad) invited .

Latest Posts..
You know how on forum software they show the new posts, and then mark them as old.. how is this usually done? Is it via cookie, with all the ids stored in an array, or some other way?

Latest SQL Data...
Is there a way to get the last row of a table and get the data out of just that row? With PHP and MySQL.... I've tryed a simple way but it seams quite unstable:

$finn_refnr = "SELECT refnr FROM ${db_records}";
$result_refnr = mysql_query($finn_refnr) or die(mysql_error());
while($row = mysql_fetch_array($result_refnr)){
$new_refnr = $row['refnr'];
echo $new_refnr;
}

Getting Latest Date
There are 3 documents 1,2 and 3

Person A completes document 1 on 1/1/2007, document 2 on 2/1/2007 and document 3 on 3/1/2007

Therefore, completes them all by 3/1/2007

Person B completes document 1 on 4/1/2007, document 2 on 2/1/2007 and document 3 on 3/1/2007

Therefore, completes them all by 4/1/2007

This is held in a table like

id person_id document_id date_completed

How do I order them by who completed them all first?

Show Latest Picture
I'm trying to show the latest picture uploaded, similar to a picture of the day, but not

I'm going to name picture files 001.gif 002.gif 003.gif etc, but what I want to do is get a list of the pictures in a directory and show the one with the highest number.

Latest PHP Patch For Windows
http://www.php.net/release_4_1_2_win32.php

As I'm very paranoid about security, I'm just checking in here to make sure I'm doing this right. I'm supposed to add:

cgi.force_redirect 0|1
cgi.redirect_status_env ENV_VAR_NAME

into php.ini. Now, it says to change 0|1 to 1 if you're running Apache, which I did, and says nothing else pertaining to that.

Retrieve Latest Result...
I need to make a query that pulls the latest added news article so whichever one has the highest id, how do I do this?

Selecting Latest Occurance Of Id In Mysqltable..
I have several mysql tables which consists of users..The mysql tables are named by the actual year, which means that a new table will be added each year...Well these users can occur one year...then dissapear...and then comeback another year... for example..

My "problem" is to get the last occurance/information from a specific user. (UID...user id..) Lets take an example....

I got these tables...
1999_tbl, 2000_tbl, 2001_tbl

UID(scott) occurs in 1999_tbl and 2000_tbl but NOT in 2001_tbl

I want the latest info from UID(scott)...which is in table 2000_tbl

I also have a table with all "yeartable"-prefixes:
ID | PREFIX
0 | 1999
1 | 2000
2 | 2001

Latest PEAR On W32, Refers To PHP5, Bug
latest PEAR on W32, refers to PHP5 inaccurately.
I've seen posts regarding path's that get messed up, by lost ini's or
registry settings.
I don't think that's my case, because I can't find PHP5 in registry or
any left-over ini's.
This is a new PC that I use on the weekends mostly.

The latest Pear version 1.4.5-I believe, also seems to have changed
(written over) include_path and extension_dir settings of php.ini
changing them to .;C:php5 and C:php5ext when there is no dir
or reference to php5 in ini's registry etc ( only C:php )

I've run into some other related obstacles using Pear and Smarty on
shared-host env.
I'd like to be able to use Smarty, even in subdirectories and not have
to hardcode path.
While a recent version of Pear (1.4.0) claims to sync local and
shared-host Pear modules, the instructions ask you to install locally
in a very non-standard area. ?

I'm running PHP 5.0.5 on WinXP - anyone else have this experience ?
What's the best way to fix it ?

Display The Latest Database Entry?

how I can display the latest database entry? PHP Code: