PHP :: Drop-down Displaying MySQL Data

Displaying Data From Mysql Database In A Drop Down List On Form.
I'm trying to do is display data from two different mysql tables from the same database in a drop down list on a html form. I have a fixtures table with the player1(userid), player2(userid), gameid, game, score1 and 2, what I want is to use the userid to get the players first name and surname from the members table (as it is a unique id), I need to do this bit before displaying it in the drop down. I think i need 2 querys to do this but when I have tried it it just echo's a blank value or the userid not the forname and surname that I want. I'm using the fetch_array function but just can't see where I'm going wrong, Code:

PHP MySQL Query Not Displaying Data
I am having some problems with a database query that I am trying to do.
I am trying to develop a way to search a database for an entry and
then edit the existing values. Upon submit, the new values are updated
in all corresponding tables (the function of the pages in question).
However, on the page that does the DB update, I also want to do some
checks on the data before performing the update.

Now, the problem that I am running into is that when I don't update the
primary key field (keyid) the page is running into the section where it
displays a message that the keyid already exists. This only happens
when the keyid on the previous page is not changed.

Upon troubleshooting the problem I found that the very first SELECT
statement does not seem to be returning any rows despite the fact that
the statement, when run on the SQL server, returns exactly one row. If
any could provide some assistance with this matter, I would be most
appreciative.

Code:
<?php
session_start();
header("Cache-control: private"); //IE 6 Fix

//continue if authenticated
if ($_SESSION['auth'] == 1)
{

//get variables from post
$keyid = $_POST['keyid'];
$username = $_POST['username'];
$corpid = $_POST['corpid'];
$usergroupname = $_POST['usergroupname'];
$oldkeyid = $_POST['oldkeyid'];
$oldcorpid = $_POST['oldcorpid'];
echo("Keyid: $keyid");

// Connect to MySQL
mysql_connect ("address.com", "user", "pass")
or die ('I cannot connect to the database because: ' .
mysql_error());
//select database on server
mysql_select_db ("seniorproject");

//SQL Statement
$sql = "SELECT keyid, corpid FROM users
WHERE keyid='".$keyid"'";

// Execute the query and put results in $result

$result = mysql_query( $sql )
or die ( 'Unable to execute query.' );
echo mysql_result($result,0);

// Count number of matches and print to screen
$numrows = mysql_numrows( $result );
if ($numrows == 1)
{echo(" SQL Query: $sql");
$result_ar = mysql_fetch_row($result);
$result_ar = mysql_fetch_row($result);
$dbkeyid = $result_ar['keyid'];
$dbcorpid = $result_ar['corpid'];
echo("<br> CorpID: $corpid <br>");
echo(" DBCorpid: $dbcorpid <br>");
echo(" DBKeyID: $dbkeyid <br>");
if($dbcorpid == $oldcorpid)
{
//If this condition is true, then KEYID has not changed. Execute
code

// Formulate the query

$sql = "UPDATE users
SET corpid = '".$corpid."', username = '".$username."',
usergroup = '".$usergroupname."', keyid = '".$keyid."'
WHERE keyid = '".$oldkeyid."'";........

Displaying Line Breaks From MySQL Data
when i retrieve data from a MySQL database which i have put line breaks in they do not show in html, i understand that this is because HTML ignores spaces unless they are formatted as   i know i could use the <pre> tag to show it as it was given, which removes all styling from the text, i need to convert the breaks in MySQL into <br> or <br><br> (to start a new paragraph)

Drop Down Not Displaying All Rows
I am having an issue with the drop down box. If there is only one row
in the table the drop down box will display nothing. once the table
has two or more rows the drop down displays the row but always one shy
of the total. here is the script i am using:

$query = "SELECT SEC_TITLE
FROM section
WHERE SEC_TITLE != ''";

$result = mysql_query($query);

while ($row = mysql_fetch_array($result, MYSQL_ASSOC))

{ echo "<form action='#'>
<select name=section>
";
while ($line = mysql_fetch_array($result, MYSQL_NUM)) {
$section = $line[0];
echo " <option value=$section>$section
";
}
}

Select Data In Drop-down Box
I have a two text box and one select box.

two text box will be filled with date values. (in the form of
dd-mm-year)
First text box is called "start date"
Second is "end date"

Thus, I want to put dates between "start date" and "end date" in select
box.

Is there any way to put automatically in option field?
My idea is, first calculate the date difference
and then, add date till startdate + difference.

Getting Data From Drop Down Menus
I have a drop down menu and I'm trying to find out which option they have pressed. I need to use the $_POST method to do this. Below is what i've written so far, but it's showing every time the messages:

"You don't have enough points" AND "you gained blah blah exp" I've blocked out some of the stuff that calls up files and stuff, just for security reasons Code:

Drop Menu Using Session Data
I've a drop menu for users to choose the month they are born from. There's an associative array, $nameOfMonths, for the months' names. E.g. [1] => 'January', $birthMonth would be an Int.

If there's session data, it will print the user selected month, followed by a list of months. If there's no session data, it will ask user to select a month, followed by a list of months. I wonder if this is the best way to code this drop menu. Code:

Passing Data From A Drop Down Menu
I have a website that I am working on. When the user logs in they see a drop down box and they are supposed to go to this box and select their name. How would I get the selection that they made to pass to the next page.

This is my code for the page with the drop down box that I have so far. It pulls from my database, and it works fine. Code:

Generic Data-based Drop Down Menu
This is becoming incredibly frustrating...

I have a pile of drop downs on one page. They are generated based upon tables in a database. After searching, the user needs to be able to refine their search. They click a form button and go back to the original search page with all the fields retaining the values of the original search. Everything works GREAT except my drop downs! They refuse to remember anything!

Saving/Loading Data In Drop-down List
I have a drop-down list with set values. When there is a corresponding value in my database when the page loads I want the drop-down list to pick the appropriate value from the list. Instead it always defaults to the first value. Can anyone give me an example of how to do this?

Also, one time after adding a variable to a classes file this symbol: ï»¿ started showing up on all of my pages. Has anyone seen this before? I can't find any information on the web about it.

Need To Create Drop Down Box With Data Pulled From Text File
Trying to make a drop down box like this example:

Dog
Cat
Bird

But sometimes I will need to update it...and add/remove names from the list. I want to do it fast and easy and I don't want to use a MySQL Database. I'd rather use a text file...can that be done? I would keep the info in a text file and then push it to a drop down box via php and html.

Displaying Data
Probably a really stupid question to ask, but if i have a number in mysql lets say 180000 and it represents a price is there a way to change that to 180,000 without saving it in the database like that?

Displaying Wrong Data?
I have a PHP script that inserts data into a database: PHP Code:

Displaying Data In Php - Like HTML?
How do I display data from MYSQL in PHP in an attractive and structured table(HTML) format? For example: I want my database to display all the reviews of a particular movie(like metacritic.com) like this: Code:

Displaying Data From Array
I have a data stored in an array. Want to display it in Internet Explorer... However,could not display it.

Inputing And Displaying Data
i have a website that will list certain shops in my area. i want these shops to be displayed in a table with 4 columns on my website, the first being the name of the shop and when you click on the shop it links you to there website. the rest of the columns will be different images depending on what service they provide.

i thought i could create a form that allows me to input the name of the shop etc, but each time i submit the form it adds the information to my database and creates a new row each time i submit a new shop on the table on my website page.

im very new to php and sql so it would be helpful if you could link me to a tutorial that sort of tells you how to do this or add me on msn.

plus i was just wondering if it is worth buying a book about php and mysql or should i stick to online tutorials. if it is, please recommend me some.

Grabbing Data From A Table And Displaying It
I'm grabbing data from a table and displaying it but I want to to spand out into multiple pages...I got it all setup but when I click on the next link nothing happens... I got it setup at "3" right now just for testing..I have 5 entries in the database.. anyways could someone help me out? here is the code:

<?php

// Get connection info
include "../connect.php";
// End connection info
if(!$rowstart) $rowstart=0;
// Start database select
$result = mysql_query(
"SELECT * FROM links WHERE notactive='$notactive' order by date desc limit $rowstart,3"); //Select Database
$result2 = mysql_query("select * from links");
if (!$result) {
echo("<P>Error performing query: " .
mysql_error() . "</P>"); exit(); }
// Display the text of each host
while ( $row = mysql_fetch_array($result) ) {
if ($notactive = 1) {
echo ("<table width=500 height=57 border=0 cellpadding=0 cellspacing=0>");
echo ("<tr>");
echo ("<td align=left valign=top><strong>" . $row["name"] . "</strong></td>");
echo ("</tr>");
echo ("<tr>");
echo ("<td align=left valign=top>" . $row["desc"] . "</td>");
echo ("</tr>");
echo ("<tr>");
echo ("<td align=left valign=top><a href=" . $row["url"] . ">". $row["url"] ."</a></td>"); }
echo ("</tr>");
echo ("<br><br>");
echo ("</table>");
}
if ($rowstart>$numrows)
{
echo "<A HREF=$php_self?rowstart=rowstart-3>";
echo "< Previous Page</A>";
}
echo "|";
$numrows=mysql_num_rows($result2);
if($rowstart+3<$numrows)
{
echo "<A HREF='$php_self?$rowstart=rowstart+3'>Next Page></A>";
}
?>

Displaying Data Based On Variable
A Newbie to PHP so please forgive any ignorance of simple methods.

I am trying to develop a site that is driven by a MYSQL database. Dependant on which domain you have been redirected from I need to somehow pickup the domain redirected from and set a global constant to only pull from relative MySQL data e.g FOO = xyz.

This site will obviously have multiple users and I need a way to ensure that user A will be browsing xyz while user B will be browsing abc.

The way I can think of but not entirely sure how to get working in PHP is to assign a Value to the session (FOO) and then on database queries uses WHERE foo_ID = (FOO).

Displaying Data In A Editing Form
I have a form that has three checkboxes. A,B&C. Data from A & C has been entered into the database by serializing it.I now want to create a edit form where the user can view his choices ticked.

<form action="postoDb.php">
A<input type="checkbox" name="Choice[]" value="A">
B<input type="checkbox" name="Choice[]" value="B">
C<input type="checkbox" name="Choice[]" value="C">
</form>

//postodb.php
$bn = serialize($Choice);
sql ="insert into table (choice) values('$bn')";

Displaying Rows Of Data As Columns?
Does anyone know of an efficient way to retreive data from a mysql db
and display the record rows as collumns?

I have been doing this:
<display headers>
<while ($row = mysql_fetch_array($result)) { >
<display row 1 of result>
<display row 2... } >
<end table>

but I want to do this instead:
<col 1 header > <row 1 col 1> <row 2 col1> <row3 col1>
<col 2 header> <row 1 col 2> <row 2 col2> <row3 col2>
<col 3 header> <row 1 col 3> <row 2 col3> <row3 col3>

Question About Displaying MySQLl Data
Could someone point in the correct direction with regards to displaying my
data in "nicer" format. I have figured out ( with the tutorials on the web,
and the O'Reilly PHP & MySQL book) how to store, retrieve, manipulate, and
display my data using PHP. But when I do display, it is the block format.
I can't seem to find a tutorial that gets into a "prettier" type of display.

Is this a HTML issue?

I am a newbie to this stuff

Displaying Data In Html Table.
I wondered if anyone could offer some guidance, I trying to write a
php script to connect to a database, and display the records in a
table.

I found the code here in a php4 text, and when I run this directly
through the php intrupeter, the line Successfully connected is display
and 4 as the number of rows.

The database table we connect to, has three field username, firstname,
surname.

When I tried a while loop, to display the data in a table. No values
were displayed.

Could anyone possibly demonstrate how to display the username,
firstname, surname values in a basic table?

<?php
function db_connect(){
global $MYSQL_ERRNO, $MYSQL_ERROR;

$link_id = mysql_connect("localhost","user04", "password04");

if(!$link_id){
$MYSQL_ERROR = "Connection failed";
return 0;
}
else if(empty($dbname) && !mysql_select_db("database04")){
$MYSQL_ERROR = mysql_error();
return 0;
}
else return $link_id;
}

!$link_id = db_connect();
if(!$link_id) die(error_message('Error connecting'));

$result = mysql_query("SELECT * FROM users");
if(!$result) error_message('Error in selection');
else echo "Successfully connected. ";

$query_data = mysql_fetch_row($result);
$total_records = $query_data[0];

if(!$total_records) error_message('No records');
else echo "$total_records.";
?>

Displaying Data From A Flat File?
ive got a flatfile database sorted out, all i need now is how to store individual flat files per user and how to paginate that data. Just to make sure i'm doing everything right, i'll post the code below Code:

Displaying All Data From Fetch Array
I'm having a problem displaying more then 1 result from a mysql_fetch_array statement in my php application.....

Combining Data From Two Fields, Displaying With Another Table
What I am trying to do is pull data from two tables and display it on one page. The code below doesn't display any errors, but all it prints is the results from the first query. Ideas? Thanks.

(I also need to somehow put a variable in for the warranty table data so it is tied to the product_status table data. You can see there is a WHERE in the first query of pid, but no WHERE in the second query. That is a seperate question, though. It would make this way too long, so I will post it separately). PHP Code:

Hide Characters When Displaying Recordset Data ?
i have a mysql and php website, some of the products have an _ underscore (epson_dx500 eg) . i display these items listed vert down the page, i want the data in the table to stay the same but when displayed on the page i want to hide the underscore.

Displaying Binary Data To Browser And Set Mime Type
I am looking for a link or help on how to set the
mime type in my php script as well as send binary data to the browser
window instead of having it ask me to save the file.

PHP-MySQL Drop-Down Box
I'm using PHP to dynamically make a drop down box of users in a table. The first row is displayed with all the users. However, on the second row non of the users are listed in the drop down box. It only works on the first.

PHP + MYSQL, Populating Drop Down Box
I am looking to generate a dropdown box from MYSQL data:

db name = h2, table = Working, Column = Home.

2 Drop Down List Using MYSQL
I have a model table with the following fields: MODEL_ID, MAKE_ID, NAME.
And a make table with the following fields:
MAKE_ID, NAME.

The first drop down list (make) to select a make of a car i.e. Chevrolet, Ford. When a make is selected, then all the model for that paticular make appear in the second drop down list.

Two Different Drop Downs From MySQL
I have a admin area for a website, and on some part the admin can
select options from two different drop down boxes. Now i thought that
it would be better to use one Query for both dropdowns. How can i get
them together?
I now have something similair twice:

<select name="series_id" class="dropdownbox" id="series_id">
<option value="0">- select -</option>

<?php
$db = mysql_connect("localhost", "name", "pass");
mysql_select_db("db_name");
$result = mysql_query("SELECT * FROM series ORDER BY name");

if ($data = mysql_fetch_array($result)){
do {
$current_id = $data['series_id'];
if($current_id == $id){$selected = "selected";};
echo "<option value="".$data['series_id'].""
".$selected.">".ucfirst($data['name'])."</option>
";
} while ($data = mysql_fetch_array($result));
} else {
echo "";
};
mysql_close($db);?>
</select>

I hope it's clear what i mean. I do not make connection twice, but the
rest is similair except for the tablename of course.

PHP And MYSQL Drop Down List
I need a script to create a drop down list which has vehicle makes. The vehicle make list should feed a vehicle model list so when I select a make all the models for the make appear in the model list. I have a make table which has MAKE_ID, make NAME and a model table which has MODEL_ID, MAKE_ID and model NAME.

I tried using a script using javascript array but wherever I have a query that pulls the NAME from the database there's always a carriage return after the NAME which messes up the javascript array. I'd rather do it all with PHP if possible. If not please give me one that works and I don't have to spend hours troubleshooting.

Generating Drop Down Boxes From Mysql
i'm using php with mysql. i need to be able to extract items out of my database and put them into a drop down box so that the data is dynamic.

In this example, i need to get all of the names from the name.office field into a drop down box to be used in further scripts/processess.

<html>
<head>
</head>
<body>

<?

$db = mysql_pconnect ("localhost", "XXXX", "XXXXX");

if (!$db)

{
echo "error: could not connect to database. please try again later";
exit;
}

mysql_select_db("XXXXXXX");
$query = "select name from office";
$result = mysql_query($query);
$num_results = mysql_num_rows($result);

$row = mysql_fetch_array($result);

?>

</body>
</html>what goes here??? i have tried many things and not been able to get it to work.

Drag And Drop Insertion In MySql
I'm developing a site in PHP/MySql that requires Ajax for this feature
: The user read an article and if ever he likes it he can drag it and
drop it into his "basket". Then when he'll consult its account he'll
found that article.

Mysql Fetch With Drop Down Menu
Im trying to do somekind of dropdown menu taking information from the database but im not quite sure how this could be done with my knowledge.

alright so lets say i have a table named: projects SELECT name FROM project. and it will put all the data found in a html drop down menu?

Drop Down Menus And Mysql Insertion
I'm looking to create drop down boxes for a user to select the date and then have it insert what they select for month, day, year, hour, minutes and insert it into mysql. It needs to be organized in a timestamp like so: 2007-02-28 21:06:40

I've created the drop down boxes and it works pretty well: Code:

Dynamic Drop Down Boxes With Mysql
I'm working on a database project for a class and am having problems populating select boxes with a field from a mysql table. The first thing that I'm trying to do is pull the field LastName from the table Faculty. Then when the user selects the faculty member in the drop down box, it will then display all of the fields related to that faculty member...first and last name, title, office number etc and allow the user to edit those fields and then resubmit them to the database.

However I can't seem to get the drop down box to populate, so if anyone has any suggestions as to how to fix this I would love to hear them. Code:

Multiple Drop Down List Problem And MySQL
I am trying to populate around 30 drop down lists via a single query to a mySQL database. The problem is how to make sure that each drop down list only display it's relevant data. for example the table is as follows:

Using Natsort On A MySQL Populated Drop Down Menu
I am trying to use the natsort function on a dropdown menu which is populated by the mysql database. Code:

Drop Down Menu Where The Items Are From A MySQL Select Query
I am just starting programming in PHP and wonder if it is possible to have a
drop down menu where the items are from a MySQL select query.

I am registering people for a weekend seminar and want to have a list of
valid weekends to choose from on the form.

Any pointers to scripts or PHP references that will help would be
appreciated.

I am using the book "Build your Own Database Driven Website Using PHP &
MySQL" to get me started, but also need recommendations for a general primer
to php.

How To Grad Information From Mysql Database With Drop Down Menu's?
i am trying to do is like this website http://www.chippeduk.com/ in the search area (bottom right), where the user selects a make (all fields grab from the database), then selects a model, then they select the fine tune, then they click search this will bring up the information about the car they have chosen. How would i do this, would i use multiple tables and then link them together via an id? i am new to PHP/MySQL but very willing to learn.

Search Drop Down Menu To Choose Mysql Table
I have currently made a simple PHP search script which, when data is entered in a text box and it's submitted, it queries a table in a MYSQL database and returns information. This works. However, I want to have a combo box beside the text box with options to query different tables depending on which item in the combo is selected.

At the moment it searches a news articles table by content and title. But I want to adapt the script so that it searches another table which holds information on people for example.

It currently gets the item from the text box with the code:

Drop-down Displaying MySQL Data

I've got this dropdown trying to call the data in...basically, I only want it to show if it is an upcoming event. Past events are hidden away in the archives.I can get it to work if I remove the if ($row['date'] etc, but not with it in. Can anyone spot any errors with my code? Code:

I've got this dropdown trying to call the data in...basically, I only want it to show if it is an upcoming event. Past events are hidden away in the archives.

I can get it to work if I remove the if ($row['date'] etc, but not with it in. Can anyone spot any errors with my code? Code: