Trying to get form to display table content on drop down

[Johnnyboy123]

Name says it all. I have a form and 2 tables that are going to have to be linked with each other. Although what I'm trying to do now is very basic and I'm struggling haha Though this is my first php project

 

Basicly I got a form with a drop down menu, which I want to display the contents of my table (course) which the user then selects from and submits to another table (student)

 

Here is my code for this part

<?php

<?php
$q = "SELECT cname FROM course ";
$result = mysql_query($q);
$course = mysql_fetch_array($result);

?>
<div id="apdiv3">

<FORM action = "demo.php" method ="post">

<p>Course name:</p>
<select name="cname"> <OPTION> <?php echo $course['cname']; ?> </option> </SELECT>

?>

 

It only displays 1 item from my course table. What am I doing wrong? And also please explain so I can learn from my mistakes

[Muddy_Funster]

You are only selecting the top level result that is returned in the array.  You will need to run through the array in order to get what you want out.

<div id="apdiv3">

<FORM action = "demo.php" method ="post">

<p>Course name:</p>
<select name="cname"> 
<?php
$q = "SELECT cname FROM course ";
$result = mysql_query($q);
WHILE ($course = mysql_fetch_array($result)){
echo "<OPTION> value='{$course['cname']}'>{$cource['cname']}</option>";
      }
echo " </SELECT>"

 

Should do it.  You also need to look into how to form <option>'s better.

[quelle]

Just add the opperator "." before "=" cuz like this you are going trough ur db results and displaying only last one !

[Muddy_Funster]

Just add the opperator "." before "=" cuz like this you are going trough ur db results and displaying only last one !

erm...now your just being silly.

[Johnnyboy123]

Ah I see. Working fine now. Thanks alot guys

[Muddy_Funster]

No worries, remember to mark as solved (bottom left of the page)