Re: <select> not populating form fields (php,mysql)
- From: David Haynes <david.haynes2@xxxxxxxxxxxx>
- Date: Fri, 28 Oct 2005 21:01:26 -0400
Ian Rutgers wrote:
Situation:
I have a form which collects a bunch of data. The form accesses two tables (users and assignment_details). The user table is accessed to get the userid (to be saved in an assignment_details record) and to display other users table info (for display only). There is other information being collected to be saved in the assignment_details table, but it is not relevant to the issue.
I am using a <select> to display user names, and extract the appropriate userid. The <select> tag uses a query to pull all the names from the users table. When the form is loaded all defined fields from the users table are displayed on the form. Selecting a different name doesn't result in extracting and displaying the result of the user fields (as in the code below): ... help!
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.
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<td><select name="adetails_employee_id">
<?php
do {
?>
<option value="<?php echo $row_rsEmployeeLookup['id_usr']?>"><?php echo ($row_rsEmployeeLookup['first_name_usr'] . ' ' . $row_rsEmployeeLookup['last_name_usr']);?></option>
<?php
} while ($row_rsEmployeeLookup = mysql_fetch_assoc($rsEmployeeLookup));
$rows = mysql_num_rows($rsEmployeeLookup);
if($rows > 0) {
mysql_data_seek($rsEmployeeLookup, 0);
$row_rsEmployeeLookup = mysql_fetch_assoc($rsEmployeeLookup);
}
?>
</select>
</td>
</tr>
<tr>
<td>mobile phone</td>
<td><?php echo $row_rsEmployeeLookup['mobile_usr']; ?></td>
</tr>
etc.
I think you are missing a fundamental concept about server-side processing which is that once the server responds to the browser (i.e. PHP has processed its response) the page is essentially static.
Try doing a right-click view page source in your browser to see what PHP has sent to the browser and then ask yourself how was the new query based on the select going to happen.
In other words, since all queries need to be resolved on the server, how is the server going to know what you selected?
-david-
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