how to pass the value of selected drop-down menu in variable to another page?

shams · #1 19-Oct-2006, 10:01

hi,
there is a php-myql script that list the mysql databases in the drop-down menu, i want the selected database from the drope down menu should pass in the $db_name to the test php, the script list the databases in the drop-down menu ok, now this is the test.php:

PHP Code:


<?php
echo 'hello'.$db_name;
?>

the $db_name should be the selected database, but the output of test.php is:
hello Db_list
this is the listDb.php:

PHP Code:


<html>
<head><title>List DB</title></head>
<body>
<form method="POST" action="test.php"><div align="center"><center><p>Select Db
<input type="hidden" name="db_name" value="Db_list">
<SELECT NAME="Db_list">
<?php
$link = mysql_connect('localhost', 'user', 'pass') or die(mysql_error());
$db_list = mysql_list_dbs($link);

while ($row = mysql_fetch_object($db_list)) {
echo "<option value=\"$row->Database\">$row->Database</option>";
}
?>
</select>
<input type="submit" value="Submit"></center></div>
</form>
</body>
</html>

admin · #2 19-Oct-2006, 15:24

Because you named the SELECT element as Db_list, the script (test.php) should be re-written so:

PHP Code:


<?php
echo 'hello ' . $_POST['Db_list']; // NOT $db_name
?>

You already have a hidden input element named db_name before the select.

shams · #3 19-Oct-2006, 18:45

thanks admin, that works great now.

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how to pass the value of selected drop-down menu in variable to another page?

Re: how to pass the value of selected drop-down menu in variable to another page?

Re: how to pass the value of selected drop-down menu in variable to another page?