mysql drop list

[jd57]

Hi

this code should allow me to list all available_images in a drop list menu, but all I get is a blank list. I've checked the row number $row[26] and that is correct.
What am I doing wrong?
I would also like to add onchange code when I click on any of the finding to go in "search" field, how can I do this?
Thanks

<?PHP
@ $db = mysql_pconnect(localhost, xxxx_xx, xxxxx);
if (!$db) {
	print "Failed to connect to the database.  Please try again later.\n";
	exit;
}
mysql_select_db(xxxx_xxx);
$sql = "SELECT * FROM `seedling` ORDER BY `Available_Images` ASC";
if (!(@ $result = mysql_query($sql))) {
	print "There was an error running your query:<p>\n". $sql . "<p>\n";
}	
print '<select name="Available_Images"><br>\n';
print '<option value="">  Choose One   ';
while ($row = mysql_fetch_row($result)) {
	$id   = $row[0];
	$Available_Images = $row[26];

print "<option value=\"$id\"";
if ($_REQUEST["Available_Images"] == $id) { print " selected"; }
print "> $Available_Images <br>\n";
}
print '</select>';

?> 

[kakki]

(just guessing)

if (!(@ $result = mysql_query($sql))) .....

i think php is "lazy", if you add a @ to a command, it sometimes wont work properly

i have copied that from php.net, maybe you can change it to sth like that

$link = mysql_connect() or die(mysql_error());
$result = mysql_query("SELECT * FROM table ");
$row = mysql_fetch_row($result);

the onChange event:
change:
print '<select name="Available_Images"><br>\n';
into

print '<form name="testForm" action=".......">';
print '<select name="Available_Images" onChange="document.testForm.submit;"><br>\n'; 
//..... rest of your code here
print '</form>';

there is a named form ( "testForm" ), which is submitted, when the status of the select field changes

[jd57]

Thank you so much, it's working fine.

[jd57]

Hi

I was able to make this script work, thanks to this forum and to the people that help me.
I still need your help.
I was wondering if it is possible to link the same value on the onChange command to another php script (SEindex.php) that is found in the same folder, and if so, how?

<?PHP

mysql_connect("localhost", "xxxx_xxxx", "xxxx") or die(mysql_error());

mysql_select_db("xxx_xxx") or die(mysql_error());
$sql = "SELECT * FROM `ihsreg` ORDER BY `Cross_Made` ASC";
if (!(@ $result = mysql_query($sql))) {
	print "There was an error running your query:<p>\n". $sql . "<p>\n";
}	
print '<select name="Pod_Name, Pollen_Name"
onChange="search.value=this.options[this.selectedIndex].text;"><br>\n';
print '<option value="">  Choose One   ';
while ($row = mysql_fetch_row($result)) {
	$id   = $row[0];
	$Pod_Name = $row[4];
	$Pollen_Name = $row[5];
    $x= x ;
	print "<option value=\"$id\"";
	if ($_REQUEST["Pod_Name,Pollen_Name"] == $id) { print " selected"; }
	print "> $Pod_Name $x $Pollen_Name <br>\n";
}
print '</select>';

?> 

[thepeccavi]

somethimes thing can be easier with an associative array... if you use

while ($row = mysql_fetch_assoc($result)) { ...

instead of

while ($row = mysql_fetch_row($result)) { ...

It may help you to debug... don't forget to change $row[0] to $row[<name of 1st field>] though! (i did that once... took me ages to figure it out lol)