Let me give this a shot :P
?
Mysql_connect(sqlserver,user,pass);
Mysql_select_db(db);
$resultset = mysql_query(select DISTINCT identifier, displayname from
table);
for($i=1;$i=mysql_num_rows($resultset);$i++)
{
$object = mysql_fetch_object($resultset);
$identifier = $object - identifier;
$displayname= $object - displayname;
print option value=\$identifier\$displayname/option;
}
?
I do believe that should do it. Correct me if I missed anything.
Ryan
-Original Message-
From: Leotta, Natalie (NCI/IMS) [mailto:[EMAIL PROTECTED]]
Sent: Friday, March 01, 2002 2:10 PM
To: PHP
Subject: RE: [PHP-DB] Display in drop-down box
Try doing a select distinct on that field and then loop through the results,
adding each one to your drop-down box as you go.
I don't have the exact code anymore because we stopped doing it that way,
but that's the basic idea of how to go about it.
I hope it helps!
-Natalie
-Original Message-
From: Alvin Ang [SMTP:[EMAIL PROTECTED]]
Sent: Friday, March 01, 2002 2:07 PM
To: PHP
Subject: [PHP-DB] Display in drop-down box
Hi there,
I was wondering if anyone would be kind enough to show me how to select a
field from a table in mysql and display it in a drop-down combo box?
I am trying to show my users a list of all the parts from a table so that
they can edit or delete it.
Thanks!
Alvin
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