Questions about the MySQL, PostgreSQL, and most other databases, as well as using it with PHP can be asked here.
Moderator: General Moderators
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Weirdan
- Moderator
- Posts: 5978
- Joined: Mon Nov 03, 2003 6:13 pm
- Location: Odessa, Ukraine
Post
by Weirdan »
and here's what should work for you:
Code: Select all
<?php require_once('../../Connections/hungryusaconnection.php'); ?>
<?php
$colname_rsmenu = "1";
if (isset($HTTP_GET_VARS['namefield'])) {
$colname_rsmenu = (get_magic_quotes_gpc()) ? $HTTP_GET_VARS['namefield'] : addslashes($HTTP_GET_VARS['namefield']);
}
mysql_select_db($database_hungryusaconnection, $hungryusaconnection);
$query_rsmenu = sprintf("SELECT * FROM menus WHERE item LIKE '%%%s%%' ORDER BY item ASC", $colname_rsmenu);
$rsmenu = mysql_query($query_rsmenu, $hungryusaconnection) or die(mysql_error());
$row_rsmenu = mysql_fetch_assoc($rsmenu);
$totalRows_rsmenu = mysql_num_rows($rsmenu);
//mysql_free_result($rsmenu);
?>
<?php
do
{
//$row_rsmenu ++;
print( '<p style="color: #000000">' );
print( $row_rsmenu ['item' ]. '<p>' );
}
while ( $row_rsmenu =mysql_fetch_assoc ($rsmenu ));
?>
note which lines are commented
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saltriver
- Forum Commoner
- Posts: 59
- Joined: Fri Mar 12, 2004 2:40 pm
- Location: Burlington, VT
Post
by saltriver »
That produced this error
Code: Select all
Parse error: parse error in /home/virtual/site110/fst/var/www/html/testing/middle/name_results.php on line 5
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Weirdan
- Moderator
- Posts: 5978
- Joined: Mon Nov 03, 2003 6:13 pm
- Location: Odessa, Ukraine
Post
by Weirdan »
it should be copy-paste error, make sure you copied source code properly. There's no syntax errors in the code I posted.
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saltriver
- Forum Commoner
- Posts: 59
- Joined: Fri Mar 12, 2004 2:40 pm
- Location: Burlington, VT
Post
by saltriver »
Boo-Ya!
There were two "E"s infront of one of the tabbed lines.
Thanks