My PHP mysql Command
| From: | Philip J. Newman | Date: | Fri, 14 Mar 2003 22:49:31 +0000 |
| Subject: | My PHP mysql Command | ||
| Groups: | php.general | ||
| Request: | Send a blank email to php-general+get-139772@lists.php.net to get a copy of this message | ||
Assuming that gAccessLevel = eather one of level-1, level-2, level-3 ...
if (!$gStart) { $gStart=0; }
$sql = "SELECT * FROM gallerys WHERE gMode = '$mode' AND ";
if ($siteAccessLevel == "level-1") {
// LEVEL-1 VIEWING ONLY.
$sql .= "WHERE (gAccessLevel = 'level-1')";
} else if ($siteAccessLevel == "level-2") {
// LEVEL-2 AND LEVEL-1 VIEWING ONLY.
$sql .= "WHERE (gAccessLevel = 'level-1' OR gAccessLevel =
'level-2')";
} else if ($siteAccessLevel == "level-3") {
// PAYING, ACCESS ALL IMAGES!
$sql .= "WHERE (gAccessLevel = 'level-1' OR gAccessLevel = 'level-2' OR
gAccessLevel = 'level-3')";
} else {
// IF ERROR SHOW LEVEL ONE.
$sql .= "WHERE (gAccessLevel = 'level-1')";
}
$sql .= " ORDER BY gId DESC LIMIT $gStart, 10";
echo $sql;
$sql_result = mysql_query($sql, $connection) or die (mysql_error());
MY ERROR MESSAGE IS ...
10You have an error in your SQL syntax near 'WHERE (gAccessLevel = 'level-1'
OR gAccessLevel = 'level-2' OR gAccessLevel = 'l' at line 1
as gAccessLevel = level-3 ...
Any Ideas why this is weong?
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