georegjlee Posted April 15, 2007 Share Posted April 15, 2007 I keep getting a message saying " Unknown column 'SN' in 'where clause' " when I select a value from the drop first drop down list. Any ideas whats up? <?php //connection to the database $dbhandle = mysql_connect("localhost", "root", "") or die("Couldn't connect to SQL Server on $myServer"); //select a database to work with $selected = mysql_select_db("waterway", $dbhandle) or die("Couldn't open database myDB"); ?> <!doctype html public "-//w3c//dtd html 3.2//en"> <html> <head> <title>Search Waterways</title> <meta name="GENERATOR" content="Arachnophilia 4.0"> <meta name="FORMATTER" content="Arachnophilia 4.0"> <SCRIPT language=JavaScript> function reload(form) { var val=form.cat.options[form.cat.options.selectedIndex].value; self.location='search.php?cat=' + val ; } </script> </head> <body> <?php @$cat=$HTTP_GET_VARS['cat']; $quer2= "SELECT DISTINCT waterway_initals FROM waterways order by waterway_initals"; $result2 = mysql_query($quer2) or DIE(mysql_error()); if(isset($cat) and strlen($cat) > 0){ $quer= "SELECT DISTINCT waterway_name FROM waterways where waterways.waterway_initals=$cat order by waterway_name"; }else{$quer= "SELECT DISTINCT waterway_name FROM waterways order by waterway_name"; } $result = mysql_query($quer) or DIE(mysql_error()); echo "<form method=post name=f1 action='search_results.php'>"; ////////// Creating the first drop down list ////////// echo "<select name='cat' onchange=\"reload(this.form)\"><option value=''>Select one</option>"; while($row = mysql_fetch_array($result2)) { if($row['waterway_initals']==@$cat){echo "<option selected value='$row[waterway_initals]'>$row[waterway_initals]</option>"."<BR>";} else{echo "<option value='$row[waterway_initals]'>$row[waterway_initals]</option>";} } echo "</select>"; ////////// Ending the first drop down list //////// ////////// Starting of second drop downlist ///////// echo "<select name='subcat'><option value=''>Select one</option>"; while($row = mysql_fetch_array($result)) { echo "<option value='$row[waterway_name]'>$row[waterway_name]</option>"; } echo "</select>"; ////////////////// This will end the second drop down list /////////// echo "<input type=submit value=Submit>"; echo "</form>"; ?> </body> </html> Link to comment https://forums.phpfreaks.com/topic/47103-solved-sql-problem/ Share on other sites More sharing options...
georegjlee Posted April 15, 2007 Author Share Posted April 15, 2007 Just had to put single quote on around the $cat variable in the SQL statement. Link to comment https://forums.phpfreaks.com/topic/47103-solved-sql-problem/#findComment-229707 Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.