Nothadoth Posted August 17, 2006 Share Posted August 17, 2006 I know that you can use the clause WHERE this_field='$this'However, is there a way you can use two WHEREs? Can someone tell me how to do this?Something like (it doesn't work i've tried it):WHERE this_field='$this' AND this_other_field='$this2'Thanks! Quote Link to comment Share on other sites More sharing options...
brown2005 Posted August 17, 2006 Share Posted August 17, 2006 WHERE this_field='$this' AND this_other_field='$this2'that is correct...... Quote Link to comment Share on other sites More sharing options...
Nothadoth Posted August 17, 2006 Author Share Posted August 17, 2006 well I am using this and it says that it isn't working.[code] $queryphones = mysql_query("SELECT * FROM phones WHERE manufacturer='$catmode' AND special='$type' ORDER by model ASC");while($phones = mysql_fetch_array($queryphones)) { }[/code]Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/noth/public_html/igbltd/browse.php on line 313http://www.finalfantasyfan.net/igbltd/browse.php?catmode=Sony%20Ericsson&parentmode=Phones&type=WalkmansPlease help Quote Link to comment Share on other sites More sharing options...
GingerRobot Posted August 17, 2006 Share Posted August 17, 2006 Try adding an or die statement. Also, separating the query out will help so you can echo the query string: $sql = "SELECT * FROM phones WHERE manufacturer='$catmode' AND special='$type' ORDER by model ASC"; $queryphones = mysql_query($sql) or die("mysql_error()<br />Sql:$sql");while($phones = mysql_fetch_array($queryphones)) { } Quote Link to comment Share on other sites More sharing options...
brown2005 Posted August 17, 2006 Share Posted August 17, 2006 <?phpsession_start();$database_host = "localhost";$database_username = "";$database_password = "";$database_name = "";$connection = mysql_connect($database_host, $database_username, $database_password) or die(mysql_error());$db = mysql_select_db($database_name, $connection);$catmode = "a";$type = "a";$queryphones = mysql_query("SELECT * FROM phones WHERE manufacturer='$catmode' AND special='$type' ORDER by model ASC");while($phones = mysql_fetch_array($queryphones)) { echo"$phones[manufacturer]";}?>that works and echos out results... Quote Link to comment Share on other sites More sharing options...
Nothadoth Posted August 17, 2006 Author Share Posted August 17, 2006 still not working.Error:[code]Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/noth/public_html/igbltd/browse.php on line 319[/code]Ill give you the whole snippet from that area. It starts at line 286 and ends at line 300[code]if ($type == "All") { $queryphones = mysql_query("SELECT * FROM phones WHERE manufacturer='$catmode' ORDER by model ASC"); $temp1 == ""; } elseif (!$type == "All") { $temp1 == $type; $sql = "SELECT * FROM phones WHERE manufacturer='$catmode' AND special='$type' ORDER by model ASC"; $queryphones = mysql_query($sql) or die("mysql_error()Sql:$sql"); }[/code] Quote Link to comment Share on other sites More sharing options...
Nothadoth Posted August 17, 2006 Author Share Posted August 17, 2006 The reason the error is after line 300 is because the while statement is further down Quote Link to comment Share on other sites More sharing options...
GingerRobot Posted August 17, 2006 Share Posted August 17, 2006 Well i would guess the error is on the other query then[code]<?phpif ($type == "All") { $sql= "SELECT * FROM phones WHERE manufacturer='$catmode' ORDER by model ASC"; $queryphones = mysql_query($sql) or die("mysql_error() Sql:$sql"); $temp1 == ""; } elseif (!$type == "All") { $temp1 == $type; $sql = "SELECT * FROM phones WHERE manufacturer='$catmode' AND special='$type' ORDER by model ASC"; $queryphones = mysql_query($sql) or die("mysql_error() Sql:$sql"); }?>[/code] Quote Link to comment Share on other sites More sharing options...
Nothadoth Posted August 17, 2006 Author Share Posted August 17, 2006 I got it working. I used what GingerRobot said.For some reason if I take out the elseif (!$type == "All") and change it to just else it works.If I leave it as elseif (!$type == "All") then it shows nothing in the while loop. Quote Link to comment Share on other sites More sharing options...
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