Fearpig Posted September 11, 2006 Share Posted September 11, 2006 Hello could someone take a look at this for me. It should just list any model where the Air Pressure Switch is 39800140 OR the Auto Air Vent is 39810370. I get a couple of error messages and I don't think I've got the "Or" bit correct.[code]$db = mysql_connect("localhost", "root", "password")or die ("Error connecting to database.");mysql_select_db("Telephonelist",$db) or die ("Couldn't select the database.");$result = mysql_query("SELECT * FROM tbl_parts WHERE Air_Pressure_Switch='39800140' || Auto_Air_Vent='39810370'",$db); $row = mysql_fetch_array($result);while ($row = mysql_fetch_array($result)) { printf("<b class='Body2'>%s</b><br>", $row["Model"]); }[/code]I get the following errors:Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in D:\Intranet v3\TEST\Spares\test.php on line 11Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in D:\Intranet v3\TEST\Spares\test.php on line 13 Quote Link to comment Share on other sites More sharing options...
obsidian Posted September 11, 2006 Share Posted September 11, 2006 in SQL, you actually have to us "OR" not the "||" operator:[code]SELECT * FROM tbl_parts WHERE Air_Pressure_Switch='39800140' OR Auto_Air_Vent='39810370'[/code] Quote Link to comment Share on other sites More sharing options...
°°Ben³ Posted September 11, 2006 Share Posted September 11, 2006 Use the mysql_error() function in future to receive a more detailed error message! i.e.[code]$resource = mysql_query($sql) or exit(mysql_error());[/code]That is really helpful while developing! Quote Link to comment Share on other sites More sharing options...
Fearpig Posted September 11, 2006 Author Share Posted September 11, 2006 Thank you Dan and Obsidian both very helpful! ;D Quote Link to comment Share on other sites More sharing options...
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