警告：mysql_result（）：提供的参数不是有效的MySQL结果资源

Question

Hi can someone explain why I'm receiving this error? Warning: mysql_result(): supplied argument is not a valid MySQL result resource in

if (mysql_result(mysql_query("SELECT count(*) FROM load_test                                                            WHERE batch_id=UCASE('".$batchid."') 
AND word='".$data[2]."',
type='".$data[3]."',
language = '".$data[4]."',
rgender = '".$data[5]."'
"), 0) == 0) {

Answer 1

Hey! You're missing AND between conditions! Don't use commas!

try this:

$query = "SELECT count(*) FROM load_test
          WHERE batch_id=UCASE('".$batchid."')
          AND word='".$data[2]."'
          AND type='".$data[3]."'
          AND language = '".$data[4]."'
          AND rgender = '".$data[5]."'";
$result = mysql_query($query) or die(mysql_error());

In this way you'll can catch the mysql error you are getting when you execute the query.