Fetch Error getting in all site,

Question

Just launching the sites but it gives error to me

 Warning: mysql_fetch_array() expects parameter 1 to be resource, Boolean given in C:\xampp\htdocs\real4u\index.php on line 203

and on many area of website , i got MySQL fetch error , please help

Index Code

<?php   if(!(isset($_GET["ps"])))        {
$setting=mysql_query("select * from setting ");
$setting1=mysql_fetch_array($setting);
?>

This is a very common question. It means your query is failing. Check for errors with `mysql_error` — datasage, Aug 06 '13 at 11:14
Your query has failed, probably with a syntax error. Echo `mysql_error` after you've submitted your query for the error message. — , Aug 06 '13 at 11:14

Conrad Lotz · Answer 1 · 2013-08-06T11:26:00.903

0

I think there is a problem is with the use of the ! operator. Try the following:

<?php   if(!isset($_GET["ps"]))        {
$setting=mysql_query("select * from `setting`");
$setting1=mysql_fetch_array($setting);
?>

To get better results troubleshooting you can add this after creating the $setting object

if (false === $setting) {
    echo mysql_error();
}

edited Aug 06 '13 at 11:26

answered Aug 06 '13 at 11:16

Conrad Lotz

7,702
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score 0 · Answer 2 · answered Aug 06 '13 at 11:17

0

Try this you will get the reason

$setting = mysql_query("select * from setting ") or die(mysql_error());

answered Aug 06 '13 at 11:17

Manoj Yadav

6,420
1
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score 0 · Answer 3 · answered Aug 06 '13 at 11:18

First, try to use mysqli or PDO, since mysql_* functions are deprecated since PHP5. Second, check the error that mysql returns by editing your mysql_query:

mysql_query("select * from setting ") or exit(mysql_error());

I guess you haven't established a MySQL connection or your setting table doesn't exists.

Fetch Error getting in all site,

3 Answers3