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Home/ Questions/Q 9179187
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T17:44:17+00:00 2026-06-17T17:44:17+00:00

I am stuck in a very basic sql query scrip. Mybe someone can notice

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I am stuck in a very basic sql query scrip. Mybe someone can notice what I am unable to see. I have checked that the sql query is working fine when it is executed from mysqladmin:

 <?
 include ('gps_db_connect.php'); 
 $query = "SELECT * from gps WHERE server_time > '20130124'";
 echo $query;
 $result = mysqli_query($connection, $query) or die(' Error getting data'); 
 echo '    After query';
 while ($row = mysql_fetch_array($result, MYSQL_ASSOC))
  {   
   echo $row['server_time'];
  }

 ?>

Here is the screen display/output:

SELECT * from gps WHERE server_time > '20130124' After                      
query
PHP Error Message 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a9440109/public_html/test.php on line 7
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-17T17:44:19+00:00Added an answer on June 17, 2026 at 5:44 pm

    try this

     $result = mysql_query($query, $connection) or die('Error');
    
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