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Asked: 2026-05-23T07:41:13+00:00

this is my code function code_exists($code){ $code = mysql_real_escape_string($code); $code_exists = mysql_query(SELECT COUNT(‘url_id’) FROM

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this is my code

function code_exists($code){
  $code = mysql_real_escape_string($code);
  $code_exists = mysql_query("SELECT COUNT('url_id') FROM 'links' WHERE 'code'='$code'");
  return (mysql_result($code_exists, 0) == 1) ? true : false;
}

it is supposed to check the db to make sure the $code is in it or not
but it keeps giving me the same error

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in 
/home/codyl/public_html/projects/tests/url/func.inc.php on line 16
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1 Answer

  1. Editorial Team

    As being said by Shakti you have an error in your sql

    $code_exists = mysql_query("SELECT COUNT(`url_id`) FROM `links` WHERE `code`='$code'");

    You need to surround field name with ticks (`) not single quotes. You should also add a statement to chech if the query was run as expected.

    $code_exists = mysql_query("SELECT COUNT(`url_id`) FROM `links` WHERE `code`='$code'") or die('Error in my query: '. mysql_error());

    This will kill your program throwing the mysql error.

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