list() query not working

ArizonaJohn · #1 June 16th, 2009, 06:16 AM

Hello,

When I run the code below when $entry = miami.com, I get the following error message:

Quote:

SELECT COUNT(*) FROM #&*+ WHERE `site` LIKE 'miami.com':You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

It looks like I'm not correctly defining $table. Any ideas how I could do that?

Thanks in advance,

John

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 $result = mysql_query("SHOW TABLES FROM feather") 

or die(mysql_error()); 
 
while(list($table)= mysql_fetch_row($result))

{

  $sqlA = "SELECT COUNT(*) FROM $table WHERE `site` LIKE '$entry'";

  $resA = mysql_query($sqlA) or die("$sqlA:".mysql_error());

  list($isThere) = mysql_fetch_row($resA);

  if ($isThere)

  {

     $table_list[] = $table;

  }

}

hsriat · #2 June 16th, 2009, 04:07 PM

I think '%' is missing.
http://dev.mysql.com/doc/refman/5.0/...-matching.html

prabirchoudhury · #3 June 16th, 2009, 10:50 PM

hey.

If no wildcard characters (%%, …, _) are used then there should be no difference in
performance between

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“… WHERE site='$entry' ” and “..WHERE site LIKE '$entry'”

the two queries . so , I think you need to check with your query and passing variable matches.

:)

list() query not working

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