mysql_fetch_row() Error

[ArizonaJohn]

Hello,

 

When I run the query below, I get this error for the line "list($isThere) = mysql_fetch_row($resA);":

 

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource

 

Any ideas why?

 

Thanks in advance,

 

John

 

<?php
$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);
  list($isThere) = mysql_fetch_row($resA);
  if ($isThere)
  {
     $table_list[] = $table;
  }
}
?>

[joel24]

check to see if that sql statement is executing properly?

change the $reA = mysql_query... line to the following... you'll want to change it back when you make the site live and put in some custom error handling

$resA = mysql_query($sqlA) or die(mysql.error());

 

[ArizonaJohn]

joel24,

 

When I follow your suggestion, I get the following error:

 

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

[MadTechie]

change

$resA = mysql_query($sqlA) or die(mysql.error());

to

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

 

But it sounds like $entry isn't being set

[ArizonaJohn]

MadTechie,

 

When I follow your suggestion, I get the following error:

 

"Fatal error: Call to undefined function error()" for the line that has "$resA = mysql_query($sqlA) or die("$sqlA:".mysql.error());"

[MadTechie]

Oh my bad..

mysql.error should be mysql_error

[ArizonaJohn]

OK, MadTechnie,

 

Now, when I enter in miami.com for $entry, I get the following error message:

 

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

[garethhall]

Are you select your DB? I am not seeing the select part in you code.

<?php
mysql_select_db(database,connection)
?>

 

[MadTechie]

Ahhh

 

i assume you entered

#&*+

while(list($entry)= mysql_fetch_row($result)) //Update this line $table to $entry

[ArizonaJohn]

MadTechnie,

 

When I follow your instructions, I get the following error:

 

SELECT COUNT(*) FROM WHERE `site` LIKE '#&*+':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 'WHERE `site` LIKE '#&*+'' at line 1

[MadTechie]

$table isn't being set.. what table are you searching ?