Unable to retreve the values from Mysql Query

[surajkandukuri]

Hi,

 

  Here is the php code that I have, Query is running properly in phpmyadmin  and is resulting 5 which is correct but when i try to get tht in php its not printing the value

 

  $sql_query= "Select Count(*) from (Select Cat_Num from cat_completiondate where `EmployeeNumber`=$employeenumber and `CategoryJobPosition` =$jobid) as temp";

      $query_result2=mysql_query($sql_query);

      $result_row2=mysql_fetch_array($query_result2);

      $value=$result_row2['temp'];

      echo($value)

 

 

  Output:- It doesnot print anything.

   

[GKWelding]

$sql_query= "Select Count(*) from (Select Cat_Num from cat_completiondate where `EmployeeNumber`=$employeenumber and `CategoryJobPosition` =$jobid) as temp";

$query_result2=mysql_query($sql_query);

while($result_row=mysql_fetch_array($query_result2){
$value = $result_row['temp'];
echo($value);
}

 

also, try using code tags.

[GKWelding]

just as a pointer, I knew the answer to this straight off, but if you didn't then try googling it first, I just googled mysql_fetch_array to test this and the first page back was http://php.net/manual/en/function.mysql-fetch-array.php which shows you exactly what I've just shown you above.

[dawsba]

function q($query,$assoc=1) {
$r = @mysql_query($query);
if( mysql_errno() ) {
  $error = 'MYSQL ERROR #'.mysql_errno().' : ' . mysql_error(). '< / small>< br>< VAR>$query< /VAR>';
  echo($error); return FALSE;
}  if( strtolower(substr($query,0,6)) != 'select' ) return array(mysql_affected_rows(),mysql_insert_id());
$count = @mysql_num_rows($r);
if( !$count ) return 0;
if( $count == 1 ) {
  if( $assoc ) $f = mysql_fetch_assoc($r);
  else $f = mysql_fetch_row($r);
  mysql_free_result($r);
  if( count($f) == 1 ) {
   list($key) = array_keys($f); 
   return $f[$key];
  } else {
   $all = array();
   $all[] = $f;
   return $all;
  }
} else {
  $all = array();
  for( $i = 0; $i < $count; $i++ ) {
   if( $assoc ) $f = mysql_fetch_assoc($r);
   else $f = mysql_fetch_row($r);
   $all[] = $f;
  }
  mysql_free_result($r);
  return $all;
}
}

add the function and call as such

$temp = q("Select Count(*) from (Select Cat_Num from cat_completiondate where `EmployeeNumber`=$employeenumber and `CategoryJobPosition` =$jobid)");

[DavidAM]

mysql_fetch_array returns an associative array or numeric array or both.  However, your query does not return a column named 'temp' so your assignment:

$value=$result_row2['temp'];

is not assigning anything to $value.

 

1. Turn on error reporting at the beginning of your script error_reporting(E_ALL);.  Then you will see the error message about referring to a non-existant key.

2. Either specifically assign the column a name or refer to the numeric index

$sql_query= "Select Count(*) AS temp from (Select Cat_Num from cat_completiondate where `EmployeeNumber`=$employeenumber and `CategoryJobPosition` =$jobid) as temp";
       $query_result2=mysql_query($sql_query);
       $result_row2=mysql_fetch_array($query_result2);
       $value=$result_row2['temp'];
       $value=$result_row2[0]; // Should return the same value
       echo($value)

 

[surajkandukuri]

Thank you DavidAM,

 

                    My problem is solved and I could see the error I made.