Array Push Problem

[ts10]

I need to build an array from the cumulative results of a sequence of mysql_query call. I'm trying to use array_push to accomplish this, as follows...

 

   $all_results = array();



   // $variable could be 1-N resulting records



   $variable = mysql(SELECT .....);

   array_push($all_results, $variable);

   .

   .

   $variable = mysql(SELECT .....);

   array_push($all_results, $variable);

   .

   .

   $variable = mysql(SELECT .....);

   array_push($all_results, $variable);

With "$allresults" finally being an array of the results of the 3 queries. But, this does not do the job. I also tried this…

     while ($one_rec = mysql_fetch_array($variable))

 {

    array_push($all_results, $one_rec); // one record at a time

 }

Any suggestions?

[devinemke]

you do not need array_push

$all_results = array();

$result = mysql_query('SELECT * FROM table') or exit(mysql_error());
while ($row = mysql_fetch_assoc($result)) {$all_results[] = $row;}

// repeat as necessary

[rmc060458]

Thanks...the array was apparantly built successfully as you suggested..."sizeof" verified that, with each new element.

However, when I return that array to the calling function, I get this:

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

Does an array that is created as a mysql_query result have some characteristic that is expected when passed to a subsequent "mysql_query" function?

[devinemke]

post your code

[rmc060458]

<?php
function &GetThisUserReferrers($uid)
{
  $L1DispNames=mysql_query("SELECT referrers.DisplayName,referrers.HierarchyKey
                                   FROM referrers
                                   WHERE referrers.registered_by = '$uid' AND
                                         referrers.level = 1
                                   ORDER BY referrers.DisplayName ASC");

  $ThisUserRefs = array();

  while ($ref=mysql_fetch_array($L1DispNames))
  {
    $Level_1 = mysql_query("SELECT referrers.*,user.username
                                   FROM referrers INNER JOIN user
                                   ON user.id = referrers.user_id
                                   WHERE referrers.registered_by = '$uid' AND
                                         referrers.DisplayName =
                                                           '$ref[DisplayName]'
                                   ORDER BY referrers.HierarchyKey");

$L1 = mysql_fetch_assoc($Level_1);
$ThisUserRefs[] = $L1;

$top = $ref[HierarchyKey] . ".";
$RefsInThisGroup = mysql_query("SELECT referrers.*,user.username
                                FROM referrers INNER JOIN user
                                ON user.id = referrers.user_id
                                WHERE referrers.registered_by = '$uid' AND
                                       referrers.HierarchyKey LIKE '$top%'
                                ORDER BY referrers.HierarchyKey");

while ($RefInThisGroup = mysql_fetch_assoc($RefsInThisGroup))
  $ThisUserRefs[] = $RefInThisGroup;
  }
  return($ThisUserRefs);
}
?>

[rmc060458]

return($ThisUserRefs);

[rmc060458]

Let the return out.....this goes before the closing PHP tag

[ts10]

Any ideas on this? I'm lost as well...

[ts10]

Does it matter if I use mysql_fetch_assoc or array? Does this have anything to do with why it's not working?

[laserlight]

Um, ts10 and rmc060458, are you the same person?

[rmc060458]

no

[laserlight]

hmm... then you should have started a new thread. I have no idea what's the problem now since there are two different help requests in a single thread.

[ts10]

We are working on the same project - the problem still persists...

[laserlight]

Oh, ok. That makes things clearer.

Anyway, you should check your:
1. Database server connection
2. Database selection
3. Queries for any errors

[ts10]

1. Database server connection
   --- Working Fine

2. Database selection
   --- Appears to be correct - did you notice any errors in the posted code?

3. Queries for any errors
   --- Same here, appears correct unless you noticed something we missed

[laserlight]

Well, there is a slight mistake in the way you use the arrays in your strings.
$ref[DisplayName] very likely should be $ref['DisplayName'], so:

$Level_1 = mysql_query("SELECT referrers.*,user.username
	FROM referrers INNER JOIN user
		ON user.id = referrers.user_id
	WHERE referrers.registered_by = '$uid'
		AND referrers.DisplayName ='{$ref['DisplayName']}'
	ORDER BY referrers.HierarchyKey");

But this isnt the cause of the warning "mysql_fetch_array(): supplied argument is not a valid MySQL result
resource".

I suggest that you use something along the lines of:

$Level_1 = mysql_query("SELECT referrers.*,user.username
	FROM referrers INNER JOIN user
		ON user.id = referrers.user_id
	WHERE referrers.registered_by = '$uid'
		AND referrers.DisplayName ='{$ref['DisplayName']}'
	ORDER BY referrers.HierarchyKey")
	OR die(mysql_error());

In order to more concisely trace the error.

[toplay]

I second laserlight and this post has my explanation about these types of errors.

FYI:

I would change this:

return($ThisUserRefs);

to not have the parenthesis around the variable.

A quote from the manual:

http://us3.php.net/manual/en/function.return.php wrote:

Note that since return() is a language construct and not a function, the parentheses surrounding its arguments are only required if the argument contains an expression. It is common to leave them out while returning a variable, and you actually should as PHP has less work to do in this case.

Note: You should never use parentheses around your return variable when returning by reference, as this will not work. You can only return variables by reference, not the result of a statement. If you use return ($a); then you're not returning a variable, but the result of the expression ($a) (which is, of course, the value of $a).