Question

array_rand() in newer php version

Asked by: planetemails

Hi ...
I got this warning when testing an old script to newer php version (4.4.7)

Warning: array_rand() [function.array-rand]: First argument has to be an array in C:\xampp\htdocs\mlm\system\randomizer.php on line 186

And below is the excerpt codes where it uses array_rand() ... I hope one can help me modify it to makes the script works on php 4.4.7 ... Thanks.

if ( $data['reff_level'] == $all )  
{ $choose = mysql_query( "SELECT * FROM member WHERE status='$status'") or error( mysql_error() ); 
} else { 
$choose = mysql_query( "SELECT * FROM member WHERE status='$status' AND level='$level'") or error( mysql_error() ); }
 
$i2r = 0;
while ( $row = mysql_fetch_array( $choose ) )
      {
      $bb2r[$i2r]=$row['username'];
      $i2r++;
      }
$bb12r    = array_rand($bb2r, 1) ;
$randomer = $bb2r[$bb12r];
 
$result = mysql_query( "SELECT * FROM member WHERE username='$randomer'" ) or error( mysql_error() );
	 
$member = mysql_fetch_array( $result );

                                  
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Asked On
2008-02-18 at 02:57:30ID23170999
Topic

PHP Scripting Language

Participating Experts
3
Points
500
Comments
11

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Answers

 

by: RoonaanPosted on 2008-02-18 at 02:59:43ID: 20918912

You have not defined $bb2r as an array in this snippet. When your query returns 0 rows, it will never be available as an array.

What does print_r($bb2r) show?

-r-

 

by: angelIIIPosted on 2008-02-18 at 02:59:45ID: 20918913

try this:

if ( $data['reff_level'] == $all )  
{ $choose = mysql_query( "SELECT * FROM member WHERE status='$status'") or error( mysql_error() ); 
} else { 
$choose = mysql_query( "SELECT * FROM member WHERE status='$status' AND level='$level'") or error( mysql_error() ); }
 
$i2r = 0;
$bb2r = array(); 
while ( $row = mysql_fetch_array( $choose ) )
      {
      $bb2r[$i2r]=$row['username'];
      $i2r++;
      }
$bb12r    = array_rand($bb2r, 1) ;
$randomer = $bb2r[$bb12r];
 
$result = mysql_query( "SELECT * FROM member WHERE username='$randomer'" ) or error( mysql_error() );
       
$member = mysql_fetch_array( $result );
                                              
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by: siliconbritPosted on 2008-02-18 at 03:05:39ID: 20918934


In your current code, $bb2r is only created and initialised the first time that mysql_fetch_array succeeds.  If it does not succeed, then not on is $bb2r not an array, it doesn't exist at all.

Try declaring the array just before your while loop starts:

$i2r = 0;
$bb2r=array();  # <--- array declared
while ( $row = mysql_fetch_array( $choose ) )

This ensures that what you pass to array_rand() is an array even if no results are returned by mysql_fetch_array.

 

by: planetemailsPosted on 2008-02-18 at 03:28:51ID: 20919015

Wouw ... it goes more complex than I thought unfortunately .. :-(

At first, as @angelIII suggest. I put:
$i2r = 0;
$bb2r=array();

But I really have no idea what/how to array $bb2r.
As @Roonaan suggest, I put this at the end:
print_r($bb2r)
.. and it only prints >> Array ( )

@siliconbrit .. by having a look at the codes, can you suggest me what to array for $bb2r

Anyone ... ?

 

by: angelIIIPosted on 2008-02-18 at 03:34:04ID: 20919037

>But I really have no idea what/how to array $bb2r.

$bb2r=array();
just sets the variables to be of type array (and empty to initialize), and hence the error should go away.

 

by: planetemailsPosted on 2008-02-18 at 03:41:07ID: 20919063

@angelIII
wouw ... never thought it's gonna be that simple .. :p
Okay I put $bb2r=array();
and retest the script but then it shows this error

Warning: array_rand() [function.array-rand]: Second argument has to be between 1 and the number of elements in the array in C:\xampp\htdocs\mlm\system\randomizer.php on line 186

Any more suggestion bro?

Here's as I add the array():

$i2r  = 0;
$bb2r = array();
while ( $row = mysql_fetch_array( $choose ) )
      {
      $bb2r[$i2r]=$row['username'];
      $i2r++;
      }
 
$bb12r    = array_rand($bb2r, 1) ;
$randomer = $bb2r[$bb12r];

                                              
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by: RoonaanPosted on 2008-02-18 at 03:50:01ID: 20919099

Can you validate that your query is actually returning any results?

 

by: planetemailsPosted on 2008-02-18 at 03:57:18ID: 20919134

@Roonaan
This script is actually working on php 4.1.0 (I tried using PHPTriad and it's totally working).

But next when I use XAMPP (php 4.4.7 which is also the version of php used in my hosting) there are several issues need to be fixed .. and this array_rand() is one of them.

So .. what/how you want me to validate the query ... sorry for all of my limitations in advance ... just trying to learn something by doing .. :-)

 

by: angelIIIPosted on 2008-02-18 at 03:59:02ID: 20919140

if your array is empty, no need to randomize into it:

$i2r  = 0;
$bb2r = array();
while ( $row = mysql_fetch_array( $choose ) )
      {
      $bb2r[$i2r]=$row['username'];
      $i2r++;
      }
 
if ($i2r>0)
{
  $bb12r    = array_rand($bb2r, 1) ;
  $randomer = $bb2r[$bb12r];
}
else
{
  // no items in array, nothing to be returned:
  $randomer = "no item returned";
}
                                              
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by: planetemailsPosted on 2008-02-18 at 04:07:15ID: 20919175

@angelIII
seems reasonable ... :-)

I'll be playing with it for a while and will let you know l8r how is it going ... Thanks.

 

by: planetemailsPosted on 2008-03-07 at 01:34:18ID: 31431743

This is it bro ... Thanks a loT.

20120131-EE-VQP-002

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