undefined offset PHP error in looping

Question

it says an error undefined offset i dont know what causes it

im trying to shuffle $numberarray without repeating the number

heres the code

$numberarray = array(1,2,3,4,5,6,7,8,9,10);
for($counter=0;$counter<=9;$counter++)
        {
        $b = $counter - 1;
        $a = $numberarray[$counter];
            $numberarray[$counter] = rand(1,10);
            do
            {
                $numberarray[$counter] = rand(1,10);

                while($b != 0)
                {
                    if($numberarray[$counter] == $numberarray[$b])
                    {

                      $numberarray[$counter] = rand(1,10);
                      $b = $counter - 1;
                      //echo $b;
                    }
                    else
                    {
                      $b--;
                    }
            }

    }while($a == $numberarray[$counter]);
    echo $numberarray[$counter].", ";
}

sample output $numberarray = {3,4,5,1,2,7,9,10,8,5}

why not use shuffle function inbuilt in php details at php.net/manual/en/function.shuffle.php — undefined_variable
– undefined_variable, Commented Sep 27, 2014 at 16:03

Adam Sinclair · Accepted Answer · 2014-09-27 16:13:12Z

1

This would be better:

(By the way you don't need an extra condition before the while since while act like a condition itself)

     $numberarray = array(1,2,3,4,5,6,7,8,9,10);
     for($counter=0;$counter<=9;$counter++)
        {
        $b = $counter - 1;
        $a = $numberarray[$counter];
            $numberarray[$counter] = rand(1,10);
            do
            {
              $numberarray[$counter] = rand(1,10);

             while($b >0)
             {
                if($numberarray[$counter] == $numberarray[$b])
                {

                    $numberarray[$counter] = rand(1,10);
                    $b = $counter - 1;
                    //echo $b;
                }
                else
                {
                    $b--;
                }
             }

        }while($a == $numberarray[$counter]);
        echo $numberarray[$counter].", ";
    }

Even if there is no more error, your code repeat numbers sometimes, so I would symply do it like this using shuffle:

      $numberarray2 = array(1,2,3,4,5,6,7,8,9,10);
      shuffle($numberarray2);
      print_r($numberarray2);

edited Sep 27, 2014 at 16:13

answered Sep 27, 2014 at 16:03

Adam Sinclair

1,64912 silver badges15 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

eman_invok05 Over a year ago

tnx for your answer sir but its not working sir... its still repeating other numbers i like to shuffle them by no repeating numbers sample output $numberarray = {3,4,1,2,7,5,6,10,8,9}

Abdessabour Mtk · Accepted Answer · 2014-09-27 16:04:31Z

0

The problem is that in the first $b is equal to -1 that's why you get the error so i think you should delete the if condition and edit the while statement to while($b!=-1).

answered Sep 27, 2014 at 16:04

Abdessabour Mtk

3,9082 gold badges16 silver badges21 bronze badges

3 Comments

Adam Sinclair Over a year ago

gosh, where did you get this idea?!

Mark Ormesher Over a year ago

@AdamSinclair As it was posted at about the same time as yours, probably from reading the question and thinking about it. Great minds sometimes think alike - doesn't necessarily mean you were copied from.

Abdessabour Mtk Over a year ago

@AdamSinclair Magic isn't it beautiful.Also My mind can't think that's why I use sorcery.

Collectives™ on Stack Overflow

undefined offset PHP error in looping

2 Answers 2

1 Comment

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related