illegal string offset foreach loop

Question

HI Following is the code , i am trying to print the array which contain values of drop down menu and then 2nd foreach loop is trying to selected the field. The output is the way i want , mean it print the all menu item, and the selected one,

foreach($results as $data): 
        $st = '';
        foreach($SelectedActor as $SelectedActor):
        if($SelectedActor['id']==$data['id']){$st='selected="selected"';}
         endforeach;
        $dd .="<option ".$st."  value=".$data['id']."> ".$data['artist_name']."    </option>";

        endforeach;

but the page also show this error

A PHP Error was encountered

Severity: Warning

Message: Illegal string offset 'id'

Filename: controllers/replik.php

Line Number: 328

Vardump for these two array are

array (size=2)
  0 => 
    array (size=2)
      'id' => string '1' (length=1)
      'artist_name' => string ' oyuncu' (length=19)
  1 => 
    array (size=2)
      'id' => string '4' (length=1)
      'artist_name' => string 'hep brabir' (length=10)

array (size=2)
  0 => 
    array (size=2)
      'id' => string '1' (length=1)
      'artist_name' => string 'oyuncu' (length=19)
  1 => 
    array (size=2)
      'id' => string '4' (length=1)
      'artist_name' => string 'hep brabir' (length=10)

Can any one help me to get the reason of this error.

var_dump($SelectedActor, $data) - one of these is not an array. — deceze
– deceze ♦, Commented Nov 18, 2013 at 12:51

Patato · Accepted Answer · 2013-11-18 12:53:22Z

2

 foreach($SelectedActor as $SelectedActor):
        if($SelectedActor['id']==$data['id']){$st='selected="selected"';}

here you use the same name of $SelectedActor, maybe

foreach($SelectedActor as $ItemSelectedActor):
            if($ItemSelectedActor['id']==$data['id']){$st='selected="selected"';}

is ok

answered Nov 18, 2013 at 12:53

Patato

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2 Comments

Dexture Over a year ago

Sorry for the stupid mistake i make here. I was doing foreach($SelectedActor as $SelectedActor): i do it like foreach($SelectedActor as $SelectedActores): just changes the second variable name and it work.

Dexture Over a year ago

but i dont unsdertand in this case why it gives error. mostly when do foreachloop way it work fines, in this case there were two foreach loop so its errror ? or thier is ome other reason.

Ramesh · Accepted Answer · 2013-11-18 16:51:34Z

0

Below line is wrong:

foreach($SelectedActor as $SelectedActor)

You need to do:

foreach($SelectedActor as $actor):
    if($actor['id']==$data['id']){
        $st='selected="selected"';
    }
endforeach;

edited Nov 18, 2013 at 16:51

answered Nov 18, 2013 at 12:52

Ramesh

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1 Comment

asprin Over a year ago

You might want to consider fixing this line if($actor'id']==$data['id']){

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