How can I loop through this array - PHP

Question

This is the var_dump($options)

array (size=4)
  0 => string '2' (length=1)
  'parent_2_children' => 
    array (size=3)
      0 => string '22' (length=2)
      1 => string '24' (length=2)
      2 => string '23' (length=2)
  1 => string '3' (length=1)
  'parent_3_children' => 
    array (size=3)
      0 => string '26' (length=2)
      1 => string '25' (length=2)
      2 => string '27' (length=2)

What I have tried up to now is

if(!is_null($options))
        {
            foreach($options as $option)
            {
                 if(!array_key_exists('parent_'.$option.'_children',$options))
                 {
                    //if position null output an error
                 }
            }   


        }

Print_r() as requested

Array
(
    [0] => 2
    [parent_2_children] => Array
        (
            [0] => 22
            [1] => 24
            [2] => 23
        )

    [1] => 3
    [parent_3_children] => Array
        (
            [0] => 26
            [1] => 25
            [2] => 27
        )

)

What do you want to do while you are looping through it?

user1726343
– user1726343

2012-10-24 07:06:09 +00:00
Commented Oct 24, 2012 at 7:06 — user1726343
– user1726343, Commented Oct 24, 2012 at 7:06
if position null output an error

Techie
– Techie

2012-10-24 07:08:55 +00:00
Commented Oct 24, 2012 at 7:08 — Techie
– Techie, Commented Oct 24, 2012 at 7:08
Your existing code should be fine for that.

user1726343
– user1726343

2012-10-24 07:12:32 +00:00
Commented Oct 24, 2012 at 7:12 — user1726343
– user1726343, Commented Oct 24, 2012 at 7:12

VDP · Accepted Answer · 2012-10-24 07:34:29Z

1

use print_r($options) in staid of var_dump it's easier to read..

check if you got a numeric key, then check if the new key exists. Throw an error.

if(!is_null($options)) {
    foreach($options as $key => $option) {
        if(is_int($key) && !array_key_exists('parent_'.$option.'_children',$options)) {
           echo 'parent_'.$option.'_children does not exist';
        }
    }   
}

Here is a working example

edited Oct 24, 2012 at 7:34

answered Oct 24, 2012 at 7:08

VDP

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3 Comments

VDP Over a year ago

what do you want it to do? I can throw an exception, or echo some text... I'll update. But note that it will only show something when the key doesn't exist

Techie Over a year ago

i'm already echo 'parent_'.$option.'_children'; this. does not work

VDP Over a year ago

the array you provided is a valid array. If you got an invalid key, it will output the message. try pushing a non existing value like $options[0] = '9999999'; before running the loop

janenz00 · Accepted Answer · 2012-10-24 07:41:07Z

0

Your code is correct. An additional check on the nature of the key will reduce the execution, since you have to do the processing only for numeric keys.

    if(!is_null($options))
    {
        foreach($options as $key => $option)
        {
            if (is_numeric($key)) {
                 if(!array_key_exists('parent_'.$option.'_children',$options))
                  {
                       print 'parent_'.$option.'_children does not exist';
                   }
             }
        }   


    }

To test the code, try the following array :

 $options = array(0 => 2, 'parent_2_children' => array ( 0 => 22, 1 => 24, 2 => 23 ), 1 => 3, 'parent_3_children' => array ( 0 => 26, 1 => 25, 2 => 27 ), 2=>4 );

whose print_r output will be :

 Array
(
[0] => 2
[parent_2_children] => Array
    (
        [0] => 22
        [1] => 24
        [2] => 23
    )

[1] => 3
[parent_3_children] => Array
    (
        [0] => 26
        [1] => 25
        [2] => 27
    )

[2] => 4

)

and, it will output :

  parent_4_children does not exist

edited Oct 24, 2012 at 7:41

answered Oct 24, 2012 at 7:16

janenz00

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1 Comment

janenz00 Over a year ago

Right now, it will not output anything, since both array keys of the form parent_'.$option.'_children exists in the array

user1726343 · Accepted Answer · 2012-10-24 07:14:28Z

0

Try this?

if(!is_null($options)){
    foreach($options as $option){
        if(!array_key_exists('parent_'.$option.'_children',$options)){
            throw new Exception("Something went wrong!");
        }
    }   
}

answered Oct 24, 2012 at 7:14

user1726343

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