php doesn't fetch end of array

Question

foreach doesn't fetch end of array. I want to use end of that in another method:

$array = array(
"Language programings"=>array("php" => 100,"js" => 200),'html'=>12);
 foreach ($array as $kk=>$val1)
   {
     echo $kk.'<br/>';
     foreach ($val1 as $key=>$val2)
      {  
         if (! end(array_keys($array)))
             echo $val2;
      }
      echo end(array_value);//must be show 12
   }

foreach will always iterate over all of the elements of an array, including the last. that's why it's called forEACH. try a regular for loop instead. — Andreas Grapentin
– Andreas Grapentin, Commented Sep 5, 2012 at 5:49

nageeb · Accepted Answer · 2012-09-05 05:56:54Z

1

If I understand correctly, in your if() statement, you're attempting to see if the pointer is at the end of the array. Unfortunately, in this case, end() will never be false and so the line echo $val2; won't ever execute.

Try replacing

if (! end(array_keys($array)))

with

if ($key <> end(array_keys($array))

also your last line should be:

echo end(array_values($array));

answered Sep 5, 2012 at 5:56

nageeb

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Comments

Littm · Accepted Answer · 2012-10-01 03:00:20Z

0

Try using below code:

$array = array("Language programings"=>array("php" => 100,"js" => 200),'html'=>12);

foreach($array as $key1 => $val1){
    if(is_array($val1)){
        echo $key1.'<br/>';
        foreach($val1 as $key2 => $val2){
            if(is_array($val2)){
                foreach($val2 as $k => $v){
                    echo $v.'<br/>';
                }
            } else {
                echo $val2.'<br/>';
            }
        }
    }
}
echo end(array_values($array));

The result will be:

Language programings  
100  
200  
12

edited Oct 1, 2012 at 3:00

Littm

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answered Sep 5, 2012 at 7:00

Pragnesh Chauhan

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1 Comment

DolDurma Over a year ago

thanks, in result of this code html is additional. and not showing

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