PHP variable in foreach loop

Question

I am trying to use a variable in a foreach loop but I am getting strange results. The first foreach loop work fine but gives a notice of a undefined variable and in the second version there is no notice but it only returns the last item in the array.

$formats = array(
    'application/x-mpegurl' => 'hls',
    'video/webm'            => 'webm',
    'video/mp4'             => 'mp4',
    'video/ogg'             => 'ogg',
    'video/flash'           => 'flash',
);

// Works perfectly but there a undefined variable $source
foreach( $formats as $format => $src ){
    if ( !empty( $src ) ) {
        $source .= '<source type="' . $format . '" src="' . $src . '">';
    }
}
echo $source;

// Returns only the last item in the variable but there is no undefined variable
foreach( $formats as $format => $src ){
    $source2 = '';
    if ( !empty( $src ) ) {
        $source2 .= '<source type="' . $format . '" src="' . $src . '">';
    }
}
echo $source2;

I have googled for solutions have not found any.

Sahil Mittal · Accepted Answer · 2014-04-13T20:26:12.487

2

Define $source an d $source1 before the loops start.

 $source = "";
 // loop starts here

Complete code:

$source = "";
foreach( $formats as $format => $src ){
   if ( !empty( $src ) ) {
     $source .= '<source type="' . $format . '" src="' . $src . '">';
   }
}
echo $source;

$source2 = '';
foreach( $formats as $format => $src ){
   if ( !empty( $src ) ) {
      $source2 .= '<source type="' . $format . '" src="' . $src . '">';
   }
}
echo $source2;

edited Apr 13 '14 at 20:26

answered Apr 13 '14 at 20:24

Sahil Mittal

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Do you know why it must be pre-defined before the loop? seems weird to me that it almost acts like a variable scope issue. – agent provocateur Nov 12 '15 at 20:50
Not the variable scope issue but on each iteration he was setting the blank value to the variable – Sahil Mittal Nov 16 '15 at 08:58

score 2 · Answer 2 · answered Apr 13 '14 at 20:24

In both cases the variables need to be defined outside the foreach loop:

$formats = array(
    'application/x-mpegurl' => 'hls',
    'video/webm'            => 'webm',
    'video/mp4'             => 'mp4',
    'video/ogg'             => 'ogg',
    'video/flash'           => 'flash',
);

// Works perfectly but there a undefined variable $source
$source = '';
foreach( $formats as $format => $src ){
    if ( !empty( $src ) ) {
        $source .= '<source type="' . $format . '" src="' . $src . '">';
    }
}
echo $source;

// Returns only the last item in the variable but there is no undefined variable
$source2 = '';
foreach( $formats as $format => $src ){
    if ( !empty( $src ) ) {
        $source2 .= '<source type="' . $format . '" src="' . $src . '">';
    }
}
echo $source2;

score 2 · Answer 3 · answered Apr 13 '14 at 20:24

First problem

Need to define the $source variable outside the loop.

Second issue

Pretty similar to first u need to define the variable outside the loop and then concate inside the loop. You are doing inside the loop which is why its getting over-written and getting the last value.

$source = '';

foreach( $formats as $format => $src ){
    if ( !empty( $src ) ) {
        $source .= '<source type="' . $format . '" src="' . $src . '">';
    }
}
echo $source;
$source2 = '';
// Returns only the last item in the variable but there is no undefined variable
foreach( $formats as $format => $src ){

    if ( !empty( $src ) ) {
        $source2 .= '<source type="' . $format . '" src="' . $src . '">';
    }
}
echo $source2;

score 1 · Answer 4 · edited May 23 '17 at 12:05

The first message undefined variable $source means that the variable named $source has not been defined yet. The code will work without defining the variable source, but it's not the way to go ;)

Although PHP does not require variable declaration, it does recommend it in order to avoid some security vulnerabilities or bugs where one would forget to give a value to a variable that he will use later in the script. What PHP does in the case of undeclared variables is issue a very low level error, E_NOTICE, one that is not even reported by default, but the Manual advises to allow during development.

(PHP: "Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset")

As for your second problem.. You're redefining $source2 every iteration of the loop. Simply move $source2 so that it is defined on the line above the foreach.

// Returns only the last item in the variable but there is no undefined variable
$source2 = '';  // MOVED THIS LINE
foreach( $formats as $format => $src ){    
    if ( !empty( $src ) ) {
        $source2 .= '<source type="' . $format . '" src="' . $src . '">';
    }
}

Read more on defining variables in the PHP manual: http://www.php.net/manual/en/language.variables.basics.php

PHP variable in foreach loop

4 Answers4