CODEWITHSUNDEEP
phploopsforeachsubstitution
user453366
user453366

Reputation: 75

PHP: foreach variable assignment and referencing: how-to?

I have an array: $aPerfparse as 2-dimensional array where index ranges from 0 to n-1, 

* aPerfparse[index]['label']         -   label of the perfdata  
*                  ['value']         -   actual perfdata  
*                  ['uom']           -   unit of measurement (might be NULL)

Need to iterate through each item and set each indexes 'value' and 'label' to a sep. variable based-on the index.

Without a loop, it would be: 

$value0 = $aPerfdata[0]['value'];  
$value1 = $aPerfdata[1]['value'];

What is correct/incorrect about this?: 

foreach ( $aPerfdata as $key => $value ) {  
    $value$key = $aPerfdata[$key]['value'];  
    $label$key = $aPerfdata[$key]['label'];   
}

Similarly, I need to take those stored $value and $label variables and reference them later in a foreach loop.

Without a loop, it would look like: 

ImageTTFText($img, $fontSize, $fontRotation, 2, $fontSize+2, $oShadow, $fontFile, $label0 . ":" . " " . $value0);  
ImageTTFText($img, $fontSize, $fontRotation, 2, $fontSize+40, $oShadow, $fontFile, $label1 . ":" . " " . $value1);

What is correct/incorrect about this?: 

foreach ( $aPerfdata as $key => $value ) {  
    ImageTTFText($img, $fontSize, $fontRotation, 2, $fontSize+$sz, $oShadow, $fontFile, $label$key . ":" . " " . $value$key);  
    sz=$sz+40;  
}

Thank you!

====

After everyone's help, I have the following working: 

foreach ( $aPerfdata as $key => $value ) 
{
    ${'label'.$key} = $aPerfdata[$key]['label'];  
    ${'value'.$key} = $aPerfdata[$key]['value'];  
}

foreach ( $aPerfdata as $key => $value )
{
    ImageTTFText($img, $fontSize, $fontRotation, 2, $fontSize+$sz, $oShadow, $fontFile, ${'label'.$key} . ":" . " " . ${'value'.$key});
    $sz=$sz+40;
}

I don't really have a need to flatten the array anymore. I tried the method mentioned by Mark, but the ImageTTFText function doesn't execute.

Upvotes: 3

Views: 2460

Answers (4)

Mark Elliot
Mark Elliot

Reputation: 77054

Why not just do

foreach ( $aPerfdata as $value ) {  
    ImageTTFText($img, $fontSize, $fontRotation, 2, $fontSize+$sz, $oShadow, $fontFile, $value['label'] . ":" . " " . $value['value']);  
    $sz=$sz+40;  
}

Among other things I think your use of variable variables is incorrect. You should really have something like:

${'value'.$key}

But really you should just be using an array rather than variable variables - and since you already have an array there's no real need to flatten it (as I show you above).

Upvotes: 0

Patrick
Patrick

Reputation: 41

It looks to me like you might be confusing foreach with a simple for loop.

If you have an array that looks like this:

Array
(
    [0] => Array
        (
            [label] => red
            [value] => 8
            [uom] => cm
        )

    [1] => Array
        (
            [label] => green
            [value] => 6
            [uom] => m
        )

    [2] => Array
        (
            [label] => blue
            [value] => 34
            [uom] => m
        )

)

You can access iterate over the keys/values as follows.


foreach($arr as $array_key => $array_value)
{
        // obviously you don't need to assign these. it's just for demonstration
        $key = $array_key;
        $label = $array_value['label'];
        $value = $array_value['value'];
        $uom = $array_value['uom'];
}

Hope that helps.

Upvotes: 1

Tilman Koester
Tilman Koester

Reputation: 1739

Firstly:

$label$key

is wrong. Should be:

$label.$key

You need to concatenate the variables with a dot(.).

Didn't understand the second part of the question. Can you just paste a var_dump of the array so we can get a clearer understanding of the structure?

Upvotes: 1

Chris Laplante
Chris Laplante

Reputation: 29658

You have a two dimensional array. The foreach statements iterates through the first dimension for you. This should work:

foreach ( $aPerfdata as $item ) {  
    $value = $item['value'];  
    $label = $item['label'];   
}

Also, I don't think you should include $ in your variable name, other than of course in front of the name.

Upvotes: 0

Related Questions