How do I return a variable instead of a string?

Question

In this example, the function validateJsonNotEmptyImage does not work as expected?

Just to clarify.

'src' => $image->src, // WORKS
'height' => $this->validateJsonNotEmptyImage('type', 'width'), // DOES NOT WORK

public function validateJsonNotEmptyImage($type, $array1) {
    if ($type, $array1) {
        if (isset($type->$array1)) {
            return $type . '->' . $array1;
        } else {
            return null;
        }
    }
}

$productsImagesToSiteArray = array();
    foreach ($this->resource->images as $image) {
        $productsImagesToSiteArray[] = array(
            'height' => $this->validateJsonNotEmptyImage('$image', 'width'),
            'src'    => $image->src,
        );
    }

your should remove single quote on $image

lighter
– lighter

2019-11-14 01:38:50 +00:00
Commented Nov 14, 2019 at 1:38 — lighter
– lighter, Commented Nov 14, 2019 at 1:38

daremachine · Accepted Answer · 2019-11-14 01:46:00Z

1

I see two mistakes

1) your method validateJsonNotEmptyImage, in body you have wrong IF it should be check with null[here1] and you should return value not string[here2].
And also you should return everytime any value or null[here3].

public function validateJsonNotEmptyImage($type, $array1) {
    if ($type != null && $array1 != null) {            <----------- here1
        if (isset($type->$array1)) {
            return $type->$array1;                     <----------- here2
        } else {
            return null;
        }
    }
    return null;                                       <----------- here3
 }

2) you should pass variable not string in your call (remove quotes from $image)

$this->validateJsonNotEmptyImage($image, 'width')

edited Nov 14, 2019 at 1:46

answered Nov 14, 2019 at 1:40

daremachine

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