Delete some data from Array in php

Question

I want to delete some data from an array in PHP. Here is the array:

array(4) {
    [0]=> array(1) { ["image"]=> string(20) "w85YrKChBGTZ9fQS.jpg" }
    [1]=> array(1) { ["image"]=> string(20) "3buahEs6rRWFdYez.jpg" }
    [2]=> array(1) { ["image"]=> string(20) "gYPtDrx3sFzkVENB.jpg" }
    [3]=> array(1) { ["image"]=> string(20) "JE3rodDvs6521cFm.jpg" }
}

Here is my method and where I am deleting:

public function deleteImage(){
    foreach (getCarImages() as $array){
        //var_dump($array).'<br>';
        $index = array_search('w85YrKChBGTZ9fQS.jpg',$array);
        if($index !== FALSE){
            var_dump($index).'<br>';
            unset($array[$index]);
        }else{
            echo '<br>else here';
        }
    }
}

And here is the result of deleteImage()

string(5) "image"

else

here

else

here

else

here

I am confused. How can I delete a nested array from the main array.

try if($index). if item not found then array_search return empty. — urfusion
– urfusion, Commented Dec 13, 2017 at 7:06
please post your getCarImages() function. What is deleteImage() actually going to do? Is there a return value? A file function? A database query? — mickmackusa
– mickmackusa ♦, Commented Dec 13, 2017 at 7:15

u_mulder · Accepted Answer · 2017-12-13 09:09:48Z

2

If you need to delete a whole subarray from array, then use array_flter function:

public function deleteImage(){
    return array_filter(getCarImages(), function ($v) {
        return $v['image'] != 'w85YrKChBGTZ9fQS.jpg';
    });
}

Update: Anonymous function doesn't know about $imageName variable. You have to use it:

public function deleteImage($imageName = null) 
{ 
    $myarray = array_filter(
        getCarImages(), 
        function ($v) use ($imageName) { return $v['image'] != $imageName; }
    );
}

edited Dec 13, 2017 at 9:09

answered Dec 13, 2017 at 7:08

u_mulder

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3 Comments

Komal khan Over a year ago

Its working fine but i want set the dynamic value like this

Komal khan Over a year ago

public function deleteImage($imageName = null){ $myarray = array_filter(getCarImages(), function ($v) { return $v['image'] != $imageName; });

Komal khan Over a year ago

but its giving me this error Message: Undefined variable: imageName

LF-DevJourney · Accepted Answer · 2017-12-13 07:35:14Z

1

You can trans the images as a reference to the function.

public function deleteImage(&$images){
    foreach ($images as $k => $array){
        //var_dump($array).'<br>';
        $index = array_search('w85YrKChBGTZ9fQS.jpg',$array);
        if($index !== FALSE){
            var_dump($index).'<br>';
            unset($images[$k]);
        }else{
            echo '<br>else here';
        }
    }
}

answered Dec 13, 2017 at 7:35

LF-DevJourney

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2 Comments

mickmackusa Over a year ago

@KrisRoofe Are we allowed to overtly ask for upvotes. I should be doing that on all of my best answers. stackoverflow.com/a/47725480/2943403

LF-DevJourney Over a year ago

@mickmackusa Yeah, you're right. I take your advise.

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