Trying to put multiple data in php/json array

Question

This is probably quite simple to do.. but I just can't think of how to do this.

I have a photo upload script, I want to return two types of data in an array, right now I just have one set of data for the photo returned.

right now I do this:

 $photos = array();

 ///Script for photo upload etc.
 array_push($photos, $newPhotoToAdd)

 //then when it's finished uploading each photo i do json_encode:
 print json_encode($photos);

all of this works perfect, however now I want to return another set of data with each photo.

I need to do something like:

 array_push($photos, ['photoID'] = $photoID, ['photoSource'] = $newPhotoToAdd)

Treffynnon · Accepted Answer · 2012-01-24 11:44:30Z

5

Original Question

You can just push an array on instead of newPhotoToAdd like the following:

$photos = array();
// perform upload
$photos[] = array(
    'photoID' => $photoID,
    'photoSource' => $newPhotoToAdd
);
print json_encode($photos);

You'll notice that I have swapped array_push() ^DOCs for the array append syntax of [] ^DOCs as it is simpler and with less overhead:

Note: If you use array_push() to add one element to the array it's better to use $array[] = because in that way there is no overhead of calling a function.

From comments: printing it out in JavaScript

You are dealing with an array of objects so you need to loop over them:

for(var i in json) {
    alert(json[i].photoID);
}

edited Jan 24, 2012 at 11:44

answered Jan 24, 2012 at 11:38

Treffynnon

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1 Comment

Dylan Cross Over a year ago

Thanks, now I'm having trouble trying to use this in javascript though. It's not working as I thought it would. var json = JSON.parse(content); alert(json.photoID); this returns undefined

Nobita · Accepted Answer · 2012-01-24 11:40:54Z

0

You want to create an array like this:

$photo = array('photoID' => $photoID, 'photoSource' => $newPhotoToAdd);

Then you want to do the push as you did before with this new array:

array_push($photos, $photo);

answered Jan 24, 2012 at 11:40

Nobita

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Comments

faq · Accepted Answer · 2012-01-24 11:41:18Z

0

do something like:

$photos = array();
array_push($photos, array('photoID'=>$photoID,'photoSource'=>$newPhotoToAdd));

foreach ($photos as $photo) {
print json_encode($photo);
}

answered Jan 24, 2012 at 11:41

faq

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