PHP in JavaScript and escaping

Question

I am using Laravel and i'm making an AJAX request. In the success event I wrote:

var name = response.filename;
var path = "<?php echo route('getImage', "name")?>";                        
document.getElementsByClassName("getImage")[0].setAttribute("src", path);

But it looks like on the second row the name variable is interpretated as text, not as variable. And the src's value is

http://localhost/bluedrive/drive/public/getimage/name;

How could I get this right?

You cannot call any PHP code from your JS. You must run the route function on the server side, return its result in the success response and set it as the image src. — Anatoliy Arkhipov
– Anatoliy Arkhipov, Commented Jan 3, 2016 at 13:37
Not strictly true @AnatoliyArkhipov if OP is putting their javascript in their blade templates. — Wader
– Wader, Commented Jan 3, 2016 at 13:54
Oh, yes, looks like I'm incorrectly understand the question. I thought that the given code is a part of some success callback in JS — Anatoliy Arkhipov
– Anatoliy Arkhipov, Commented Jan 3, 2016 at 17:13

Wader · Accepted Answer · 2016-01-03 13:52:25Z

1

You could do the following, assuming the javascript is in your blade templates.

var name = response.filename;
var path = "{{ route('getImage', ['image_name']) }}";
document.getElementsByClassName("getImage")[0].setAttribute("src", path.replace('image_name', name));

I think your best option however is to return the full image src back in your ajax response. So your controller may look something like this

Input::file('image')->move('/my/filepath', $filename);

return response()->json([
    'image_url' => url('/my/filepath/' . $filename),
]);

This returns the url for the image which you can then use in your javascript as below

.success(function(response) {
    document.getElementsByClassName("getImage")[0].setAttribute("src", response.image_url);
});

answered Jan 3, 2016 at 13:52

Wader

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