CODEWITHSUNDEEP
javascriptbookmarklet
Andrew
Andrew

Reputation: 3999

Javascript bookmarket unexpected identifier?

It would be hugely appreciated if you could just run through this code and see if you spot an "unexpected identifier". Thanks so much guys. 

javascript:(function () {
    url = document.location.href;
    if (url.match('youtube.com/watch?')) {
        var s=document.createElement('script');
        s.setAttribute('src','http://jquery.com/src/jquery-latest.js');
        document.getElementsByTagName('body')[0].appendChild(s);
        dataString = 'url=' url;
        $.ajax({
            type: 'POST',
            url: '/create/',
            data: dataString,
            success: function(data){ console.log(data); }
        });
    }
    else {
        alert('This is not a youtube video.')
    }
    })();

Upvotes: 0

Views: 563

Answers (4)

mVChr
mVChr

Reputation: 50205

Just to be more different...

dataString = 'url=' url;

should be:

var urleq = 'url=',
    dataString = [urleq, eval(urleq.substr(0,urleq.length-1))].join('');

Upvotes: 0

alex
alex

Reputation: 490479

Just to be different....

dataString = 'url='.concat(url);

Upvotes: 1

David Tang
David Tang

Reputation: 93694

This:

dataString = 'url=' url;

Should be:

dataString = 'url=' + url;

Mind you, you can avoid awkward string concats by giving $.ajax an object to work with - especially useful if the number of params increases:

$.ajax({
    data: {url: url}
});

Upvotes: 7

Mike Lewis
Mike Lewis

Reputation: 64177

You assignment of dataString is not concatenating properly:

dataString = 'url=' url;

should be:

dataString = 'url=' + url;

Upvotes: 2

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