CODEWITHSUNDEEP
javascriptjquerytwitter
Sharon
Sharon

Reputation: 3909

How can I create a tweet from JavaScript/jQuery?

I'm trying to allow users to tweet a link from within my jQuery app. Basically, they create their own page, and then they select a number of social networks to share it on. On clicking 'Share', the app loops over the social network checkboxes, and posts to each, as selected.

I'm struggling to find a good way to do this for Twitter - everywhere I look online, I seem to get conflicting advice.

So, basically, what should happen is this:

It doesn't sound very complicated, but I can't find a way of doing it!

Edit

Thanks to the responses below, I've now got the following code:

var serverUrl = 'https://twitter.com/intent/tweet?url="www.rocketreel.com"&text="Test"';

$.ajax({
  type: "GET",
  url: serverUrl,
  success: function(data) {
},

error: function(data) {
  var obj = jQuery.parseJSON(data);
  alert(obj.error);
  }
});

Now when I run it, I get (in the console):

XMLHttpRequest cannot load https://twitter.com/intent/tweet?url=%22www.mysite.com%22&text=%22Test%22. Origin null is not allowed by Access-Control-Allow-Origin.

I'm not able to run it from the actual site at the moment, so am using localhost - is that why I'm getting this?

Upvotes: 0

Views: 2965

Answers (1)

VDP
VDP

Reputation: 6420

Have you tried reading the api documentation?

https://dev.twitter.com/docs/tweet-button

https://twitter.com/about/resources/buttons#tweet

You can check the code of the tweet button and execute the code behind the tweet button. Pretty straight forward...

Here is the url that's being called: 

https://twitter.com/intent/tweet?original_referer=yoururl.com&source=tweetbutton&url=http://urltotweet.com&text=[text to tweet]

UPDATE:

You are running into a cross origin issue. You can't load data from an other origin through ajax unless you use some workarounds (CORS, proxy,... since you can't modify the twitter api, a proxy is your only option. But that too can come with it's complications..) you probably don't even need.

Just open up a new window with the location... window.open(...) or create an iframe with the provided url in staid of loading a complete page through ajax.

Upvotes: 1

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