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How do I do a simple Ajax request?

in Using jQuery • 10 years ago

I'm scattered in my knowledge of web development. So this is quite frustrating.
I've used Ajax once or twice before in normal javascript. I didn't fully understand what was going on.

but I digress.

Shouldn't this code make newContent = a string that's printed by getNewPosts.php
It should also give me a thing that says "done".

$(document).ready(function() {
var newContent = $.ajax({
url: "getNewPosts.php"
success: function(){
alert("done");
}
}
).responseText;
});

Replies(2)

andrewgsw

Re: How do I do a simple Ajax request?

10 years ago

$(document).ready(function() {
var newContent = $.ajax({
          url: "getNewPosts.php" , // missing this comma
          success: function(){
             alert("done");
          }

      }
      ).responseText;
});

keysle

Re: Re: How do I do a simple Ajax request?

10 years ago

Even with it. I still don't get an alert pop messages saying "DONE"
In order for alert to work, do I need something else?

----edit----
i just tried alert("TEST!"); on the document load... and nothing came up.
There must be something wrong.

-----edit----

var newContent = $.ajax({
url: 'getNewPosts.php',
success: function(){
$('.posts').prepend(newContent);
}
}
).responseText;

even with this I still get no responseText
http://k8r.us/test is the testing page. all the jquery is there.
The success function.
http://k8r.us/test/getNewPosts.php exist :/
the jQuery include is there.

This is all very troubling. I can't figure it out

----------------edit--------------------------

I figured it out. The jQuery had an error because some of the targeted objects did not exist.
I thought Javascript ignored this. jQuery may not.

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