jquery从url获取id并发布到具有动态id的页面

Question

I need to make a like/dislike section using jquery and html so I have this code:

    $('#like').mousedown(function() {
       var name = $('#like').val();
       $.post('php/vote_movie.php', { like: name }, function(data) {
          $('#like_feedback').html(data);
       });
    });    

all works well if I manually type the url like vote_movie.php/movie_id=22 for example. but I need the url to update itself because in php I use $_GET['movie_id'] url for the vote page so jquery needs to take that id from url and match-it with $_GET['movie_id'] from vote_movie.php

basically I need something like this:

$('#like').mousedown(function() {
   var name = $('#like').val();
   $.post('php/vote_movie.php?movie_id=<$_GET['movie_id']>', { like: name }, function(data) {
      $('#like_feedback').html(data);
   });
});    

and I have no idea how to do it because I'm a newbie in both php and java.

Any help on how to achieve this will be greatly appreciated.

Thank you.

Update: to a better understanding here are my files: likes.js http://pastebin.com/FnVFrDZX watch_movie.php http://pastebin.com/QVScXZCd vote_movie.php http://pastebin.com/g70xbvsf

Answer 1

Insert "data-id" to like button.

Example:

   <span class="likebutton" data-id="1234567" id="like"></span>

 $('#like').mousedown(function() {
       var name = $('#like').val();
       var dataid =  $('#like').data('id');

       $.post('php/vote_movie.php?id='+dataid+'', { like: name }, function(data) {
          $('#like_feedback').html(data);
       });
    });