使用AJAX显示SQL查询结果

Question

I have a news feed on my site, onto which users can post posts. These posts can then be liked by other users (just like facebook for example).

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The Problem
I would like to display the users, who liked a post using ajax. Whenever a certain element is hovered.
At the moment the users are correctly displayed but below every post, not just the one which contains the hovered element.

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My attempt to solve the problem

 <!-- HOVER THIS --> <span class="likers small link"></span> <!-- DISPLAY LIKERS HERE --> <small class="displayLikers"></small> <!-- AJAX --> <script> $(function() { $(".likers").hover(function(){ $.ajax({ url: "get/likers", type: "GET", success: function(response) { $(this).closest('.displayLikers').html(response); } }); }); }); </script> 

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I would be very thankful for any kind of help!

Answer 1

Inside the $.ajax , $(this) does not refer to $(".likers") just add $(this) to a variable and use it in the ajax response function;

$(function() {
    $(".likers").hover(function(){
        var likes = $(this);
        $.ajax({
            url: "get/likers",
            type: "GET",
            success: function(response) {
                likes.closest('.displayLikers').html(response);
            }
       });
    });
});

Answer 2

In your example the .displayLikers is a sibling so you should probably use next() . Moreover this will refer to the actual context of the success function, so you have to create a reference before.

$(function() {
    $(".likers").hover(function(){
        var self = $(this);
        $.ajax({
            url: "get/likers",
            type: "GET",
            success: function(response) {
                self.next('.displayLikers').html(response);
            }
       });
    });
});