我正在使用json作为按钮。当用户点击它时，正确地增加了类似的计数，并在数据库中存储了相应的id，但是它显示了浏览器中所有id的增量是错误的。我想显示只有用户喜欢的ID，但它显示所有。json递增一个ID但显示所有ID

indexp.php文件

include('/home/xxx/con.php');

$query="Select * from users_jokes order by id desc limit 10";

$result=mysql_query($query);

?>

var ajaxSubmit = function(formEl) {

var url = $(formEl).attr('action');

var comment=document.getElementById("jokes_comment").value;

var joke_id=document.getElementById("joke_id_hidden").value;

$.ajax({

url: url,

data:{

'action':'addComment',

'comment':comment,

'joke_id':joke_id

},

dataType: 'json',

type:'POST',

success: function(result) {

console.log(result);

$.ajax({

url: url,

data:{

'action':'getLastComment',

'joke_id':joke_id

},

dataType: 'json',

type:'POST',

success: function(result) {

$('#jokes_comment').val("");

console.log(result[0].description);

$("#header ul").append('

'+result[0].description+'');

},

error: function(){

alert('failure');

}

});

},

error: function(){

alert('failure');

}

});

// return false so the form does not actually

// submit to the page

return false;

}

var ajaxLike=function()

{

var joke_id=document.getElementById("joke_id_hidden").value;

// setup the ajax request

$.ajax(

{

url: 'likeskk.php',

data:{

'action':'increment_like',

'joke_id':joke_id

},

dataType: 'json',

type:'POST',

success: function(result)

{

$.ajax(

{

url: 'likeskk.php',

data:{

'action':'display_like',

'joke_id':joke_id

},

dataType: 'json',

type:'POST',

success: function(result)

{

console.log(result);

$("label[for='like_counter']").text(result.likes);

},

error: function(result)

{

alert("error 2");

},

});

},

error: function()

{

alert('failure');

}

});

return false;

}

commnet list

?>

while($data = mysql_fetch_array($result)){

$id = $data['id'];

$cont = $data['content'];

$likes = $data['likes'];

?>

Like

?>