Javascript $.parseJSON意外字符

我试图从

span

元素上的html

data

属性发送数据，并使用Ajax接收数据，然后使用php和mysql处理数据，并将新值返回到html中的数据属性，但我收到一个错误，错误是“$.parseJSON意外字符”，有没有人能看一下我的代码，看看我是否正确地处理了数据，因为我刚刚开始使用JSON

HTML/PHP

<span data-object=
'{"art_id":"<?php echo $row['art_id'];?>",
"art_featured":"<?php echo $row['art_featured'];?>"}' 
class="icon-small star-color"></span>
<!-- art_id and art_featured are both int and art_featured will be either 1 or 0 -->

PHP/mySQL

if($_POST['art_featured']==1) {
        $sql_articles = "UPDATE `app_articles` SET `art_featured` = 0 WHERE `art_id` =".$_POST['art_id'];

        $result = array('art_id' => $_POST['art_id'], 'art_featured' => 0);
        echo json_encode($result);
    }
    else if($_POST['art_featured']==0){
        $sql_articles = "UPDATE `app_articles` SET `art_featured` = 1 WHERE `art_id` =".$_POST['art_id'];

        $result = array('art_id' => $_POST['art_id'], 'art_featured' => 1);
        echo json_encode($result);
    }

    if(query($sql_articles)) {

    }
    else {

    }

---

您不需要使用

$.parseJSON

，jQuery会为您这样做

$("span[class*='star']").on('click', function () {
    var data = $(this).data('object');
    var $this = $(this);

    $.ajax({
        type: "POST",
        url : "ajax-feature.php",
        data: {art_id: data.art_id,art_featured: data.art_featured}
    }).done(function(result) {
        data.art_featured = result;
        $this.data('object', data);
    });

});

---

您以后也不需要对其进行字符串化。

$sql\u articles=“UPDATE app\u articles SET art\u featured=0，其中art\u id=“.$\u POST['art\u id'”您需要修复此查询（以及第二个查询），它们都对SQL注入非常开放。谢谢，是的，我还没有完成SQL页面，我只是在保护它之前让它工作，但是谢谢！

$("span[class*='star']").on('click', function () {
    var data = $(this).data('object');
    var $this = $(this);

    $.ajax({
        type: "POST",
        url : "ajax-feature.php",
        data: {art_id: data.art_id,art_featured: data.art_featured}
    }).done(function(result) {
        data.art_featured = result;
        $this.data('object', data);
    });

});