我运行json_encode
PHP函数并获得以下输出:
[{"id":"1","size":"124","name":"Team1","picture":1},
{"id":"5","size":"76","name":"Team 4","picture":2},
{"id":"3","size":"25","name":"Team2","picture":3},
{"id":"4","size":"17","name":"Team3","picture":4}]
现在我想使用JQUERY解析它,并将它添加到我的网页中。我运行以下脚本:
<script>
$(function() {
$('#myButton2').click(function(e) {
$.get("http://localhost:99/result/getBestOne", function(data) {
alert(data+"");
var obj = JSON.parse(data);
alert(obj+"");
});
});
});
</script>
我的问题是第一个警报正在工作,但在第二个警报中,我得到错误:**Ucought SyntaxError: Unexpected Token**
问题出在哪里?
问题是它不是有效的JSON。。缺少最后2个对象的起始大括号。。
{"id":"1","size":"124","name":"Team1","picture":1},
{"id":"5","size":"76","name":"Team 4","picture":2},
{"id":"3","size":"25","name":"Team2","picture":3},
{"id":"4","size":"17","name":"Team3","picture":4}
哪里是大括号"{对于这两条线是""
"id":"3〃"尺寸":"25〃"name":"Team2"图片":3} ,
"id":"4〃"尺寸":"17〃"name":"Team3"图片":4}
您试图解析的文本不是有效的JSON,因此会出现语法错误。
如果你不确定哪里出了问题,我建议你将JSON代码复制并粘贴到这个网站:
http://jsonlint.com/