JSon解析我的代码返回错误

Question

<script type="text/javascript">
$(document).ready(function(){
var obj = "";
var param = [{ "name": "id", value: 9}];   
$.ajax({
  type: 'POST',
  url: '2.php',
  data: param,
  success: function(dataFetch) {
var obj = JSON.parse(dataFetch);
alert(obj[0].name);
 },
  dataType: 'json',
  async:false
});
});
</script>

above is my code. The return result of 2.php is

{"name":"Halford Tee","type":"client","mobile":"1234567","location":"","description":""}

It give me an error

Uncaught SyntaxError: Unexpected end of input

When I remove this part

var obj = JSON.parse(dataFetch);
alert(obj[0].name);

The error is gone. Is there anything wrong with my code?

---

After looking at valuable guide from fellow programer

My 2.php returns

{"name":"David Tan","type":"leads","mobile":"91234567","location":"","description":""}

I tried the following

$(document).ready(function(){
var obj = "";
var param = [{ "name": "eventId", value: 9}];   
$.ajax({
  type: 'POST',
  url: '2.php',
  data: param,
  success: function(dataFetch) {
alert(dataFetch[0].name);
 },
  dataType: 'json',
  async:false
});
});

It give me an error

cannot read property 'name' of undefined.

Answer 1

when you use dataType: 'json' in your ajax then you don't need to use JSON.parse. your data will automatically be parsed in json format.so you can directly use it.

 use : `alert(dataFetch.name);`// you do not have any array so directly use it.your data is an object not array of object.

note: i think you should use async:false only when your calling has to complete before the next statement in your function can be called.