JSon parse return error on my code

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.

`dataType: 'json'` will make jQuery parse the response automatically. What happens if you only use `alert(dataFetch[0].name);`? — Felix Kling, Jan 08 '15 at 04:48
The json string returned by 2.php doesn't have an array. Try `alert(dataFetch.name);` — zahirdhada, Jan 08 '15 at 05:22
Object {name: "David Tan", type: "leads", mobile: "91234567", location: "", description: ""} — BUN, Jan 08 '15 at 05:23
Manage to fetch using dataFetch.name instead of dataFetch[0].name Thanks! But I got another question, what if my return json is more than 2 rows, because for this case I only got 1 row. — BUN, Jan 08 '15 at 05:24
in that case loop through your data using $.each .read about it. — Suchit kumar, Jan 08 '15 at 05:26
see this post:http://stackoverflow.com/questions/3030321/jquery-each-loop-with-json-array — Suchit kumar, Jan 08 '15 at 05:28

Suchit kumar · Accepted Answer · 2015-01-08T05:23:40.517

0

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.

edited Jan 08 '15 at 05:23

answered Jan 08 '15 at 05:11

Suchit kumar

11,809
3
22
44

I updated my question and it seems got issue still. yes it has to be complete before the next statement can be called. – BUN Jan 08 '15 at 05:18

score 0 · Answer 2 · answered Jan 08 '15 at 05:18

0

After: var obj = JSON.parse(dataFetch);

alert(JSON.stringify(obj));

it might help you to rectify your problem.

answered Jan 08 '15 at 05:18

Gaurav Dave

6,838
9
25
39

After doing alert you will find keys and values related to that json you are working with. – Gaurav Dave Jan 08 '15 at 05:20

score 0 · Answer 3 · answered Jan 08 '15 at 06:37

(This answer is based on your comment that there might be more than one row.)

I think this is one of those always-1 vs 0-or-more vs 1-or-more problems.

You have to decide, exactly what does 2.php return?

Always a single row? {"name":"David Tan",...}
Always an array of zero-or-more rows? [{"name":"David Tan",...}, {"name":"Clark Kent",...},...]
Or simply one-or-more rows, but not bracketed as an array? {"name":"David Tan",...}{"name":"Clark Kent",...}

In case (1), dataFetch.name is "David Tan".

In case (2), dataFetch[0].name is "David Tan".

In case (3), your output is not valid JSON. If you have the ability to change the PHP script, then change the output to Case (2) format <<GOOD SOLUTION. If you must deal with someone else's PHP script, then you need to change your ajax call to dataType: "text", then clean it up to make it valid JSON, then parse the JSON client-side. You do not want to omit dataType and have jQuery default to "Intelligent Guess", because results will vary depending on how many rows there are in the output.

JSon parse return error on my code

3 Answers3