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Krystal
Krystal

Reputation: 93

Number error in parsing json

I have a php class( ajax called) which returns json_encode data as

["2016-02-08 09:00:00.000","2016-02-15 09:00:00.000"]

I'm trying to do jquery.parseJSON(data) and it's giving me error "Unexpected number", what am I doing wrong?

Upvotes: 0

Views: 107

Answers (3)

Ruan Mendes
Ruan Mendes

Reputation: 92294

You are trying to parse an array, not a string. JSON.parse (and any other JSON parser) expects a string, therefore Array.toString is called and your call becomes jquery.parseJSON("2016-02-08 09:00:00.000,2016-02-15 09:00:00.000")

//Error "Unexpected number", toString is called on the input array
JSON.parse(["2016-02-08 09:00:00.000","2016-02-15 09:00:00.000"]) 
// Returns an array object
JSON.parse('["2016-02-08 09:00:00.000","2016-02-15 09:00:00.000"]') // OK

If you are using the return of json_encode inline, you don't need to parse it, just assign it to a variable, JavaScript will do the parsing.

var dates = <?= json_encode($dates) ?>;

If you are using jQuery the data will typically already be parsed into JSON in the callback, if it doesn't, you can force it using dataType: 'json'

Upvotes: 2

trincot
trincot

Reputation: 350750

jQuery does the decoding for you when you perform the AJAX call with the JSON dataType:

$.ajax({
  url: 'mypage.php',
  data: mydata,
  dataType: 'json'
})
.done(function(response) {
   // Response is an array, not a JSON string, jQuery decoded it.
   // Demo:
   console.log(response[0]); // "2016-02-08 09:00:00.000"
   console.log(response[1]); // "2016-02-15 09:00:00.000"
}

This is explained in the jQuery docs:

dataType

...
"json": Evaluates the response as JSON and returns a JavaScript object.

So, don't use jquery.parseJSON on the result. It has already been done for you.

Upvotes: 0

Nechemya Kanelsky
Nechemya Kanelsky

Reputation: 673

var x = '["2016-02-08 09:00:00.000","2016-02-15 09:00:00.000"]';
$.parseJSON(x) // return an array

Upvotes: 0

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