jQuery / Ajax: How to loop through array as part of Ajax success function

Question

I have an Ajax call that returns an array and need to do something with each of the values from this array.

So far I have the below but this returns the following error:

Uncaught TypeError: Cannot use 'in' operator to search for 'length' in array(5)...

Can someone tell me what I am doing wrong here resp. how I can do something with each of the values ?

My Ajax:

$.ajax({        
    type: "post",   
    url: "ajax.php",
    cache: "false",
    data: {
        node: 'fetchValues',
        itemIDs: itemIDs
    },
    success: function(data){
        console.log(data);  // for testing only
        jQuery.each(data, function(index, value){
            console.log(value);
        });
    }
});

Example for "data" (from console log):

array(5) {
  [1]=>
  string(6) "Value1"
  [2]=>
  string(6) "Value2"
  [3]=>
  string(6) "Value3"
  [4]=>
  string(6) "Value4"
  [5]=>
  string(6) "Value5"
}

Many thanks in advance for any help.

Have you tried a simple for? var x=0; for(x;x<data.length;x++) { var value=data[x]; } — Ricardo Pontual
– Ricardo Pontual, Commented Jul 20, 2015 at 11:10
try this one for(var x in data){ console.log(x + ' :' + data[x]); — Monah
– Monah, Commented Jul 20, 2015 at 11:13
@RicardoPontual: Yes, I tried this before but it returns an error referring to the length. — TaneMahuta
– TaneMahuta, Commented Jul 20, 2015 at 11:32
@HadiHassan: Thanks for this. Wouldn't I need to use this with a loop ? — TaneMahuta
– TaneMahuta, Commented Jul 20, 2015 at 11:32
@TaneMahuta I didn't get what you mean, but the comment i provided above will loop to each property in the returned object via ajax — Monah
– Monah, Commented Jul 20, 2015 at 11:34

Anto S · Accepted Answer · 2015-07-20 13:51:02Z

5

seems like your array not parsed properly

from php side before sending response

echo json_encode($result); // REPLACE $result WITH  YOUR OUTPUT ARRAY

On jquery side:

$.ajax({        
    type: "post",   
    url: "ajax.php",
    dataType : 'JSON',
    cache: "false",
    data: {
        node: 'fetchValues',
        itemIDs: itemIDs
    },
    success: function(data){
        console.log(data);  // for testing only
       var data=$.parseJSON(data);
        jQuery.each(data, function(index, value){
            console.log(value);
        });
    }
});

Ref: http://api.jquery.com/jquery.parsejson/

edited Jul 20, 2015 at 13:51

answered Jul 20, 2015 at 11:22

Anto S

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13 Comments

TaneMahuta Over a year ago

Thanks for this. I had tried something similar before but this returns the following error: "Uncaught SyntaxError: Unexpected token a... "

Anto S Over a year ago

@TaneMahuta before sending response are use using json_encode() ? Whats your server side script? Is it PHP?

TaneMahuta Over a year ago

Not here. Should I try that ?

Anto S Over a year ago

Whats your server side script? Is it PHP?

TaneMahuta Over a year ago

Yes - currently I am just dumping the array there.

|

Community · Accepted Answer · 2017-05-23 10:24:19Z

1

You need to parse the string in your data variable to an object:

success: function(data){
    console.log(data);  // for testing only
    var res=$.parseJSON(data);
    jQuery.each(res, function(index, value){
        console.log(value);
    });
}

You can either use

var res=$.parseJSON(data)

Or plain JS:

var res=JSON.parse(data)

Source

edited May 23, 2017 at 10:24

CommunityBot

11 silver badge

answered Jul 20, 2015 at 11:18

Guruprasad J Rao

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5 Comments

TaneMahuta Over a year ago

Thanks for this. I had tried something similar before but this returns the following error: "Uncaught SyntaxError: Unexpected token a... " - same as for the answer above.

Guruprasad J Rao Over a year ago

Try giving $(data) in $.each instead of plaint data

TaneMahuta Over a year ago

Thanks - you mean to change this in your parseJSON line and to use $.each instead of jQuery.each ?

TaneMahuta Over a year ago

Thanks - there is no data in your jQuery.each. :)

Guruprasad J Rao Over a year ago

ok. just console.log(res) and console.log($(res)); using my method and let me know what you are getting?

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