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Below ajax code is receiving a json array from php. Rest details i have written as comments: 

 $.ajax({
     type: "POST",
     url: "dbtryout2_2.php",
     data: datastr,
     cache: false,
     //dataType: 'json',
     success: function (arrayphp) {
         //"arrayphp" is receiving the json array from php.
         //below code is working where iam displaying the array directly
         //This code displays the array in raw format.       
         $(".searchby .searchlist").append(arrayphp);
     }

 });

FRIENDS CONCENTRATE ON THIS SECTION.NOW I WILL MAKE U THE PROBLEM MORE CLEAR AND EXACT: 1)success function is having two code 2)one is uncommented and other is commnented 3)the commneted code works if i comment code "dataType: "json"", 4)but the uncommneted code does not work with the situation which the below code currently has

 $.ajax({
     type: "POST",
     url: "dbtryout2_2.php",
     data: datastr,
     dataType: "json",
     cache: false,
     success: function (arrayphp) {
         $.each(arrayphp, function (i, v) {
             alert(i + "--" + v);
         });
         /*$(".searchby .searchlist").append(arrayphp);*/
     },
     error: function (xhr) {
         console.log(xhr.responseText);
     }

 });

BELOW IS THE PHP CODE SNIPPET RESPONSIBLE FOR RETURNING JSON ARRAY:

 $arrayphp = array();
 //$result is containing the list of albums
 //iam one by one taking the album names and assigning it to $row
 //then from $element iam pushing the elements in $arrayphp
 //after pushing all the elements iam returning the json encoded array to ajax code.  
 while ($row = mysql_fetch_array($result)) {
     $element = $row['cat_name'];
     array_push($arrayphp, $element);
 }
 echo json_encode($arrayphp);
 }

THE ARRAY RETURNED IS A LIST OF ALBUM NAMES:

["album1", "album2", "album5", "album4", "album6", "album7", "album8", "album9", "album10", "album11"]

THE EXACT ABOVE ARRAY IS GETTING RETURNED.

WOULD ANYBODY FIGURE OUT WAT THE PROBLEM IS WITH MY CODE?

share|improve this question
 
check firebug or some other browser debugger - sounds like the problem is the way your trying to loop the json .. can you add the returned data to your question ? –  ManseUK Aug 31 '13 at 19:17
1  
could you give a little detail on the structure of the return? –  james emanon Aug 31 '13 at 19:17
 
pleas, post the json it retrieves –  dscdsc Aug 31 '13 at 19:18
1  
Use jQuery.parseJSON() api.jquery.com/jQuery.parseJSON –  Vicky Thakor Aug 31 '13 at 19:21
 
I don't think you need to parseJSON from an $.ajax call. You should try console.log(arrayphp); in your success method, and see what comes out in the console under developer tools. You should also add this after success to see if you are getting an error: error: function(xhr){ console.log(xhr.responseText); } –  Jake Aug 31 '13 at 19:34
show 5 more comments

2 Answers

up vote 0 down vote

try this in your ajax success:

var i = 0;
$.each(arrayphp, function () {
      alert(arrayphp[i]);
      i++
 });
share|improve this answer
add comment
up vote 0 down vote

I tried your function and it worked fine for me when I left out this line: contentType: "application/json", and when I added it in, I started getting a failed response of "0".

In the manual, it says to use this content type for most cases: 'application/x-www-form-urlencoded; charset=UTF-8' (which is the default). So unless you have a good reason, I think you should remove that line.

In the first function, it doesn't exist and you turned off 'json' dataType, so you are probably alerting a really nice looking string that looks like the data you want. If you make sure that you have dataType: 'json' and remove contentType: application/json', you should be good to go.

EDIT: This is exactly what I used, and it worked great. Not sure what else to suggest. What is datastr in your js?

$.ajax({
    url:      'dbtryout2_2.php',
    type:     'post',
    async:    true,
    cache:    false,
    dataType: 'json',

    success: function( response )
    {
        // response is returned as json object: ["album1","album2","album3","album4","album5"]
        //console.log(response);

        if ( response )
        {
            $.each(response, function (i, v) {
                alert(i + "--" + v);
            });
        }
    },

    error: function( xhr )
    {
        console.log(xhr.responseText);
    }
});

Note: If you have specified "dataType:'json'", you don't have to $.parseJSON, because the response is already returned as JSON.

share|improve this answer
 
@ jake still its not working.see the code above . i have clarified my problem again. –  Nitish kumar thakur Sep 1 '13 at 8:30
 
well finally the above code is working.U know actually the problem was in the data that i was returning from php. I was echoing two things from php one was a "message" that i had written initially for checking whether php code is running or not and the other was the "array" that actually i wanted to return.But i was not at all concentrating on the string that was getting returned with the array. –  Nitish kumar thakur Sep 3 '13 at 16:21
1  
well thanks a lot to all of you for your suggestions –  Nitish kumar thakur Sep 3 '13 at 16:24
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