使用JQuery从PHP获取JSON响应

Question

I have accumulated and chopped up about 5 or 6 different tutorials of this now, and I still can't find out what's wrong!

Using JQuery Mobile (phonegap) to send and receive data to a PHP server. I cannot seem to get the JQuery script to pick up a response. PHP on server:

<?php

// Set up associative array
$data = array('success'=> true,'message'=>'Success message: worked!');

// JSON encode and send back to the server
echo json_encode($data);

?>

JQuery Function (Which is being called):

<script type="text/javascript">
$(function () {
    $('#inp').keyup(function () {
        var postData = $('#inp').val();
        $.ajax({
            type: "POST",
            dataType: "json",
            data: postData,
            beforeSend: function (x) {
                if (x && x.overrideMimeType) {
                    x.overrideMimeType("application/json;charset=UTF-8");
                }
            },
            url: 'removedmyurlfromhere',
            success: function (data) {
                // 'data' is a JSON object which we can access directly.
                // Evaluate the data.success member and do something appropriate...
                if (data.success == true) {
                    alert('result');
                    $('.results').html(data.message);
                } else {
                    $('.results').html(data.message);
                }
            }
        });
    });
});
</script>

Sorry for the formatting, it all went pair shaped copying it across. Removed my url.

I know that the function is firing, because if I remove the alert('here'); and put it above the ajax call it displays.

Anybody know why the success function isn't calling? nothing shows in the div with class results on any browser.

Thanks!

Answer 1

In PHP, add the JSON content-type header:

header('Content-Type: application/json');

also, try listening for complete as well to see if you do get any response back:

$.ajax({

 // ....
 complete: function (xhr, status) {
   $('.results').append('Complete fired with status: ' + status + '<br />');
   $('.results').append('XHR: ' + (xhr ? xhr.responseText : '(undefined)'));
 }
 // ...

});

Answer 2

Keep same domain origins in mind here, use jsonp with a callback has always been helpful for me, that does mean adding a $_GET['callback'] to your php script and wrap the json_encode with '('.json_encode($Array).')' for proper formatting.

http://api.jquery.com/jQuery.getJSON/

The last demo on that page is the best example I have found.

Answer 3

Hey i had the same problem

Firstly i used $.getJSON blah blah.... but didn't worked for some reason...

But the normal ajax call worked.. MY Page returns a json output as you.. try removing the "datatype"

I Used code copied from JQUERY site which is below i pasted

$.ajax({
  type: "POST",
  url: "some.php",
  data: { name: "John", location: "Boston" }
})
  .done(function( msg ) {
    alert( "Data Saved: " + msg );
  });

Below is the code how i used and worked with json output i got...

$.ajax({
          type: "GET",
          url: "json.php",
          data: { searchword: q, projid:prid }
        })
          .done(function( jsonResult ) {
            var str='';
        for(var i=0; i<jsonResult.length;i++)
          {
              //do something here.. with jsonResult."yournameof "
            alert(jsonResult[i].id+jsonResult[i].boqitem);
          }
});

Answer 4

Looks like cross domain request.

Try it by changing :

dataType : "json"

to

dataType : "jsonp"

Answer 5

I know it is very late. But it can simply be handled like

var postData = {'inp': $('#inp').val()};
$.ajax({
    type: "POST",
    data: postData,
    url: 'removedmyurlfromhere', // your url
    success: function (response) {
        // use exception handling for non json response
        try {
            response = $.parseJSON(response);
            console.log(response); // or whatever you want do with that
        } catch(e) {}
    },
    error: function( jqXHR, textStatus, errorThrown ) {},
    complete: function(  jqXHR, textStatus ) {}
});