1

Environment: PHP 5.3.5 MySQL Server 5.5.8

Created my first ever ajax call and after much determination I have got results. The results appear to be returning as a string. for when I request

alert(result[0]);

I am only returned a single character.

If I try to use a field name, I have a value of undefined returned.

Here is the javascript for my AJAX:

$.ajax({ 
    type: "POST", 
    url: "models/ajaxHandler.php", 
    data: "handler=this&stu=21", 
    success: function(result){ 
            if (result !== null) 
            { 
                    alert(result + " " + result[0] +" " result['firstname']);  
                    var obj =  JSON && JSON.parse(result) || $.parseJSON(result);
                    alert(obj + " " + obj[0] + " " + obj['firstname']);
            } 
            ShowDialog(false); 
            e.preventDefault(); 
    }, 
    error: function()
    { 
            alert("ajax failure"); 
    } 
});

So once the result is captured using a PDO connection in PHP to mysql with a fetchAll(). I return the results from the php like so:

echo json_encode($results);

The alerts for the results comeback like so:

[{"firstname":" Test","0":" Test","lastname":" One","1":" One","id":"2","2":"2","st_usage_id":null,"3":null},{"firstname":" Mr","0":" Mr","lastname":" Two","1":" Two","id":"3","2":"3","st_usage_id":null,"3":null},{"firstname":" Mr","0":" Mr","lastname":" Three","1":" Three","id":"5","2":"5","st_usage_id":null,"3":null}] [ undefined

The parse of the JSON returns all results as object.

I have also tried returning the results in php using:

print_r(json_encode($results));

It displays the same string above. Lastly I used 

$var_dump(results);

This returned the following:

array(3) {
  [0]=>
  array(8) {
    ["firstname"]=>
    string(7) " DrTest"
    [0]=>
    string(7) " DrTest"
    ["lastname"]=>
    string(4) " One"
    [1]=>
    string(4) " One"
    ["user_public_info_id"]=>
    string(1) "2"
    [2]=>
    string(1) "2"
    ["st_usage_id"]=>
    NULL
    [3]=>
    NULL
  }
  [1]=>
  array(8) {
    ["firstname"]=>
    string(3) " Dr"
    [0]=>
    string(3) " Dr"
    ["lastname"]=>
    string(4) " Two"
    [1]=>
    string(4) " Two"
    ["user_public_info_id"]=>
    string(1) "3"
    [2]=>
    string(1) "3"
    ["st_usage_id"]=>
    NULL
    [3]=>
    NULL
  }
  [2]=>
  array(8) {
    ["firstname"]=>
    string(3) " Dr"
    [0]=>
    string(3) " Dr"
    ["lastname"]=>
    string(6) " Three"
    [1]=>
    string(6) " Three"
    ["user_public_info_id"]=>
    string(1) "5"
    [2]=>
    string(1) "5"
    ["st_usage_id"]=>
    NULL
    [3]=>
    NULL
  }
}
 a undefined

I am not sure what it is I am missing, and I'm sure it is something simple. I feel like I have tried everything. Please could someone tell me where I went wrong?

4

4 回答 4

4

指定 json 数据类型。

dataType: "json"

请注意，结果永远不会出现null在最新版本的 jQuery 中，而是在没有返回任何内容的情况下会出错。

于 2013-09-11T19:52:04.167 回答
2

如果您想接受答案：

尝试：

alert(obj[0]['firstname']);
于 2013-09-12T16:52:06.940 回答
1

添加“结果 = $.parseJSON(result);” 到你成功调用的开始，它应该把它变成一个 Javascript 对象。

于 2013-09-11T19:53:02.577 回答
0

PHP 文档http://php.net/manual/en/function.json-decode.php指出，当 json_decode 的第二个参数为 true 时，“返回的对象将被转换为关联数组”。

这应该为你做：

$results = json_decode(json_encode($results), true);
print_r($results);
于 2013-09-11T19:56:08.523 回答