2012-06-15 58 views
0

如何通过jQuery迭代JSON返回并返回每个对象的索引?一些下面我的代码如何从multidimention迭代json数组

例子:

foreach (x = somevalue ; x < length of array ; x +++ 
{ 
      that x must be the index 
     [0] => Array 

      $('#actual_'+x).text(data.actual+" hrs."); 
      $('#total_'+x).text(data.total+" hrs."); 
      $('#regular_'+x).text(data.regular+" hrs."); 



} 

    (
[0] => Array 
    (
     [actual] => 9 
     [total] => 10 
     [regular] => 0 
     [over] => 11 
     [total_h] => 11 
     [eng_pay] => 148.5 
     [rate] => 9 
    ) 

[1] => Array 
    (
     [actual] => -1 
     [total] => 0 
     [regular] => -1 
     [over] => 2 
     [total_h] => 1 
     [eng_pay] => 18 
     [rate] => 9 
    ) 

我想通过这个jQuery AJAX控件的成功功能遍历:

$.ajax({ 
    type:'POST', 
    url: 'cal_grid.php', 
    dataType: 'json', 
    cache: false, 

     data:$('#grid_frm').serialize(), 
     success: function(data) 


     { 


     alert(data); 



     } // response call back ends 

    });//ajax call ends 
+1

'var x;对于(x = 0; x DaveRandom

+0

对不起... –

回答

1

未经测试,但应该做的招。把它投入你的成功功能。

$.each($(data), function(i, obj){ 
    console.log(i); //spits out your index into console 
});