php – 使用JSON.stringify时不需要的数组索引

我使用JSON.stringify方法将数组传递给服务器.

我有一个有4个元素的数组：

arr[10] = 1;
arr[20] = 1;
arr[30] = 1;
arr[40] = 1;

然后我这样做：

arr = JSON.stringify(arr);

然后将其发送到服务器：

                jQuery.ajax({
                    type: "post",
                    url: baseurl+"profile/mprofile/action/ratings/add_ratings",
                    data:{"checkBox":checkBox,"review":review,"speciality":speciality,"arr":arr},
                    success: function(data, status) {  

                        jQuery('#header-error').html(data);
                    } 
                }); 

我在PHP中获取数组：

    $arr = $this->ci->input->post('arr');
    $arr = json_decode($arr);
    print_r($arr) ; die ;

结果是

Array
(
    [0] => 
    [1] => 
    [2] => 
    [3] => 
    [4] => 
    [5] => 
    [6] => 
    [7] => 
    [8] => 
    [9] => 
    [10] => 1
    [11] => 
    [12] => 
    [13] => 
    [14] => 
    [15] => 
    [16] => 
    [17] => 
    [18] => 
    [19] => 
    [20] => 1
    [21] => 
    [22] => 
    [23] => 
    [24] => 
    [25] => 
    [26] => 
    [27] => 
    [28] => 
    [29] => 
    [30] => 1
    [31] => 
    [32] => 
    [33] => 
    [34] => 
    [35] => 
    [36] => 
    [37] => 
    [38] => 
    [39] => 
    [40] => 1
)

为什么会这样？

解决方法:

这就是javascript数组的工作方式.将值设置为超出范围的索引时,将展开数组,成功设置值并将所有“缺失”值设置为undefined.

您被PHP行为所破坏,这种行为混合了数组和字典(或“关联数组”,PHP谈话)之间的差异.

要避免此行为,只需在javascript而不是数组中创建字典.

var arr = {};
arr[10] = 1;
arr[20] = 2;

此外,在PHP端,您应该将true作为第二个参数传递给json_decode.

json_decode($arr, true)

这将使它返回数组而不是stdclass.