日期:2014-05-17  浏览次数:20669 次

如何把JSON的多维数组转换成JS的多维数组
PHP code

<?php
        $arr1=array("a","b","c");
        $arr2=array(
                array("a1","a2","a3"),
                array("b1","b2","b3"),
                array("c1","c2","c3")
        );
        $arr3=array(
                 array(
                    array("a11","a12","a13"),
                    array("a21","a22","a23"),
                    array("a31","a32","a33")
                 ),
                 array(
                    array("b11","b12","b13"),
                    array("b21","b22","b23"),
                    array("b31","b32","b33")
                 ),
                 array(
                    array("c11","c12","c13"),
                    array("c21","c22","c23"),
                    array("c31","c32","c33")
                 )
        );
    $parm=$_GET["parm"];
    switch ($parm){
        case "arr1":
            echo json_encode($arr1);
            break;
        case "arr2":
            echo json_encode($arr2);
            break;
        case "arr3":
            echo json_encode($arr3);
            break;        
    }

    
?>



JScript code

$.getJSON("../php/json.php?parm=arr2", function(json){
//中间应该怎么写
});



最终JS的如果是
JScript code

var arr1=["a","b","c"];
        var arr2=[
                ["a1","a2","a3"],
                ["b1","b2","b3"],
                ["c1","c2","c3"]
        ];
        var arr3=[
                 [
                    ["a11","a12","a13"],
                    ["a21","a22","a23"],
                    ["a31","a32","a33"]
                 ],
                 [
                    ["b11","b12","b13"],
                    ["b21","b22","b23"],
                    ["b31","b32","b33"]
                 ],
                 [
                    ["c11","c12","c13"],
                    ["c21","c22","c23"],
                    ["c31","c32","c33"]
                 ]
        ];



------解决方案--------------------
对js 不是那么熟悉,所以我想知道的是竟然都传递json 给js 了,还要还原成数组做什么呢?
------解决方案--------------------
jquery 的 getJSON 方法已经将接受到得数据转换成了 json 对象数组
就是说 function(json) 中的 json 是json对象数组
你已经可以直接使用了 json.name
在通常情况下,用完就结束了。

看你的描述,似乎是要从其中构造出全局变量
你可以用 for(k in json) 遍历出其中的每个成员,使用 eval 函数将其导入全局变量中。
当然变量名你得自己构造,因为你的 php 代码没有传出