Question

您好我有一个使用JSON.Stringify

输出的对象

{"0":["test1","ttttt","","","","","","","",""],"1":["test2","ghjgjhgjh","","","","","","","",""]}

我希望得到这样的输出。

[["test1","ttttt","","","","","","","",""],["test2","ghjgjhgjh","","","","","","","",""]]

我试过这个删除＆＃34; 0＆＃34;和＆＃34; 1＆＃34;使用.map

var itemjson = $.map(cleanedGridData, function (n) {
        return n;
    });

然而，这给出了（下面）的输出，它已经变平了。

["test1", "ttttt", "", "", "", "", "", "", "", "", "test2", "ghjgjhgjh", "", "", "", "", "", "", "", ""]

Answer 1

您可以使用它来提取值：

var res = {"0":["test1","ttttt","","","","","","","",""],"1":["test2","ghjgjhgjh","","","","","","","",""]}
Object.keys(res).map(function(key) {
    return res[key];
});

Object.keys将列出初始对象中的所有键。然后，您可以使用map迭代这些键，并在该函数中拉出值。

Answer 2

更改return语句，将n放入另一个数组中。

return [n];

这是因为jQuery的$.map会将返回的数组展平为结果，因此需要将其包装在外部数组中。

或者只是使用for in循环。

var itemjson = [];
for (var key in cleanedGridData) {
  itemjson.push(cleanedGridData[key]);
}

Answer 3

一旦您的环境（例如，如果您希望支持所有浏览器的浏览器）有Object.values（和spread operator），您就可以选择

[...Object.values(cleanedGridData)]

Javascript对象图（展平）

3 个答案: