隐藏JSON.stringify（）输出中的某些值 [英] Hide certain values in output from JSON.stringify()

问题描述

是否可以排除某些字段包含在json字符串中？

Is it possible to exclude certain fields from being included in the json string?

这是一些伪代码

var x = {
    x:0,
    y:0,
    divID:"xyz",
    privateProperty1: 'foo',
    privateProperty2: 'bar'
}

我想排除privateProperty1和privateproperty2出现在json字符串中

I want to exclude privateProperty1 and privateproperty2 from appearing in the json string

所以我想，我可以使用stringify替换函数

So I thought, I can use the stringify replacer function

function replacer(key,value)
{
    if (key=="privateProperty1") then retun "none";
    else if (key=="privateProperty2") then retun "none";
    else return value;
}

和stringify

and in the stringify

var jsonString = json.stringify(x,replacer);

但是在jsonString中我仍然认为它是

But in the jsonString I still see it as

{...privateProperty1:value..., privateProperty2:value }

我想要没有privateproperties的字符串。

I would like to the string without the privateproperties in them.

推荐答案

Mozilla docs 说返回 undefined （而不是none）：

http://jsfiddle.net/userdude/rZ5Px/

function replacer(key,value)
{
    if (key=="privateProperty1") return undefined;
    else if (key=="privateProperty2") return undefined;
    else return value;
}

var x = {
    x:0,
    y:0,
    divID:"xyz",
    privateProperty1: 'foo',
    privateProperty2: 'bar'
};

alert(JSON.stringify(x, replacer));

这是一种复制方法，以防您决定走这条路线（根据您的评论）。

Here is a duplication method, in case you decide to go that route (as per your comment).

http://jsfiddle.net/userdude/ 644sJ /

function omitKeys(obj, keys)
{
    var dup = {};
    for (var key in obj) {
        if (keys.indexOf(key) == -1) {
            dup[key] = obj[key];
        }
    }
    return dup;
}

var x = {
    x:0,
    y:0,
    divID:"xyz",
    privateProperty1: 'foo',
    privateProperty2: 'bar'
};

alert(JSON.stringify(omitKeys(x, ['privateProperty1','privateProperty2'])));

编辑 - 我更改了底部功能中的功能键以保留它从混乱中解脱出来。

EDIT - I changed the function key in the bottom function to keep it from being confusing.

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