精简JSON，同时保留节点结构和数组

Question

我如何从JSON中修剪一切，除了我在不同级别指定的几个属性，同时保持我的节点结构和数组结构?

我研究了一下Underscore.js，它似乎没有那么多的细粒度控制来保留节点结构。在下面的例子中，理想情况下，我希望能够指定'_id', 'revisions[0]._id', 'revisions[0]._clientHasViewed'作为参数来保持这些属性。

肯定有一个简单的方法来做到这一点。这是我要找的:

原来

{
    "_id": "50cbf5214ffaee8f0400000a",
    "_user": "50b1a966c12ef0c426000007",
    "expenses": [],
    "name": "Untitled Project",
    "payments": [],
    "revisions": [
        {
            "_id": "50cbfae65c9d160506000007",
            "clientHasViewed": false,
            "comments": [],
            "dateCreated": "2012-12-15T04:21:58.605Z"
        },
        {
            "_id": "50cbfae65c9d160506000008",
            "clientHasViewed": false,
            "comments": [],
            "dateCreated": "2012-12-15T04:21:58.605Z"
        }
    ],
    "status": "Revised",
    "thumbURL": "/50cd3107845d90ab28000007/thumb.jpg"
}

修剪

{
    "_id": "50cbf5214ffaee8f0400000a",
    "revisions": [
        {
            "_id": "50cbfae65c9d160506000007",
            "clientHasViewed": false,
        },
    ],
}

Answer 1

ExtJs有一个copyTo函数(只有一个级别)，但你可以用AngularJs创建类似的东西(angular有angular. js)。复制，但复制整个对象):

var copyTo = function(dest, source, names){
    names = names.split(/[,;s]/);
    angular.forEach(names, function(name){
        if(source.hasOwnProperty(name)){
             dest[name] = source[name];
         }
     });
     return dest;
};

。

var trimmed = copyTo({}, original, '_id,');
    trimmed.revisions = [{}];
    trimmed = copyTo(trimmed.revisions[0], original.revisions[0], '_id,_clientHasViewed,');