Question

我在变量中有一个json结构，说“data”就像这样

{
"SearchWithMasterDataDIdAndScandefinitionDAO": [
    {
        "dateDm_id": 20120602,
        "issueValue": "ELTDIWKZ",
        "scanName": "Company Stored as Person (Given Name)",
        "severityCode": 1,
        "severityName": "High",
        "totalDiscovered": 1
    },
    {
        "dateDm_id": 20120602,
        "issueValue": "LTDFPVOI",
        "scanName": "Company Stored as Person (Given Name)",
        "severityCode": 1,
        "severityName": "High",
        "totalDiscovered": 1
    },
    {
        "dateDm_id": 20121005,
        "issueValue": "LTDILWGY",
        "scanName": "Company Stored as Person (Given Name)",
        "severityCode": 1,
        "severityName": "High",
        "totalDiscovered": 1
    },
    {
        "dateDm_id": 20121005,
        "issueValue": "YMORCLTD",
        "scanName": "Company Stored as Person (Given Name)",
        "severityCode": 1,
        "severityName": "High",
        "totalDiscovered": 1
    }
]
}

现在我想将此结构更改为这样的结构其中每个对象的datedm_id成为新json结构的根，其中数组为值新的通缉结构：

{
"20120602": [
    {
        "issueValue": "ELTDIWKZ",
        "scanName": "Company Stored as Person (Given Name)",
        "severityCode": 1,
        "severityName": "High",
        "totalDiscovered": 1
    },
    {
        "issueValue": "LTDFPVOI",
        "scanName": "Company Stored as Person (Given Name)",
        "severityCode": 1,
        "severityName": "High",
        "totalDiscovered": 1
    }
],
"20121005": [
    {
        "issueValue": "YMORCLTD",
        "scanName": "Company Stored as Person (Given Name)",
        "severityCode": 1,
        "severityName": "High",
        "totalDiscovered": 1
    },
    {
        "issueValue": "COOPER",
        "scanName": "Company Stored as Person (Given Name)",
        "severityCode": 1,
        "severityName": "High",
        "totalDiscovered": 15
    }
]
}

请帮我解决这个问题

Answer 1

请参阅小提琴：http://jsfiddle.net/YVB2Y/

总之，您需要在for循环中创建一个对象。

var returnVar = {}
b.SearchWithMasterDataDIdAndScandefinitionDAO.forEach(function(item){
    var thisItem;
    if(returnVar[item.dateDm_id] == undefined){
        thisItem = [];
        returnVar[item.dateDm_id]  = thisItem;
    }
    else {
        thisItem = returnVar[item.dateDm_id];
    }

    var obj = {};

    obj.issueValue = item.issueValue;
    //and so on..

    thisItem.push(obj);

});

console.log(returnVar);

Answer 2

underscore.js非常适合处理复杂的数据结构：

var json = { ... };
var result = _.groupBy(json['SearchWithMasterDataDIdAndScandefinitionDAO'], 
    function (iterator) {
        var key = iterator['dateDm_id'];
        delete iterator['dateDm_id'];

        return key;
    });

console.log(result);

使用现有的json结构在javascript中构造json格式

2 个答案: