從Angular中的物件陣列中洗掉物件-有解無憂

我有這種型別的 JSON 和我想要的，從物件陣列中洗掉StartGeotag的完整物件。

[{
    CreatedDate: "2022-02-17T10:30:07.0442288Z"
    DeletedDate: null ProjectId: "05b76d03-8c4b-47f4-7c20-08d9e2990812"
    StartGeotag: {
        Type: 'Point',
        Latitude: '33.6607231',
        CreatedDate: '2022-02- :34:46.5389961Z'
    }
    StartTime: "2022-02-17T10:30:05.828Z"
}]

uj5u.com熱心網友回復：

通過使用 ES6 擴展運算子，您可以實作：無論您想洗掉什么，在引數中給出該鍵，然后...rest回傳其余引數。

var data = [{ CreatedDate: "2022-02-17T10:30:07.0442288Z", 
DeletedDate: null, 
ProjectId: "05b76d03-8c4b-47f4-7c20-08d9e2990812",
StartGeotag: {Type: 'Point', Latitude: '33.6607231', 
CreatedDate: '2022-02- :34:46.5389961Z'},
StartTime: "2022-02-17T10:30:05.828Z"}]

const res= data.map(({StartGeotag, ...rest}) => ({...rest})); 

console.log(res);

uj5u.com熱心網友回復：

你可以通過解構你的物件來做到這一點。

[{ CreatedDate: "2022-02-17T10:30:07.0442288Z" , DeletedDate: null, ProjectId: "05b76d03-8c4b-47f4-7c20-08d9e2990812",
    StartGeotag: {Type: 'Point', Latitude: '33.6607231', CreatedDate: '2022-02- :34:46.5389961Z'},
    StartTime: "2022-02-17T10:30:05.828Z"}].map(({StartGeotag, ...item}) => item)

如果您只想獲取 StartGeoTag 物件，則可以通過以下方式進行：

 [{ CreatedDate: "2022-02-17T10:30:07.0442288Z" , DeletedDate: null, ProjectId: "05b76d03-8c4b-47f4-7c20-08d9e2990812",
    StartGeotag: {Type: 'Point', Latitude: '33.6607231', CreatedDate: '2022-02- :34:46.5389961Z'},
    StartTime: "2022-02-17T10:30:05.828Z"}].map(({StartGeotag}) => StartGeotag)

uj5u.com熱心網友回復：

您可以使用ES6中的物件解構賦值。

作業演示：

let jsonObj = [{
    CreatedDate: "2022-02-17T10:30:07.0442288Z",
    DeletedDate: null,
    ProjectId: "05b76d03-8c4b-47f4-7c20-08d9e2990812",
    StartGeotag: {
        Type: 'Point',
        Latitude: '33.6607231',
        CreatedDate: '2022-02- :34:46.5389961Z'
    },
    StartTime: "2022-02-17T10:30:05.828Z"
}];

let res = jsonObj.map(({StartGeotag, ...remainingItems}) => remainingItems)

console.log(res);

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標籤：数组 json 有角度的打字稿排序

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