Question

在我的情况下，是否有一种简单的方法仅返回几个键，例如，我只想返回标题名称，字段和类型（如果存在）。我知道我可以通过

删除密钥

for (i = 0; i < obj.length; i++) {
  delete obj[i]['type']
  delete obj[i]['hide']
  delete obj[i]['position']
}

这意味着我将不得不遍历数组中的所有文档并删除键，但不确定是否有更简单的更快方法来实现。就我而言，该文档永远不会庞大，因为它仅存储列定义以导出数据

[   {
             "headerName": "assessed_combined_value",
             "field": "assessed_combined_value",
             "hide": false,
             "position": 8,
             "type": "money"
        },
        {
             "headerName": "assessee1",
             "field": "assessee1",
             "hide": false,
             "position": 1
        },
        {
             "headerName": "assessee2",
             "field": "assessee2",
             "hide": false,
             "position": 2
        },
        {
             "headerName": "bathrooms",
             "field": "bathrooms",
             "hide": false,
             "position": 5
        }
        ]

Answer 1

您可以使用解构

let data = [{"headerName": "assessed_combined_value","field": "assessed_combined_value","hide": false,"position": 8,"type": "money"},{"headerName": "assessee1","field": "assessee1","hide": false,"position": 1},{"headerName": "assessee2","field": "assessee2","hide": false,"position": 2},{"headerName": "bathrooms","field": "bathrooms","hide": false,"position": 5}]

let final = data.map(({field, headerName}) => ({headerName, field}))

console.log(final)

如果情况相反，您只想保留一些属性，并选择所有其他属性，则可以使用spread syntax，即，如果您只想保留field键

let final = data.map(({field, ...rest}) => rest)

Answer 2

您可以将传播运算符与先前发布的答案结合使用，以仅显示当前迭代对象中存在的那些字段。

let info = [{
    "headerName": "assessed_combined_value",
    "field": "assessed_combined_value",
    "hide": false,
    "position": 8,
    "type": "money"
},
{
    "headerName": "assessee1",
    "field": "assessee1",
    "hide": false,
    "position": 1
},
{
    "headerName": "assessee2",
    "field": "assessee2",
    "hide": false,
    "position": 2
},
{
    "headerName": "bathrooms",
    "field": "bathrooms",
    "hide": false,
    "position": 5
}]

const result = info.map(({field, headerName, type}) => ({
    ...(headerName && {"headerName": headerName}),
    ...(field && {"field": field}),
    ...(type && {"type": type})
}));
console.log(result);

如何删除一系列文档中的键

2 个答案: