Question

我有一个有趣的情况，我试图包围我的头。我可以访问JSON输出，其中的键名称是_22，_155等数字。我创建了一个映射，其中包含所有这些键名称的转换，但我有很难想出如何更新json并将原始键名替换为地图中的键名。

以下是我的json字符串的示例，其中包含实际字段名称：

FieldID对应于源json数据，没有下划线。

[
{
    "FieldID": "7",
    "FieldName": "Emp ID",
    "IsSortableInput": "0",
    "IsFilterableInput": "0",
    "ViewAlias": "EmpID",
    "LogicOperator": "IN",
    "AllowMultipleValues": "1",
    "IsViewable": "1",
    "PrimaryOnly": "0",
    "DisplayOrder": "1"
},
{
    "FieldID": "3",
    "FieldName": "QID",
    "InputTypeID": "1",
    "FieldPlaceholder": "Enter Employee Name",
    "IsSortableInput": "1",
    "IsFilterableInput": "1",
    "ViewAlias": "QID",
    "LogicOperator": "IN",
    "AllowMultipleValues": "1",
    "IsViewable": "1",
    "PrimaryOnly": "0",
    "DisplayOrder": "2",
    "FieldTypeDesc": "PeopleSearch"
},
{
    "FieldID": "6",
    "FieldName": "NTID",
    "IsSortableInput": "1",
    "IsFilterableInput": "0",
    "ViewAlias": "NTID",
    "LogicOperator": "IN",
    "AllowMultipleValues": "1",
    "IsViewable": "1",
    "PrimaryOnly": "0",
    "DisplayOrder": "3"
}

我的目标是遍历源json字符串，在地图中找到键名（定义为json）并重命名它们。最终在我的源json数据中用_7替换Emp ID。

来源JSON：

[
{
    "_3": "Q123",
    "_7": "E12345"
},
{
    "_3": "X123",
    "_7": "E34567"
}
]

Desired Output是改变源JSON，将其转换为：

[
{
    "QID": "Q123",
    "Emp ID": "E12345"
},
{
    "QID": "X123",
    "Emp ID": "E34567"
}
]

我很遗憾甚至从这个方法开始。不确定是否可以重命名密钥名称，或者是否需要在流程等中删除密钥名称。

寻找解决此问题的一些建议。

Answer 1

您可以为旧密钥和新密钥构建引用，并迭代数组和密钥。然后分配给新属性并删除旧属性。

＆＃13;

var keynames = [{ FieldID: 7, FieldName: "Emp ID", IsSortableInput: 0, IsFilterableInput: 0, ViewAlias: "EmpID", LogicOperator: "IN", AllowMultipleValues: 1, IsViewable: 1, PrimaryOnly: 0, DisplayOrder: 1 }, { FieldID: 3, FieldName: "QID", InputTypeID: 1, FieldPlaceholder: "Enter Employee Name", IsSortableInput: 1, IsFilterableInput: 1, ViewAlias: "QID", LogicOperator: "IN", AllowMultipleValues: 1, IsViewable: 1, PrimaryOnly: 0, DisplayOrder: 2, FieldTypeDesc: "PeopleSearch" }, { FieldID: 6, FieldName: "NTID", IsSortableInput: 1, IsFilterableInput: 0, ViewAlias: "NTID", LogicOperator: "IN", AllowMultipleValues: 1, IsViewable: 1, PrimaryOnly: 0, DisplayOrder: 3 }],
    source = [{ _3: "Q123", _7: "E12345" }, { _3: "X123", _7: "E34567" }],
    reference = {};

keynames.forEach(function (o) {
    reference['_' + o.FieldID] = o.FieldName;
});

source.forEach(function (o) {
    Object.keys(o).forEach(function (k) {
        o[reference[k]] = o[k];
        delete o[k];
    });
});

console.log(source);

＆＃13;

.as-console-wrapper { max-height: 100% !important; top: 0; }

＆＃13;

Answer 2

解析完JSON文件后，我会使用：

删除key.name

然后

重新创建一个具有所需值的键，希望这会有所帮助。

Answer 3

yum install condor-all

Answer 4

let mapping = [
{
    "FieldID": "7",
    "FieldName": "Emp ID",
    "IsSortableInput": "0",
    "IsFilterableInput": "0",
    "ViewAlias": "EmpID",
    "LogicOperator": "IN",
    "AllowMultipleValues": "1",
    "IsViewable": "1",
    "PrimaryOnly": "0",
    "DisplayOrder": "1"
},
{
    "FieldID": "3",
    "FieldName": "QID",
    "InputTypeID": "1",
    "FieldPlaceholder": "Enter Employee Name",
    "IsSortableInput": "1",
    "IsFilterableInput": "1",
    "ViewAlias": "QID",
    "LogicOperator": "IN",
    "AllowMultipleValues": "1",
    "IsViewable": "1",
    "PrimaryOnly": "0",
    "DisplayOrder": "2",
    "FieldTypeDesc": "PeopleSearch"
},
{
    "FieldID": "6",
    "FieldName": "NTID",
    "IsSortableInput": "1",
    "IsFilterableInput": "0",
    "ViewAlias": "NTID",
    "LogicOperator": "IN",
    "AllowMultipleValues": "1",
    "IsViewable": "1",
    "PrimaryOnly": "0",
    "DisplayOrder": "3"
}
];

let source = [
{
    "_3": "Q123",
    "_7": "E12345"
},
{
    "_3": "X123",
    "_7": "E34567"
}
];

source.map(ele => {
    let ret = {};
    Object.keys(ele).forEach(key => {
        let keyNum = key.replace('_','');
        let match = mapping.find(ele => {
            return ele.FieldID === keyNum;
        });
        ret[match.FieldName] = ele[key];
    });
    return ret;
});

更新动态JSON密钥名称

4 个答案: