将列表转换为具有新属性名称的另一个新列表

Posted 2023-03-16

【中文标题】将列表转换为具有新属性名称的另一个新列表

【英文标题】：covert list into another new list with new property name

【发布时间】：2021-10-11 01:41:46

【问题描述】：

我有一个列表看起来像所有设置。我想将该列表转换为新列表。由于我是新手，所以我没有太多想法，我尝试按照以下方式进行操作，但新列表中的第一项始终为空。

const [mySetting, setMySet] = useState([]);
  useEffect(() => 
    const allSettings = [
       name: "Setting1", value: true, label: 1 ,
       name: "Setting2", value: true, label: 2 ,
       name: "Setting3", value: true, label: 3 ,
       name: "Setting4", value: false, label: 4 ,
       name: "Setting5", value: true, label: 5 ,
       name: "Setting6", value: true, label: 6 ,
       name: "Setting7", value: true, label: 7 
    ];
    const settings = [];
    const allSettingsMap = allSettings.reduce((resMap, current) => 
      settings.push(resMap);
      return 
        ...resMap,
        SettingID: current.label,
        Name: current.name,
        value: current.value
      ;
    , );
    setMySet(settings);
  , []);

//我想要这样的新列表：

const newSettings = [
           name: "Setting1", value: true, SettingID: 1 ,
          name: "Setting2", value: true, SettingID: 2 ,
           name: "Setting3", value: true, SettingID: 3 ,
           name: "Setting4", value: false, SettingID: 4 ,
           name: "Setting5", value: true, SettingID: 5 ,
           name: "Setting6", value: true, SettingID: 6 ,
           name: "Setting7", value: true, SettingID: 7 
        ];

【参考方案1】：

您可以使用array#map 重命名对象中的键并保持其他键不变。对于每个对象，您可以选择 label 并将其重命名为 SettingID 并保持其他键值相同。

const allSettings = [  name: "Setting1", value: true, label: 1 ,  name: "Setting2", value: true, label: 2 ,  name: "Setting3", value: true, label: 3 ,  name: "Setting4", value: false, label: 4 ,  name: "Setting5", value: true, label: 5 ,  name: "Setting6", value: true, label: 6 ,  name: "Setting7", value: true, label: 7  ],
      result = allSettings.map((label, ...other) => (...other, SettingID: label));
console.log(result);

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【参考方案2】：

用这个 map 方法替换你的 allSettings.reduce 方法

const settings = [];

allSettings.map((setting) => 
  const item =  ...setting ;
  const tempValue = item?.label;
  // delete item label
  delete item["label"];

  item["SettingID"] = tempValue;

  settings.push(item);
);

【参考方案3】：

您可以通过数组上的映射轻松做到这一点

const allSettings = [
       name: "Setting1", value: true, label: 1 ,
       name: "Setting2", value: true, label: 2 ,
       name: "Setting3", value: true, label: 3 ,
       name: "Setting4", value: false, label: 4 ,
       name: "Setting5", value: true, label: 5 ,
       name: "Setting6", value: true, label: 6 ,
       name: "Setting7", value: true, label: 7 
    ];
   const newAllSettings = allSettings.map(item => 
      return 
        name: item.name,
        value: item.value,
        SettingID:item.label
      ;
    );
    console.log(newAllSettings)