如何替换嵌套对象中的键

3回答

明月笑刀无情

你的代码有错别字。某些变量显示为displaName而不是 displayName。您需要递归调用函数才能按预期工作。您没有使用newKeys变量进行重命名。你只是硬编码它像newKey = 'label'. 但是这个问题与问题无关。const resp = {  data: {    id: '1',    displayName: 'A',    children: [{      id: '2',      displayName: 'B',      children: [{        id: '3',        displayName: 'C',        children: [          //More nested array here        ]      }]    }]  }}const newKeys = {  displayName: "label"};const renamedObj = this.renameKeys(resp.data, newKeys);console.log(renamedObj);function renameKeys(obj, newKeys) {  const keyValues = Object.keys(obj).map(key => {    let newKey = null    if (key === 'displayName') {      newKey = newKeys.displayName    } else {      newKey = key    }    if (key === 'children') {      obj[key] = obj[key].map(obj => renameKeys(obj, newKeys));        }    return {      [newKey]: obj[key]    };  });  return Object.assign({}, ...keyValues);}

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红颜莎娜

这是一个非常通用的解决方案，它将遍历一个对象并更新给定对象字面量 ( keysToUpdate)中的任何对象键。const orig = {  id: '1',  displayName: 'A',  children: [{    id: '2',    displayName: 'B',    children: [{      id: '3',      displayName: 'C',      children: [        //More nested array here      ]    }]  }]};const updateDisplayNameToLabel = (val, keysMap) => {  if (val == null) return null;  if (Array.isArray(val)) {    return val.map(item => updateDisplayNameToLabel(item, keysMap));  } else if (typeof val == "object") {    return Object.keys(val).reduce((obj, key) => {      const propKey = updateDisplayNameToLabel(key, keysMap);      const propVal = updateDisplayNameToLabel(val[key], keysMap);      obj[propKey] = propVal;      return obj;    }, {});  } else if (typeof val === "string") {    return keysMap[val] || val;  }  return val;}const keysToUpdate = {  displayName: 'label',  children: 'items'};const updated = updateDisplayNameToLabel(orig, keysToUpdate);console.log(updated);

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幕布斯6054654

现有的答案很好，但我无法抗拒添加一个JSON.stringify版本：const data = { id: '1', displayName: 'A', children: [ { id: '2', displayName: 'B', children: [ { id: '3', displayName: 'C', children: [] }] }] };const result = JSON.parse(JSON.stringify(data).replace(/"displayName":/g, '"value":'));console.log(result);显然，如果您有一个看起来像键的值，这将不起作用，因此它假设您有可预测数据的保证。如果您有多个替代品，您可以使用const data = { id: '1', displayName: 'A', children: [ { id: '2', displayName: 'B', children: [ { id: '3', displayName: 'C', children: [] }] }] };const swaps = {displayName: "foo", children: "baz", id: "corge"};const pattern = new RegExp(  Object.keys(swaps).map(e => `(?:"(${e})":)`).join("|"), "g");const result = JSON.parse(  JSON.stringify(data).replace(pattern, m => `"${swaps[m.slice(1,-2)]}":`));console.log(result);更传统的递归选项可能如下（仍然假定/硬编码children）：const changeKey = (node, keySubs) =>   Object.entries(keySubs).reduce((a, [oldKey, newKey]) => {    a[newKey] = a[oldKey];    delete a[oldKey];    return a;  }, {...node, children: node.children.map(e => changeKey(e, keySubs))});const data = { id: '1', displayName: 'A', children: [ { id: '2', displayName: 'B', children: [ { id: '3', displayName: 'C', children: [] }] }] };const swaps = {displayName: "label", id: "better id"};console.log(changeKey(data, swaps));迭代：const changeKey = (node, keySubs) => {  const result = {children: []};  const stack = [[node, result]];  while (stack.length) {    const [curr, parent] = stack.pop();    const child = Object.entries(keySubs)      .reduce((a, [oldKey, newKey]) => {        a[newKey] = a[oldKey];        delete a[oldKey];        return a;      }, {...curr, children: []})    ;    parent.children.push(child);    stack.push(...curr.children.map(e => [e, child]));  }  return result;};const data = { id: '1', displayName: 'A', children: [ { id: '2', displayName: 'B', children: [ { id: '3', displayName: 'C', children: [] }] }] };const swaps = {displayName: "label", id: "better id"};console.log(changeKey(data, swaps));

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