如果物件中的子項中存在特定鍵，則將子值應用于父鍵

const obj = {
  uid: "893212",
  a: {name: "Down here!", uid: "1231"},
  b: {
    c: {uid: "5965"},
    name: "bud name",
  }, 
  d: {name: "doodle name"},
  e: {name: "alexa name"},
  f: ["kk", "jj"],
  g: [
   {
     h: {uid: "47895"},
     i: {uid: "4785"}
   },
   {
     j: {uid: "4895"}
   }
  ]
};

在上述物件中，如果“uid”存在，則其值應作為值應用于其父物件。結果應該如下

var result = {
  uid: "893212",
  a: "1231",
  b: {c: "5965", name: "bud name"},
  d: {name: "doodle name"},
  e: {name: "alexa name"},
  f: ["kk", "jj"],
  g: [
   {h: "47895", i: "4795"},
   {j: "4895"}
  ]
}

我嘗試使用遞回函式來操作物件。

const mapObj = (obj = {}) => {
  if (isObject(obj)) {
    const entries = Object.entries(obj);

    for (let i = 0; i < entries.length; i  = 1) {
      const [objK, objV] = entries[i];
      
      if (isObject(objV) && 'uid' in objV) {
        obj[objK] = objV['uid'];          
      } else if(isObject(objV)){
        findNestedObject(objV);
      } else if(isArray(objV)) {
        objV.forEach(val => {
          findNestedObject(val);
        })
      }
    } 
  }
};

有什么簡單的方法可以做到這一點并轉換物件陣列

uj5u.com熱心網友回復：

嘗試這個：

const obj = {
  uid: "893212",
  a: {
    name: "Down here!",
    uid: "1231",
  },
  b: {
    c: {
     uid: "5965",
     },
     name: "bud name",
  }, 
  d: {
      name: "doodle name"
  },
  e: {
      name: "alexa name"
  },
  f: ["kk", "jj"],
};

const process = (anObject) => {
  const toReturn = {};
  Object.keys(anObject).forEach((key, index) => {
    const currentProperty = anObject[key];  
    if(Array.isArray(currentProperty)) {  
      toReturn[key] = currentProperty;
    }
    else if(typeof(currentProperty) === 'object') {
      const uid = currentProperty?.uid;
      if(uid) {
        toReturn[key] = uid;
      } 
      else {
        toReturn[key] = process(currentProperty);
      }         
    } else {
      toReturn[key] = currentProperty;
    }
  });
  return toReturn;
}

const target = process(obj);
console.log(JSON.stringify(target, null, 2));

uj5u.com熱心網友回復：

在對子級呼叫遞回時將指標傳遞給父級的遞回。

const obj = {uid:"893212",a:{name:"Down here!",uid:"1231"},b:{c:{uid:"5965"},name:"bud name"},d:{name:"doodle name"},e:{name:"alexa name"},f:["kk","jj"],g:[{h:{uid:"47895"},i:{uid:"4785"}},{j:{uid:"4895"}}]};

function do_obj(obj, parent, parent_key) {
  Object.keys(obj).forEach(function(key) {
    var item = obj[key];
    if (key === 'uid' && parent) {
      parent[parent_key] = item;
    }
    if (typeof item === 'object' && item !== null) {
      do_obj(item, obj, key);
    }
  })
}

do_obj(obj)
console.log(obj)

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

uj5u.com熱心網友回復：

試試看：

const obj = { uid: "893212", a: {name: "Down here!", uid: "1231"}, b: { c: {uid: "5965"}, name: "bud name", }, d: {name: "doodle name"}, e: {name: "alexa name"}, f: ["kk", "jj"], g: [ { h: {uid: "47895"}, i: {uid: "4785"} }, { j: {uid: "4895"} } ] };

const refactor = (ins, depth) => {
  if (Array.isArray(ins)) {
    return ins.map(it => refactor(it, depth 1));
  }
  
  if (typeof ins === "object" && ins !== null) {
    if (ins["uid"] && depth !== 0) return ins.uid;
    return Object.entries(ins).reduce((a, [k, v]) => ({
      ...a,
      [k]: refactor(v, depth 1)
    }), {});
  }
  return ins;
}

console.log(refactor(obj, 0));

標籤：javascript目的ecmascript-6ecmascript-5

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