我怀疑答案很简单，但我搜索了这个网站和其他人，但没找到。我有一个双嵌套的数据结构，无法弄清楚如何迭代最内层。我猜它可能涉及forEach（）或map（）方法，但我尝试过的都没有。
背景：
我已经简化了这个问题的数据。数据（粘贴在下面）存储在包含2个零售商店对象的数组中。每个商店对象都有一个brew install python属性，其值是一个访问对象数组。每次访问（对象）由访问日期标识（假设在日期B最多可以访问商店A）。每个访问对象都包含一个visits属性，其值是访问时进行的一系列事务（购买或退货）。在实际的完整数据中，每个商店每个日期的交易数量变化很大。
我需要帮助的任务：
（a）将财产values重命名为key，
（b）将属性visitDate重命名为values，
（c）删除8个冗余属性（从transactions到storeID），但保留storeVisitDate和action属性，以及
（d）将财产dollarAmount重命名为dollarAmount。
任何帮助都将非常感激。感谢。
dollars答案 0 :(得分：3)
你是对的！您可以使用.map（）完成大部分操作。新的ES6标准使这很容易，并且使用以下功能，您甚至不会修改任何原始数据！：
array.map(store => {
//return a new object that takes all the store info, then reassigns the visits key in a new object
return Object.assign({}, store, {
//map over visits, and reassign the key key to visitDate
visits: store.visits.map(({ key: visitDate, values }) => {
return {
//return an obj with visit date
visitDate,
// do destructuring again to create objects of action,dollars
transactions: values.map(({ action, dollarAmount: dollars }) => ({ action, dollars }))
};
})
});
});
答案 1 :(得分：1)
（working example on jsFiddle here - 只需打开JS控制台即可查看已转换的数据集）
关于以下解决方案的几点意见：
它大量使用“map”，而不是手动迭代数组。我发现它更具可读性。
// Go over all stores
stores.map(function(store) {
// In each store, go over all visits.
store.visits.map(function(visit) {
// In each visit, copy 'key' to 'visitDate'
// and 'values' to 'transactions'.
// Then delete old names ('key' and 'values').
visit.visitDate = visit.key;
visit.transactions = visit.values;
delete visit.key;
delete visit.values;
// For each transaction, replace it with a simple
// map with only 'action' and 'dollars'.
visit.transactions = visit.transactions.map(function(tx) {
return {
action: tx.action,
dollars: tx.dollarAmount
};
});
});
});
请注意，IE9及更高版本支持map()。
答案 2 :(得分：0)
要重命名属性，您可以创建一个具有相同值的新属性，然后创建旧值{。}}。
你可以将它用于所有部分，但是看看你的结构，通过使用delete创建一个新的＆＃34;优化的＆＃34;数组，将b，c，d组合在一起看起来更整洁。事务并将其分配给新的transactions属性。
// (I put this in a function so the logic can be at the top of the snippet)
function fixData(stores) {
// loop stores and their visits
for (var iStore = 0; iStore < stores.length; iStore++) {
var store = stores[iStore];
for (var iVisit = 0; iVisit < store.visits.length; iVisit++) {
var visit = store.visits[iVisit];
// (a) rename property key to visitDate
// add a new property with the same value then delete the old property
visit.visitDate = visit.key;
delete visit.key;
// (b) rename property values to transactions
// add a new property with the same value then delete the old property
// (c) delete the 8 redundant properties (from storeID to storeVisitDate)
// we could delete keys but quicker to map a new object
// (d) rename property dollarAmount to dollars.
// just give the new object property a different name
visit.transactions = visit.values.map(function(trans) {
return {
action: trans.action,
dollars: trans.dollarAmount
}
});
delete visit.values;
}
}
console.log(stores);
}
var stores = [{
"storeName": "Ye Olde Candy Shoppe",
"address": "1313 Vampire Lane, Cityville NY 99999",
"zipCode": "99999",
"storeSize": "large",
"visits": [{
"key": "5/3/12",
"values": [{
"storeID": "53454447",
"storeName": "Ye Olde Candy Shoppe",
"city": "Cityville",
"building": "1313",
"street": "Vampire Lane",
"zipcode": "99999",
"storeSize": "large",
"storeVisitDate": "5/3/12",
"action": "Return",
"dollarAmount": "65.43"
}, {
"storeID": "53454447",
"storeName": "Ye Olde Candy Shoppe",
"city": "Cityville",
"building": "1313",
"street": "Vampire Lane",
"zipcode": "99999",
"storeSize": "large",
"storeVisitDate": "5/3/12",
"action": "Purchase",
"dollarAmount": "12.43"
}, {
"storeID": "53454447",
"storeName": "Ye Olde Candy Shoppe",
"city": "Cityville",
"building": "1313",
"street": "Vampire Lane",
"zipcode": "99999",
"storeSize": "large",
"storeVisitDate": "5/3/12",
"action": "Purchase",
"dollarAmount": "5.43"
}]
}, {
"key": "12/31/12",
"values": [{
"storeID": "53454447",
"storeName": "Ye Olde Candy Shoppe",
"city": "Cityville",
"building": "1313",
"street": "Vampire Lane",
"zipcode": "99999",
"storeSize": "large",
"storeVisitDate": "12/31/12",
"action": "Purchase",
"dollarAmount": "2.53"
}]
}, {
"key": "1/24/13",
"values": [{
"storeID": "53454447",
"storeName": "Ye Olde Candy Shoppe",
"city": "Cityville",
"building": "1313",
"street": "Vampire Lane",
"zipcode": "99999",
"storeSize": "large",
"storeVisitDate": "1/24/13",
"action": "Return",
"dollarAmount": "2.53"
}, {
"storeID": "53454447",
"storeName": "Ye Olde Candy Shoppe",
"city": "Cityville",
"building": "1313",
"street": "Vampire Lane",
"zipcode": "99999",
"storeSize": "large",
"storeVisitDate": "1/24/13",
"action": "Return",
"dollarAmount": "64.22"
}]
}]
}, {
"storeName": "Mike's Bikes",
"address": "2626 Aardvark Circle, Townsville NY 88888",
"zipCode": "88888",
"storeSize": "small",
"visits": [{
"key": "8/8/14",
"values": [{
"storeID": "24335234",
"storeName": "Mike's Bikes",
"city": "Townsville",
"building": "2626",
"street": "Aardvark Circle",
"zipcode": "88888",
"storeSize": "small",
"storeVisitDate": "8/8/14",
"action": "Purchase",
"dollarAmount": "443.55"
}, {
"storeID": "24335234",
"storeName": "Mike's Bikes",
"city": "Townsville",
"building": "2626",
"street": "Aardvark Circle",
"zipcode": "88888",
"storeSize": "small",
"storeVisitDate": "8/8/14",
"action": "Purchase",
"dollarAmount": "34"
}, {
"storeID": "24335234",
"storeName": "Mike's Bikes",
"city": "Townsville",
"building": "2626",
"street": "Aardvark Circle",
"zipcode": "88888",
"storeSize": "small",
"storeVisitDate": "8/8/14",
"action": "Purchase",
"dollarAmount": "12.32"
}]
}, {
"key": "10/3/15",
"values": [{
"storeID": "24335234",
"storeName": "Mike's Bikes",
"city": "Townsville",
"building": "2626",
"street": "Aardvark Circle",
"zipcode": "88888",
"storeSize": "small",
"storeVisitDate": "10/3/15",
"action": "Purchase",
"dollarAmount": "233.1"
}, {
"storeID": "24335234",
"storeName": "Mike's Bikes",
"city": "Townsville",
"building": "2626",
"street": "Aardvark Circle",
"zipcode": "88888",
"storeSize": "small",
"storeVisitDate": "10/3/15",
"action": "Return",
"dollarAmount": "44.99"
}]
}]
}];
fixData(stores);
答案 3 :(得分：0)
如果我们假设你的数组被称为arr，我建议：
arr.forEach(function(currentValue, index, array) {
currentValue.visits = currentValue.visits.map(function(currentValue, index, array) {
currentValue.visitDate = currentValue.key;
delete currentValue.key;
currentValue.transactions = currentValue.values.map(function(currentValue, index, array) {
currentValue = {action: currentValue.action, dollars: currentValue.dollarAmount};
return currentValue;
});
delete currentValue.values;
return currentValue;
});
});
答案 4 :(得分：0)
好吧，因为这是数据＆＃39;而不是一些内存中的对象，我猜你是通过使用JSON.parse从JSON获得的。 （如果没有，您仍然可以首先使用JSON.strigify）
您知道JSON.parse是否接受控制此类内容的功能？
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/JSON/parse
...所以
fixed=JSON.parse(dataString, function(key,value){
if(key=="key"){this.visitDate=value;return}
if(key=="values"){this.transactions=value;return}
if(key=="somethingYouDontWant"){return;}
//etc...
return value;
})
答案 5 :(得分：0)
它的纯javascript没有任何库，并使用简单的地图：
var y = x.map(function(xs){
xs.visits = xs.visits.map(function (item){
return {
visitDate: item.key,
transactions: item.values.map(function(v){
return {
action: v.action,
dollars: v.dollarAmount
};
})
};
});
return xs;
});
答案 6 :(得分：0)
使用Array.map()和delete运算符的灵活解决方案：
var delete_props = ["storeID","storeName","city","building","street", "zipcode","storeSize", "storeVisitDate"];
// arr is your initial array
arr.map(function(obj){
obj['visits'].map(function(inner_obj){
inner_obj['visitDate'] = inner_obj['key'];
delete inner_obj['key'];
inner_obj['values'].map(function(values_obj){
values_obj['dollars'] = values_obj['dollarAmount'];
delete values_obj['dollarAmount'];
delete_props.forEach(function(v){
delete values_obj[v];
});
});
inner_obj['transactions '] = inner_obj['values'];
delete inner_obj['values'];
});
});
答案 7 :(得分：0)
这是另一个解决方案，代码非常简单，但没有优化。
#topbar { background:rgba(0, 0, 0, 0.65); }