使用 es6 解构对象数组
Destructuring an array of objects with es6
我正在尝试将具有三个属性的对象数组解构为三个单独的数组,例如
可能像
const {wlAddresses,wlTokens,wlTickets} = Object.map()
或
const [wlAddresses,wlTokens,wlTickets] = Object.map()
哪里 Object.map() returns 像这样
[{ wlAddresses: '23',
wlTokens: 1,
wlTickets: 3 },
{ wlAddresses: '24',
wlTokens: 1,
wlTickets: 2 },
{ wlAddresses: '25',
wlTokens: 1,
wlTickets: 3 }]
我试过那个方法,它只 returns 第一个对象,而不是后面的对象。我知道我可以通过映射对象并将所有内容作为单独的数组返回来解决这个问题,但也许我可以使用解构来做到这一点。
注意:只是能不能做的问题,不强答
您可以使用嵌套循环和 Object.entries()
let o = [{"wlAddresses":"23","wlTokens":1,"wlTickets":3},{"wlAddresses":"24","wlTokens":1,"wlTickets":2},{"wlAddresses":"25","wlTokens":1,"wlTickets":3}];
let entries = Object.entries(o);
let res = Array.from({length:entries.length}).fill([]);
for (let [, {wlAddresses,wlTokens,wlTickets}] of entries) {
Object.entries({wlAddresses,wlTokens,wlTickets})
.forEach(([, prop], index) => {
res[index] = [...res[index], prop]
})
}
let [wlAddresses,wlTokens,wlTickets] = res;
console.log(wlAddresses,wlTokens,wlTickets);
单独使用解构赋值
let o = [{"wlAddresses":"23","wlTokens":1,"wlTickets":3},{"wlAddresses":"24","wlTokens":1,"wlTickets":2},{"wlAddresses":"25","wlTokens":1,"wlTickets":3}];
let len = o.length;
let [wlAddresses,wlTokens,wlTickets] = Array.from({length:len}, () => []);
let i = 0;
if (i < len) {
({wlAddresses:wlAddresses[wlAddresses.length]
, wlTokens:wlTokens[wlTokens.length]
, wlTickets:wlTickets[wlTickets.length]} = o[i]);
console.log(wlAddresses,wlTokens,wlTickets);
++i;
} else {
console.log("done");
console.log(wlAddresses,wlTokens,wlTickets);
}
if (i < len) {
({wlAddresses:wlAddresses[wlAddresses.length]
, wlTokens:wlTokens[wlTokens.length]
, wlTickets:wlTickets[wlTickets.length]} = o[i]);
console.log(wlAddresses,wlTokens,wlTickets);
++i;
} else {
console.log("done");
console.log(wlAddresses,wlTokens,wlTickets);
}
if (i < len) {
({wlAddresses:wlAddresses[wlAddresses.length]
, wlTokens:wlTokens[wlTokens.length]
, wlTickets:wlTickets[wlTickets.length]} = o[i]);
console.log(wlAddresses,wlTokens,wlTickets);
++i;
} else {
console.log("done");
console.log(wlAddresses,wlTokens,wlTickets);
}
if (i < len) {
({wlAddresses:wlAddresses[wlAddresses.length]
, wlTokens:wlTokens[wlTokens.length]
, wlTickets:wlTickets[wlTickets.length]} = o[i]);
console.log(wlAddresses,wlTokens,wlTickets);
++i;
} else {
console.log("done");
console.log(wlAddresses,wlTokens,wlTickets);
}
在第二个硬编码片段中使用代码循环
let o = [{"wlAddresses":"23","wlTokens":1,"wlTickets":3},{"wlAddresses":"24","wlTokens":1,"wlTickets":2},{"wlAddresses":"25","wlTokens":1,"wlTickets":3}];
let len = o.length;
let [wlAddresses,wlTokens,wlTickets] = [[], [], []];
let i = 0;
while (i < len) {
({wlAddresses:wlAddresses[wlAddresses.length]
, wlTokens:wlTokens[wlTokens.length]
, wlTickets:wlTickets[wlTickets.length]} = o[i]);
++i;
}
console.log("done");
console.log(wlAddresses,wlTokens,wlTickets);
当你有一个对象或数组并且需要提取特定的项目时,解构是很好的,但这里不是这种情况。您的数据不是一个简单的对象或数组——它是一个对象数组。您将无法通过简单的作业来完成此操作。您需要将数据转换成您想要的格式。例如这样的事情:
let arr = [{
wlAddresses: '23',
wlTokens: 1,
wlTickets: 3
},
{
wlAddresses: '24',
wlTokens: 1,
wlTickets: 2
},
{
wlAddresses: '25',
wlTokens: 1,
wlTickets: 3
}
]
let r = arr.reduce((acc, curr) => {
for ([key, value] of Object.entries(curr)) {
if (! acc[key]) acc[key] = []
acc[key].push( curr[key])
}
return acc
}, {})
const {wlAddresses,wlTokens,wlTickets} = r
console.log(wlAddresses,wlTokens,wlTickets)
假设数组中的每个对象都具有完全相同的相同的键,您可以直接映射它。
const data = [{wlAddresses:'23',wlTokens:1,wlTickets:3},{wlAddresses:'24',wlTokens:1,wlTickets:2},{wlAddresses:'25',wlTokens:1,wlTickets:3}];
const r = Object.keys(data[0]).map((v) => ({ [v]: data.map((c) => c[v]) }));
console.log(JSON.stringify(r));
我正在尝试将具有三个属性的对象数组解构为三个单独的数组,例如
可能像
const {wlAddresses,wlTokens,wlTickets} = Object.map()
或
const [wlAddresses,wlTokens,wlTickets] = Object.map()
哪里 Object.map() returns 像这样
[{ wlAddresses: '23',
wlTokens: 1,
wlTickets: 3 },
{ wlAddresses: '24',
wlTokens: 1,
wlTickets: 2 },
{ wlAddresses: '25',
wlTokens: 1,
wlTickets: 3 }]
我试过那个方法,它只 returns 第一个对象,而不是后面的对象。我知道我可以通过映射对象并将所有内容作为单独的数组返回来解决这个问题,但也许我可以使用解构来做到这一点。
注意:只是能不能做的问题,不强答
您可以使用嵌套循环和 Object.entries()
let o = [{"wlAddresses":"23","wlTokens":1,"wlTickets":3},{"wlAddresses":"24","wlTokens":1,"wlTickets":2},{"wlAddresses":"25","wlTokens":1,"wlTickets":3}];
let entries = Object.entries(o);
let res = Array.from({length:entries.length}).fill([]);
for (let [, {wlAddresses,wlTokens,wlTickets}] of entries) {
Object.entries({wlAddresses,wlTokens,wlTickets})
.forEach(([, prop], index) => {
res[index] = [...res[index], prop]
})
}
let [wlAddresses,wlTokens,wlTickets] = res;
console.log(wlAddresses,wlTokens,wlTickets);
单独使用解构赋值
let o = [{"wlAddresses":"23","wlTokens":1,"wlTickets":3},{"wlAddresses":"24","wlTokens":1,"wlTickets":2},{"wlAddresses":"25","wlTokens":1,"wlTickets":3}];
let len = o.length;
let [wlAddresses,wlTokens,wlTickets] = Array.from({length:len}, () => []);
let i = 0;
if (i < len) {
({wlAddresses:wlAddresses[wlAddresses.length]
, wlTokens:wlTokens[wlTokens.length]
, wlTickets:wlTickets[wlTickets.length]} = o[i]);
console.log(wlAddresses,wlTokens,wlTickets);
++i;
} else {
console.log("done");
console.log(wlAddresses,wlTokens,wlTickets);
}
if (i < len) {
({wlAddresses:wlAddresses[wlAddresses.length]
, wlTokens:wlTokens[wlTokens.length]
, wlTickets:wlTickets[wlTickets.length]} = o[i]);
console.log(wlAddresses,wlTokens,wlTickets);
++i;
} else {
console.log("done");
console.log(wlAddresses,wlTokens,wlTickets);
}
if (i < len) {
({wlAddresses:wlAddresses[wlAddresses.length]
, wlTokens:wlTokens[wlTokens.length]
, wlTickets:wlTickets[wlTickets.length]} = o[i]);
console.log(wlAddresses,wlTokens,wlTickets);
++i;
} else {
console.log("done");
console.log(wlAddresses,wlTokens,wlTickets);
}
if (i < len) {
({wlAddresses:wlAddresses[wlAddresses.length]
, wlTokens:wlTokens[wlTokens.length]
, wlTickets:wlTickets[wlTickets.length]} = o[i]);
console.log(wlAddresses,wlTokens,wlTickets);
++i;
} else {
console.log("done");
console.log(wlAddresses,wlTokens,wlTickets);
}
在第二个硬编码片段中使用代码循环
let o = [{"wlAddresses":"23","wlTokens":1,"wlTickets":3},{"wlAddresses":"24","wlTokens":1,"wlTickets":2},{"wlAddresses":"25","wlTokens":1,"wlTickets":3}];
let len = o.length;
let [wlAddresses,wlTokens,wlTickets] = [[], [], []];
let i = 0;
while (i < len) {
({wlAddresses:wlAddresses[wlAddresses.length]
, wlTokens:wlTokens[wlTokens.length]
, wlTickets:wlTickets[wlTickets.length]} = o[i]);
++i;
}
console.log("done");
console.log(wlAddresses,wlTokens,wlTickets);
当你有一个对象或数组并且需要提取特定的项目时,解构是很好的,但这里不是这种情况。您的数据不是一个简单的对象或数组——它是一个对象数组。您将无法通过简单的作业来完成此操作。您需要将数据转换成您想要的格式。例如这样的事情:
let arr = [{
wlAddresses: '23',
wlTokens: 1,
wlTickets: 3
},
{
wlAddresses: '24',
wlTokens: 1,
wlTickets: 2
},
{
wlAddresses: '25',
wlTokens: 1,
wlTickets: 3
}
]
let r = arr.reduce((acc, curr) => {
for ([key, value] of Object.entries(curr)) {
if (! acc[key]) acc[key] = []
acc[key].push( curr[key])
}
return acc
}, {})
const {wlAddresses,wlTokens,wlTickets} = r
console.log(wlAddresses,wlTokens,wlTickets)
假设数组中的每个对象都具有完全相同的相同的键,您可以直接映射它。
const data = [{wlAddresses:'23',wlTokens:1,wlTickets:3},{wlAddresses:'24',wlTokens:1,wlTickets:2},{wlAddresses:'25',wlTokens:1,wlTickets:3}];
const r = Object.keys(data[0]).map((v) => ({ [v]: data.map((c) => c[v]) }));
console.log(JSON.stringify(r));