数组推送行为

Question

虽然我有一些基础的JavaScript背景，但我写下了以下代码：

var data=[{"_id":"57b3e7ec9b209674f1459f36","fName":"Tom","lName":"Moody","email":"[e-mail]","age":30},{"_id":"57b3e8079b209674f1459f37","fName":"Pat","lName":"Smith","email":"[e-mail]","age":32},{"_id":"57b3e8209b209674f1459f38","fName":"Sam","lName":"Dawn","email":"[e-mail]","age":28},{"_id":"57b3e8219b209674f1459f39","fName":"Sam","lName":"Dawn","email":"[e-mail]","age":28}]
var tempArr=[];
var table=[];
var dataArr = Object.keys(data).map(function(k) { return data[k] });
dataArr.forEach(function(user) {
  tempArr[0]=user.fName;
  tempArr[1]=user.lName;
  tempArr[2]=user.email;
  tempArr[3]=user.age;
  table.push(tempArr);
  console.log('table'+JSON.stringify(table));
 });

在最后一个循环中，我期望table包含Tom、Pat和Sam的数组。然而，实际情况是这样的：

table[["Tom","Moody","[e-mail]",30]]
table[["Pat","Smith","[e-mail]",32],["Pat","Smith","[e-mail]",32]]
table[["Sam","Dawn","[e-mail]",28],["Sam","Dawn","[e-mail]",28],["Sam","Dawn","[e-mail]",28]]
table[["Sam","Dawn","[e-mail]",28],["Sam","Dawn","[e-mail]",28],["Sam","Dawn","[e-mail]",28],["Sam","Dawn","[e-mail]",28]]

为什么 push() 会替换表格中的先前条目？非常感谢您的帮助。

- user2693135

---

tempArr从未改变，因此您一直在将相同的引用推入表格。将tempArr设为内部函数中的局部变量，以在forEach调用中使用。 - nolan

Answer 1

其他人已经指出了您代码中的问题。

但是，您也使事情变得比必要的更加复杂。您只需执行以下操作即可：

```python #在需要使用时导入模块 import module #调用函数 module.function() ```

这样可以避免不必要的重复和冗余代码。

var data=[{"_id":"57b3e7ec9b209674f1459f36","fName":"Tom","lName":"Moody","email":"[e-mail]","age":30},{"_id":"57b3e8079b209674f1459f37","fName":"Pat","lName":"Smith","email":"[e-mail]","age":32},{"_id":"57b3e8209b209674f1459f38","fName":"Sam","lName":"Dawn","email":"[e-mail]","age":28},{"_id":"57b3e8219b209674f1459f39","fName":"Sam","lName":"Dawn","email":"[e-mail]","age":28}];

var table = data.map(function(user) {
    return [
      user.fName,
      user.lName,
      user.email,
      user.age,
    ];
});
      
console.log(table);

如果您使用ES6：

var table = data.map(user => [ user.fName, user.lName, user.email, user.age ];

- str

Answer 2

您不需要手动编写所有样板代码。使用适当的数组迭代器（在您的情况下是map）。

var table = data.map(function(user) {
   return [user.fName, user.lName, user.email, user.age];
});

- Yury Tarabanko

Answer 3

显然，为了实现功能式方法，地图是必须的，但如果您喜欢命令式样式，则一个简单的方法是使用以下的for of循环。

var data = [{"_id":"57b3e7ec9b209674f1459f36","fName":"Tom","lName":"Moody","email":"[e-mail]","age":30},{"_id":"57b3e8079b209674f1459f37","fName":"Pat","lName":"Smith","email":"[e-mail]","age":32},{"_id":"57b3e8209b209674f1459f38","fName":"Sam","lName":"Dawn","email":"[e-mail]","age":28},{"_id":"57b3e8219b209674f1459f39","fName":"Sam","lName":"Dawn","email":"[e-mail]","age":28}],
   table = [];
for (var user of data) table.push([user.fName,user.lName,user.email,user.age]);
console.log(table);

- Redu

Answer 4

var data = [{"_id": "57b3e7ec9b209674f1459f36","fName": "Tom","lName": "Moody","email": "[e-mail]","age": 30}, {"_id": "57b3e8079b209674f1459f37","fName": "Pat","lName": "Smith","email": "[e-mail]","age": 32}, {"_id": "57b3e8209b209674f1459f38","fName": "Sam","lName": "Dawn","email": "[e-mail]","age": 28}, {"_id": "57b3e8219b209674f1459f39","fName": "Sam","lName": "Dawn","email": "[e-mail]","age": 28}],
    table = [];

data.forEach(function(user) {
    table.push([user.fName, user.lName, user.email, user.age]);
});

console.log(table);

- Yosvel Quintero

Answer 5

这里的问题不在于push。在Javascript中，变量存储对数组的引用。而在表格中，您正在将引用推入到同一个数组tempArr中。在推送之前，您需要创建新的数组或创建数组的深拷贝。

例如

var data=[{"_id":"57b3e7ec9b209674f1459f36","fName":"Tom","lName":"Moody","email":"[e-mail]","age":30},{"_id":"57b3e8079b209674f1459f37","fName":"Pat","lName":"Smith","email":"[e-mail]","age":32},{"_id":"57b3e8209b209674f1459f38","fName":"Sam","lName":"Dawn","email":"[e-mail]","age":28},{"_id":"57b3e8219b209674f1459f39","fName":"Sam","lName":"Dawn","email":"[e-mail]","age":28}]
var table=[];
var dataArr = Object.keys(data).map(function(k) { return data[k] });
dataArr.forEach(function(user) {

  var tempArr=[];
  tempArr[0]=user.fName;
  tempArr[1]=user.lName;
  tempArr[2]=user.email;
  tempArr[3]=user.age;
  table.push(tempArr);
  console.log('table'+JSON.stringify(table));
 });

- Himanshu Pandey

Answer 6

与许多其他编程语言不同，JavaScript 通过引用传递所有内容。这意味着当你执行 table.push(tempArr); 时，你实际上并没有推送 tempArr 的值，而是推送了对 tempArr 的引用。因此，如果你这样做：

var a = 'a';

var table = [];
table.push(a);
a = 'b';
console.log(table[0]);

你将会得到输出 b。你需要做的是定义一个新变量来进行push操作，就像这样：

var data=[{"_id":"57b3e7ec9b209674f1459f36","fName":"Tom","lName":"Moody","email":"[e-mail]","age":30},{"_id":"57b3e8079b209674f1459f37","fName":"Pat","lName":"Smith","email":"[e-mail]","age":32},{"_id":"57b3e8209b209674f1459f38","fName":"Sam","lName":"Dawn","email":"[e-mail]","age":28},{"_id":"57b3e8219b209674f1459f39","fName":"Sam","lName":"Dawn","email":"[e-mail]","age":28}]

var table=[];
var dataArr = Object.keys(data).map(function(k) { return data[k] });
dataArr.forEach(function(user) {
    var tempArr=[];
    tempArr[0]=user.fName;
    tempArr[1]=user.lName;
    tempArr[2]=user.email;
    tempArr[3]=user.age;
    table.push(tempArr);
 });

 console.log('table'+JSON.stringify(table));

- Gareth Parker