Question

虽然我有一些基本的JavaScript背景，但我偶然发现了我写的这段代码：

var data=[{"_id":"57b3e7ec9b209674f1459f36","fName":"Tom","lName":"Moody","email":"Tom@example.com","age":30},{"_id":"57b3e8079b209674f1459f37","fName":"Pat","lName":"Smith","email":"pat@example.com","age":32},{"_id":"57b3e8209b209674f1459f38","fName":"Sam","lName":"Dawn","email":"sam@example.com","age":28},{"_id":"57b3e8219b209674f1459f39","fName":"Sam","lName":"Dawn","email":"sam@example.com","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，我希望表包含Tom，Pat和Sam的数组。相反，这就是我得到的：

table[["Tom","Moody","Tom@example.com",30]]
table[["Pat","Smith","pat@example.com",32],["Pat","Smith","pat@example.com",32]]
table[["Sam","Dawn","sam@example.com",28],["Sam","Dawn","sam@example.com",28],["Sam","Dawn","sam@example.com",28]]
table[["Sam","Dawn","sam@example.com",28],["Sam","Dawn","sam@example.com",28],["Sam","Dawn","sam@example.com",28],["Sam","Dawn","sam@example.com",28]]

为什么push（）替换表中的上一个条目？任何帮助将受到高度赞赏。

Answer 1

其他人已经在你的代码中指出了问题。

然而，你也使事情变得更加复杂。你可以这样做：

＆＃13;

var data=[{"_id":"57b3e7ec9b209674f1459f36","fName":"Tom","lName":"Moody","email":"Tom@example.com","age":30},{"_id":"57b3e8079b209674f1459f37","fName":"Pat","lName":"Smith","email":"pat@example.com","age":32},{"_id":"57b3e8209b209674f1459f38","fName":"Sam","lName":"Dawn","email":"sam@example.com","age":28},{"_id":"57b3e8219b209674f1459f39","fName":"Sam","lName":"Dawn","email":"sam@example.com","age":28}];

var table = data.map(function(user) {
    return [
      user.fName,
      user.lName,
      user.email,
      user.age,
    ];
});
      
console.log(table);

＆＃13;

或者如果您使用ES6：

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

Answer 2

您不需要手动编写所有样板代码。使用正确的数组迭代器（在您的情况下为map）。

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

Answer 3

显然，map是为了功能性方法而去的方式，但是如果你喜欢命令式样式，一种简单的方法可能是使用for of循环，如下所示。

＆＃13;

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

＆＃13;

Answer 4

这里的问题不在于推动。 javascript中的变量存储对数组的引用。在表中，您正在推送对相同数组tempArr的引用。您需要在推送之前创建新数组，或者在推送之前创建数组的深层副本。

示例

var data=[{"_id":"57b3e7ec9b209674f1459f36","fName":"Tom","lName":"Moody","email":"Tom@example.com","age":30},{"_id":"57b3e8079b209674f1459f37","fName":"Pat","lName":"Smith","email":"pat@example.com","age":32},{"_id":"57b3e8209b209674f1459f38","fName":"Sam","lName":"Dawn","email":"sam@example.com","age":28},{"_id":"57b3e8219b209674f1459f39","fName":"Sam","lName":"Dawn","email":"sam@example.com","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));
 });

Answer 5

嗯，与许多其他语言不同，JavaScript已经通过引用传递了所有内容。这意味着当您table.push(tempArr);时，您实际上并没有推动tempArr的值，而是将引用推送到{{1 }}。所以，如果你这样做：

var a ='a';

tempArr

您将获得var table = []; table.push(a); a = 'b'; console.log(table[0]);作为输出。你想要做的是定义一个要推送的新变量，比如这个

Answer 6

var data = [{"_id": "57b3e7ec9b209674f1459f36","fName": "Tom","lName": "Moody","email": "Tom@example.com","age": 30}, {"_id": "57b3e8079b209674f1459f37","fName": "Pat","lName": "Smith","email": "pat@example.com","age": 32}, {"_id": "57b3e8209b209674f1459f38","fName": "Sam","lName": "Dawn","email": "sam@example.com","age": 28}, {"_id": "57b3e8219b209674f1459f39","fName": "Sam","lName": "Dawn","email": "sam@example.com","age": 28}],
    table = [];

data.forEach(function(user) {
    table.push([user.fName, user.lName, user.email, user.age]);
});

console.log(table);

数组推送行为

6 个答案: