对象数组内部连接对象数组

Question

假设我有一个对象数组，如：

const bookDetails = [
    {
        "bookId": "1235",
        "emailId": "samplek@gmail.com",
        "bookIssue": [{"Book not properly aligned": true, "some issue1": true}]
    },
    {
        "bookId": "1235",
        "emailId": "s@gmampleail.com",
        "bookIssue": [{"some issues with book": true, "some issue2": true}]
    }]

我要交单的内容如下：

[
    {"bookId": "1235", "emailId": "samplek@gmail.com", "bookIssue": "Book not properly aligned,some issue1"},
    {"bookId": "1235", "emailId": "s@gmampleail.com", "bookIssue": "some issues with book,some issue2"}
]

为此我试过，

bookDetails.map((i) => i.bookIssue = Object.keys(i.bookIssue[0]).join(","))

它根据需要给出O/p，但它开始给出价值，

[{"bookId":"1235","emailId":"samplek@gmail.com","bookIssue":"0"},
{"bookId":"1235","emailId":"s@gmampleail.com","bookIssue":"0"}]

有什么可以解决的问题，还有什么其他的方法可以实现吗？

Answer 1

请参阅my comment和georg，您的代码工作正常(除了不正确地使用map )，前提是您想要修改适当的对象。

如果您想要在一个新数组中创建新对象(例如，正确使用map )，您可以做您要做的事情来获取键，但是创建一个新对象，结果如下：

const result = bookDetails.map(entry => {
    // Grab the keys from the first entry and join them
    const bookIssue = Object.keys(entry.bookIssue[0]).join(",");
    // Build the new object to return
    return {...entry, bookIssue};
});

活生生的例子：

​

const bookDetails = [
    {"bookId":"1235","emailId":"samplek@gmail.com","bookIssue":[{"Book not properly aligned": true,"some issue1":true}]},
    {"bookId":"1235","emailId":"s@gmampleail.com","bookIssue":[{"some issues with book": true, "some issue2":true }]}
];

const result = bookDetails.map(entry => {
    // Grab the keys from the first entry and join them
    const bookIssue = Object.keys(entry.bookIssue[0]).join(",");
    // Build the return object
    return {...entry, bookIssue};
});

console.log(result);

​

如果bookIssue可以有多个条目(如果不能，为什么它是一个数组？)您希望将的bookIssue中的所有条目连接在一起，您可以使用bookIssue上的map从其对象中获取所有键并将它们连接起来，然后加入结果数组：

const result = bookDetails.map(entry => {
    const bookIssue = entry.bookIssue
        .map(entry => Object.keys(entry).join(","))
        .join(",");
    // Build the return object
    return {...entry, bookIssue};
});

活生生的例子：

​

const bookDetails = [
    {"bookId":"1235","emailId":"samplek@gmail.com","bookIssue":[
        {"Book not properly aligned": true,"some issue1":true},
        {"another issue": true,"yet another issue":true}
    ]},
    {"bookId":"1235","emailId":"s@gmampleail.com","bookIssue":[{"some issues with book": true, "some issue2":true }]}
];

const result = bookDetails.map(entry => {
    const bookIssue = entry.bookIssue
        .map(entry => Object.keys(entry).join(","))
        .join(",");
    // Build the return object
    return {...entry, bookIssue};
});

console.log(result);

​

当然，如果只有一个条目，这也是可行的。

Answer 2

您的代码也能工作，因为您正在更改对象bookIssue (与bookDetails中的引用相同)。下面是将属性bookIssue映射到所需值的另一种方法，方法是在映射中返回新对象。

​

const bookDetails = [
    {
        "bookId": "1235",
        "emailId": "samplek@gmail.com",
        "bookIssue": [{"Book not properly aligned": true, "some issue1": true}]
    },
    {
        "bookId": "1235",
        "emailId": "s@gmampleail.com",
        "bookIssue": [{"some issues with book": true, "some issue2": true}]
    }];
    
const output = bookDetails.map(book => {
  return {...book, bookIssue: Object.keys(book.bookIssue[0]).join(',')}
});

console.log(output);

​

Answer 3

只需使用forEach更新您的Map函数，并查看console.log(bookDetails)。您正在检查地图输出，而不是检查bookDetails。这就是你困惑的原因。也可以在代码中检查console.log(bookDetails)。它将如预期的那样工作：bookDetails.forEach((i) => i.bookIssue = Object.keys(i.bookIssue).join("，“)；