如何将多个For循环重构为一个？

如何解决如何将多个For循环重构为一个？

如何重构此功能？我觉得嵌套的for循环会起作用，但是我还没有弄清楚如何使其起作用。

功能如下：

class Librarian {
  constructor(name,library) {
    this.name = name;
    this.library = library;
  }
findBook(bookTitle) {
    for (var i = 0; i < this.library.shelves.fantasy.length; i++){
      if (bookTitle === this.library.shelves.fantasy[i].title){
        this.library.shelves.fantasy.splice(i,1);
        return `Yes,we have ${bookTitle}`;
      }
    } for (var i = 0; i < this.library.shelves.fiction.length; i++){
      if (bookTitle === this.library.shelves.fiction[i].title){
        this.library.shelves.fiction.splice(i,we have ${bookTitle}`;
      }
    } for (var i = 0; i < this.library.shelves.nonFiction.length; i++){
      if (bookTitle === this.library.shelves.nonFiction[i].title){
        this.library.shelves.nonFiction.splice(i,we have ${bookTitle}`;
      }
    } return `Sorry,we do not have ${bookTitle}`;
  } 
}

这是我的尝试：

findBook(bookTitle) {
    for (var j = 0; j < this.libray.shelves.length; j++) {
      for (var i = 0; i < this.library.shelves[i].length; i++){
        if (bookTitle === this.library.shelves[j].title){
          this.library.shelves[j].splice(i,1);
          return `Yes,we have ${bookTitle}`;
        }
      }
    } return `Sorry,we do not have ${bookTitle}`;
}

解决方法

获取流派的名称，然后遍历它们：

findBook(bookTitle) {
  const genres = Object.keys(this.library.shelves);
  for (var i = 0; i < genres.length; i++) {
    const genre = genres[i];
    for (var j = 0; j < this.library.shelves[genre].length; j++) {
      if (bookTitle === this.library.shelves[genre][j].title) {
        this.library.shelves[genre].splice(i,1);
        return `Yes,we have ${bookTitle}`;
      }
    }
  }
  return `Sorry,we do not have ${bookTitle}`;
}

,

看起来您真的很想让它工作，您只需要在第二个for循环中迭代this.library.shelves[j].length而不是i。然后，您需要访问第j个架子的第i个成员。

findBook(bookTitle) {
   for (var j = 0; j < this.libray.shelves.length; j++) {
      for (var i = 0; i < this.library.shelves[j].length; i++){
         if (bookTitle === this.library.shelves[j][i].title){
            this.library.shelves[j].splice(i,1);
            return `Yes,we have ${bookTitle}`;
         }
      }
   } 
   return `Sorry,we do not have ${bookTitle}`;
}