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Script Considerations脚本注意事项

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The results of database queries cannot be passed inside the following contexts:数据库查询的结果不能在以下上下文中传递：

  • Class constructor functions类构造函数函数
  • Non-async generator functions非异步生成器函数
  • Callbacks to .sort() on an array回调到数组上的.sort()

To access to the results of database calls, use async functions, async generator functions, or .map().要访问数据库调用的结果，请使用异步函数、异步生成器函数或.map()。

Constructors构造函数

The following constructors do not work:以下构造函数不起作用：

// This code will fail
class FindResults {
   constructor() {
      this.value = db.students.find();
   }
}

// This code will fail
function listEntries() { return db.students.find(); }
class FindResults {
   constructor() {
      this.value = listEntries();
   }
}

Use an async function instead:请改用async函数：

class FindResults {
   constructor() {
      this.value = ( async() => {
         return db.students.find();
      } )();
   }
}
Note

You can also create a method that performs a database operation inside a class as an alternative to working with asynchronous JavaScript.您还可以创建一个在类中执行数据库操作的方法，作为使用异步JavaScript的替代方法。

class FindResults {
   constructor() { }

   init() { this.value = db.students.find(); }
 }

To use this class, first construct a class instance then call the .init() method.要使用这个类，首先构造一个类实例，然后调用.init()方法。

Generator Functions生成器函数

The following generator functions do not work:以下生成器函数不起作用：

// This code will fail
function* FindResults() {
   yield db.students.findMany();
}

// This code will fail
function listEntries() { return db.students.findMany(); }
function* findResults() {
   yield listEntries();
}

Use an async generator function instead:请改用async generator function：

function listEntries() { return db.students.findMany(); }
async function* findResults() {
   yield listEntries();
}

Array Sort数组排序

The following array sort does not work:以下数组排序不起作用：

// This code will fail
db.getCollectionNames().sort( ( collectionOne, collectionTwo ) => {
  return db[ collectionOne ].estimatedDocumentCount() - db[ collectionOne ].estimatedDocumentCount() )
} );

Use .map() instead.请改用.map()。

db.getCollectionNames().map( collectionName => {
   return { collectionName, size: db[ collectionName ].estimatedDocumentCount() };
} ).sort( ( collectionOne, collectionTwo ) => {
   return collectionOne.size - collectionTwo.size;
} ).map( collection => collection.collectionName);

This approach to array sort is often more performant than the equivalent unsupported code.这种数组排序方法通常比等效的不受支持的代码更具性能。