Question

我具有以下涉及猫鼬的功能：

  let username = userObject.username;
  Purchase.find({
    account: username,
    fufilled: true 
  })
    .populate("keys")
    .exec(function(err, foundPurchases) {
      if (err) {
        return inStockItems;
      } else {
        if (foundPurchases.length === 0) {
          return inStockItems;
        } else {
          // these lists will be a list of IDs of game detail entries in the database
          let listOfReceivedIds = foundPurchases.keys.map(obj => obj.game);

          for (let i = 0; i < inStockItems.length; i++) {
            if (inStockItems.length <= minimum) {
              return inStockItems;
            }

            let currentProductAnalysing = inStockItems[i];
            if (listOfReceivedIds.includes(currentProductAnalysing._id)) {
              console.log("removing product");
              inStockItems.splice(i, 1);
            }
          }
          return inStockItems;
        }
      }
    });

我正在运行以下函数，该函数返回undefined

inStockItems = function_name(inStockItems, userObject, amount);

如何重写函数，以便函数返回inStockItems而不是undefined的值。谢谢。

Answer 1

如果您不给它回调，则Mongoose中的

.exec会返回查询结果的承诺。这里的函数需要调用.exec()并更改其返回值，以返回与该函数不同的东西：

async function function_name(inStockItems, userObject, amount) {
  const foundPurchases = await Purchase.find({
      account: username,
      fufilled: true
    })
    .populate("keys")
    .exec();
  if (foundPurchases.length === 0) {
    return inStockItems;
  } else {
    let listOfReceivedIds = foundPurchases.keys.map(obj => obj.game);

    for (let i = 0; i < inStockItems.length; i++) {
      if (inStockItems.length <= minimum) {
        return inStockItems;
      }

      let currentProductAnalysing = inStockItems[i];
      if (listOfReceivedIds.includes(currentProductAnalysing._id)) {
        console.log("removing product");
        inStockItems.splice(i, 1);
      }
    }
    return inStockItems;
  }
}

另一种选择是将回调参数传递给函数，然后从exec()回调中调用它，但诺言通常更干净。

返回前如何等待猫鼬.exec（）函数完成？

1 个答案: