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[英]Combining result of two different Queries from two different Model MongoDB
[英]MongoDB and Mongoose - id from two different collection queries not matching
我正在構建一個應用程序,人們可以在其中制作便餐並邀請其他用戶參加他們的便餐。 我有一個似乎很簡單的問題,但我無法讓它工作:
我想創建一個邀請路線來檢查 potluck.attendees 以查看他們之前是否被邀請過(並根據他們是 0-pending、1-attending還是重新邀請他們,如果他們之前被邀請過,發送不同的錯誤)並且他們的狀態為 2 拒絕),如果不是,則將受邀者放入 potluck.attendees 對象數組中,並將potluck 的 _id 放入受邀者的 user.potluks 對象數組中。
以下是這兩個模型的超級修剪版本:
簡化的聚餐模式
const PotluckSchema = new Schema({
attendees: [
{
attendeeId: {
type: mongoose.Schema.Types.ObjectId,
ref: 'User'
},
status: Number,
enums: [
0, //'pending',
1, //'attending',
2 //'declined'
]
}
]
});
簡化的用戶模型
const UserSchema = new Schema({
potlucks: [
{
potluck: {
type: mongoose.Schema.Types.ObjectId,
ref: 'Potluck'
}
}
]
});
到目前為止,我在這條路線上所擁有的是:
router.put('/attendees/invite/:potluckId/:inviteeId', async (req, res) => {
try {
const currPotluck = await db.Potluck.findOne({
_id: req.params.potluckId,
createdBy: req.user._id
});
// Makes sure potluck belongs to user before allowing to invite
if (!currPotluck)
return res.status(401).json({
msg: 'You are not authorized to invite people to this potluck.'
});
const invitee = await db.User.findOne({ _id: req.params.inviteeId });
console.log(currPotluck);
console.log(invitee);
for (let i = 0; i < currPotluck.attendees.length; i++) {
// Checks if invitee already exists in potluck.attendees
// and if their status is 0 or 1 (pending or attending)
// It will stop function
if (
currPotluck.attendees[i].attendeeId == invitee._id &&
currPotluck.attendees[i].status == 0 ||
currPotluck.attendees[i].attendeeId == invitee._id &&
currPotluck.attendees[i].status == 1
) {
return res.status(401).send({
error: 'This member has already been invited to your potluck'
});
} else if (
currPotluck.attendees[i].attendeeId == invitee._id &&
currPotluck.attendees[i].status == 2
) {
// if their status is 2 (declined)
// it will edit their existing object in the attendees array to pending
// and re-insert potluck in invitee's user.potlucks model
await db.Potluck.findOneAndUpdate(
{ _id: currPotluck._id },
{ $set: { 'attendees.$[el].status': 0 } },
{ arrayFilters: [{ 'el.attendeeId': invitee._id }] }
);
await db.User.findOneAndUpdate(
{ _id: invitee._id },
{ $push: { potlucks: { potluck: currPotluck._id } } }
);
res.send(`This user has been re-invited to your potluck!`);
}
}
// If they don't exist already in potluck.attendees, create new object
// in potlucks.attendees and user.potlucks for invitee
await db.Potluck.findOneAndUpdate(
{ _id: currPotluck._id },
{ $push: { attendees: { attendeeId: invitee._id, status: 0 } } }
);
await db.User.findOneAndUpdate(
{ _id: invitee._id },
{ $push: { potlucks: { potluck: currPotluck._id } } }
);
res.send(`This user has been invited to your potluck!`);
} catch (err) {
console.error(err.message);
res.status(500).send('Server Error');
}
});
現在來看問題:
如果我在郵遞員中運行此代碼,它將運行在 for 循環之后的兩個“findOneAndUpdate”,無論是否匹配。 在嘗試調試時,我在 console.logged 中同時記錄了invitee._id
和currPotluck.attendees[i].attendeeId
(對於我知道受邀者已經存在於數組中的測試),它們都顯示為相同的 ID。
但是,如果我嘗試使用console.log (currPootluck.attendees[i].attendeeId == invitee._id)
每次都會出現false
。 我為兩者做了一個“類型”,它們作為對象出現,它們在 console.logs 中似乎是相同的類型 - 可能是字符串?
我知道解決方案必須非常簡單,但我無法弄清楚,任何幫助將不勝感激!
currPootluck.attendees[i].attendeeId
和invitee._id
都是ObjectIds
,為了檢查它們是否相同,您必須先將其轉換為字符串。
這應該可以完成這項工作: currPootluck.attendees[i].attendeeId.toString() == invitee._id.toString()
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