Mongoose - 获取和删除子记录
Mongoose - Get and Delete a subrecord
我有一个这样定义的模型:
const mongoose = require('mongoose');
const Schema = mongoose.Schema;
const feedbackSchema = new Schema({
Name: {
type: String,
required: true,
},
Email: {
type: String,
required: true,
},
Project: {
type: String,
required: true,
},
Wonder: {
type: String,
required: true,
},
Share: {
type: String,
required: true,
},
Delight: {
type: String,
required: true,
},
Suggestions: {
type: String,
required: true,
},
Rating: {
type: String,
required: true,
},
dateCreated: {
type: Date,
default: Date.now(),
},
user: {
type: Schema.Types.ObjectId,
ref: 'User'
}
});
const UserSchema = new Schema({
googleId: {
type: String
},
displayName: {
type: String
},
firstName: {
type: String
},
lastName: {
type: String
},
image: {
type: String
},
createdAt: {
type: Date,
default: Date.now(),
},
feedback: [feedbackSchema],
})
module.exports = mongoose.model("User", UserSchema);
示例文档:
{
_id: ObjectId('60b9dc728a516a4669b40dbc'),
createdAt: ISODate('2021-06-04T07:42:01.992Z'),
googleId: '2342987239823908423492837',
displayName: 'User Name',
firstName: 'User',
lastName: 'Name',
image: 'https://lh3.googleusercontent.com/a-/89wf323wefiuhh3f9hwerfiu23f29h34f',
feedback: [
{
dateCreated: ISODate('2021-06-04T07:42:01.988Z'),
_id: ObjectId('60b9dc858a516a4669b40dbd'),
Name: 'Joe Bloggs',
Email: 'joe@bloggs.com',
Project: 'Some Project',
Suggestions: 'Here are some suggestions',
Rating: '10'
},
{
dateCreated: ISODate('2021-06-04T08:06:44.625Z'),
_id: ObjectId('60b9df29641ab05db7aa2264'),
Name: 'Mr Bungle',
Email: 'mr@bungle',
Project: 'The Bungle Project',
Suggestions: 'Wharghable',
Rating: '8'
},
{
dateCreated: ISODate('2021-06-04T08:08:30.958Z'),
_id: ObjectId('60b9df917e85eb6066049eed'),
Name: 'Mike Patton',
Email: 'mike@patton.com',
Project: 'No More Faith',
Suggestions: 'Find the faith',
Rating: '10'
},
],
__v: 0
}
我定义了两条路线,当用户单击 UI 上反馈项上的按钮时调用第一个路线,将用户带到“您确定要删除此记录吗”类型的页面,显示 selected 反馈记录中的一些信息。
第二条路线,当用户单击 'confirm' 时,子记录将从文档中删除。
我遇到的问题是我似乎无法从用户那里获取反馈以便通过 id select 文档,这是我到目前为止的确认路线:
router.get('/delete', ensureAuth, async (req, res) => {
try {
var url = require('url');
var url_parts = url.parse(req.url, true);
var feedbackId = url_parts.query.id;
const allFeedback = await User.feedback;
const feedbackToDelete = await allFeedback.find({ _id: feedbackId });
console.log(feedbackToDelete);
res.render('delete', {
imgSrc: user.image,
displayName: user.firstName,
feedbackToDelete
});
} catch (error) {
console.log(error);
}
})
非常感谢帮助
更新
您应该能够做到这一点:
const feedbackToDelete = await User.feedback.find({ _id: feedbackId });
或者如果 feedbackId
只是一个字符串,看起来是这样,您可能必须执行以下操作:
// Create an actual _id object
// That is why in your sample doc you see ObjectId('foobarbaz')
const feedbackId = new mongoose.Types.ObjectId(url_parts.query.id);
const feedbackToDelete = await User.feedback.find({ _id: feedbackId });
原创
不应该这样:
const allFeedback = await User.feedback;
(一个字段)
是这样的:
const allFeedback = await User.feedback();
(一个method/function)
?
我有一个这样定义的模型:
const mongoose = require('mongoose');
const Schema = mongoose.Schema;
const feedbackSchema = new Schema({
Name: {
type: String,
required: true,
},
Email: {
type: String,
required: true,
},
Project: {
type: String,
required: true,
},
Wonder: {
type: String,
required: true,
},
Share: {
type: String,
required: true,
},
Delight: {
type: String,
required: true,
},
Suggestions: {
type: String,
required: true,
},
Rating: {
type: String,
required: true,
},
dateCreated: {
type: Date,
default: Date.now(),
},
user: {
type: Schema.Types.ObjectId,
ref: 'User'
}
});
const UserSchema = new Schema({
googleId: {
type: String
},
displayName: {
type: String
},
firstName: {
type: String
},
lastName: {
type: String
},
image: {
type: String
},
createdAt: {
type: Date,
default: Date.now(),
},
feedback: [feedbackSchema],
})
module.exports = mongoose.model("User", UserSchema);
示例文档:
{
_id: ObjectId('60b9dc728a516a4669b40dbc'),
createdAt: ISODate('2021-06-04T07:42:01.992Z'),
googleId: '2342987239823908423492837',
displayName: 'User Name',
firstName: 'User',
lastName: 'Name',
image: 'https://lh3.googleusercontent.com/a-/89wf323wefiuhh3f9hwerfiu23f29h34f',
feedback: [
{
dateCreated: ISODate('2021-06-04T07:42:01.988Z'),
_id: ObjectId('60b9dc858a516a4669b40dbd'),
Name: 'Joe Bloggs',
Email: 'joe@bloggs.com',
Project: 'Some Project',
Suggestions: 'Here are some suggestions',
Rating: '10'
},
{
dateCreated: ISODate('2021-06-04T08:06:44.625Z'),
_id: ObjectId('60b9df29641ab05db7aa2264'),
Name: 'Mr Bungle',
Email: 'mr@bungle',
Project: 'The Bungle Project',
Suggestions: 'Wharghable',
Rating: '8'
},
{
dateCreated: ISODate('2021-06-04T08:08:30.958Z'),
_id: ObjectId('60b9df917e85eb6066049eed'),
Name: 'Mike Patton',
Email: 'mike@patton.com',
Project: 'No More Faith',
Suggestions: 'Find the faith',
Rating: '10'
},
],
__v: 0
}
我定义了两条路线,当用户单击 UI 上反馈项上的按钮时调用第一个路线,将用户带到“您确定要删除此记录吗”类型的页面,显示 selected 反馈记录中的一些信息。
第二条路线,当用户单击 'confirm' 时,子记录将从文档中删除。
我遇到的问题是我似乎无法从用户那里获取反馈以便通过 id select 文档,这是我到目前为止的确认路线:
router.get('/delete', ensureAuth, async (req, res) => {
try {
var url = require('url');
var url_parts = url.parse(req.url, true);
var feedbackId = url_parts.query.id;
const allFeedback = await User.feedback;
const feedbackToDelete = await allFeedback.find({ _id: feedbackId });
console.log(feedbackToDelete);
res.render('delete', {
imgSrc: user.image,
displayName: user.firstName,
feedbackToDelete
});
} catch (error) {
console.log(error);
}
})
非常感谢帮助
更新
您应该能够做到这一点:
const feedbackToDelete = await User.feedback.find({ _id: feedbackId });
或者如果 feedbackId
只是一个字符串,看起来是这样,您可能必须执行以下操作:
// Create an actual _id object
// That is why in your sample doc you see ObjectId('foobarbaz')
const feedbackId = new mongoose.Types.ObjectId(url_parts.query.id);
const feedbackToDelete = await User.feedback.find({ _id: feedbackId });
原创
不应该这样:
const allFeedback = await User.feedback;
(一个字段)
是这样的:
const allFeedback = await User.feedback();
(一个method/function)
?