錯誤資訊：
err: {
"type": "MongooseError",
"message": "Query was already executed: guilds.findOne({ name: 'Stream Squire' })",
代碼：
async function lookUpDB(guildName, lookFor) {
const query = {name: guildName}
const guildModel = await model('guilds', GuildSchema)
const guildFind = guildModel.findOne(
query,
(err, doc) => {
console.log(doc || err)
}
)
return guildFind
}
資料庫資料：
目標：我正在嘗試創建一個函式，該函式可以在我需要在資料庫中查找某些內容時隨時使用。當我嘗試查找前綴時，當前將錯誤放在首位。我對 mongoose 或 mongodb 不太熟悉，這是一個不和諧的機器人，我正在嘗試學習如何使用它
更新：公會架構
const { model, Schema } = require('mongoose')
const GuildSchema = new Schema(
{},
{
strict: false,
versionKey: false,
}
)
module.exports = { GuildSchema, model }
資料庫：
const mongoose = require('mongoose')
const { logger } = require('../config/pino')
const { config } = require('../config/dotenv')
mongoose.connect(
//removed info
{
useNewUrlParser: true,
useUnifiedTopology: true,
}
)
const { connection: db } = mongoose
db.on('connected', () => {
logger.info('Database connected')
})
db.on('disconnected', () => {
logger.info('Database disconnected')
})
db.on('error', (err) => {
logger.error(err)
})
module.exports = { db }
uj5u.com熱心網友回復：
好的，按照慣例，如果您還沒有宣告創建一個架構檔案，例如，
const mongoose = require('mongoose');
const Schema = mongoose.Schema;
const schemaDefinition = {
name: {
type: String,
required: true,
},
botName: {
type: String,
required: true,
},
Users:[{
type: String
}],
//add the rest
}
const GuildSchema = new Schema(schemaDefinition);
module.exports = mongoose.model('guilds', GuildSchema);
然后在你的 API 邏輯檔案中，匯出這個模式模型，
const GuildsModel = require('path to file');
async function lookUpDB(guildName, lookFor) {
const query = {name: guildName}
const guildFind = GuildsModel.findOne(
query,
(err, doc) => {
console.log(doc || err)
}
)
return guildFind
}
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標籤：javascript 猫鼬