MySQL - 无法运行查询答案

【问题标题】：MySQL - Trouble to run a queryMySQL - 无法运行查询
【发布时间】：2020-09-04 05:42:03
【问题描述】：

我有一个查询，我选择了一些列，然后我选择了left join。我在MySQL Workbench 使用MariaDB。

我尝试运行的查询

sequelize.query(`SELECT
        distinct av.idAnnouncementVehicles as id, 
        av.price, 
        CONCAT("https://autoparanaiba.s3-sa-east-1.amazonaws.com/" , SUBSTRING_INDEX(ap.image, '.', 1) , '-355x180.' , SUBSTRING_INDEX(ap.image, '.', -1)) as  image,
        am.worked_hours, 
        am.year AS ano,
    md.description AS modelo, 
        bn.description AS marca, 
        fu.description AS combustivel,
        pw.description AS potência,
        tc.description AS tração
        FROM announcement_vehicles AS av 
        left join announcemment_photos AS ap on ap.annoucements_id = av.idAnnouncementVehicles
        left join persons AS pe on pe.id = av.personId 
        left join agricultural_machine AS am on am.agricultural_machine_id = av.itemId 
        left join itens AS it on it.id = av.itemId
        left join traction AS tc on tc.id = trac.traction_id
        left join power AS pw on pw.id = trac.power_id
        left join fuel AS fu on fu.id = trac.fuel_id 
        where it.type=${type} and (av.plans_id IS NOT NULL or pe.type = 'jud')
        group by av.idAnnouncementVehicles order by av.plans_id, av.idAnnouncementVehicles desc limit 8`, { type: sequelize.QueryTypes.SELECT })
        .then((result) => {
            return res.json({ success: true, result: result })
        }).catch((err) => {
            return res.status(400).json(err)
        })

此查询正在向我返回此错误。但是我在代码中有一个类似的查询不会抛出这个错误。

错误

{
  "name": "SequelizeDatabaseError",
  "parent": {
    "code": "ER_BAD_FIELD_ERROR",
    "errno": 1054,
    "sqlState": "42S22",
    "sqlMessage": "Unknown column 'md.description' in 'field list'",

我有类似的查询

    sequelize.query(`SELECT
        distinct av.idAnnouncementVehicles as id, 
        av.price, 
        CONCAT("https://autoparanaiba.s3-sa-east-1.amazonaws.com/" , SUBSTRING_INDEX(ap.image, '.', 1) , '-355x180.' , SUBSTRING_INDEX(ap.image, '.', -1)) as  image,
        ve.mileage, 
        fi.marca, 
        fi.name, 
    fi.ano, 
        fi.ano_modelo as modelo, 
        fu.description as combustivel  
        FROM announcement_vehicles as av 
        left join announcemment_photos as ap on ap.annoucements_id = av.idAnnouncementVehicles
        left join persons as pe on pe.id = av.personId 
        left join vehicles as ve on ve.item_id = av.itemId 
        left join itens as it on it.id = av.itemId
        left join fipe as fi on fi.id = ve.fipe_id 
        left join fuel as fu on fu.id = ve.fuel_id 
        left join color as co on co.id = ve.color_id 
        where it.type=${type} and (av.plans_id IS NOT NULL or pe.type = 'jud')
        group by av.idAnnouncementVehicles order by av.plans_id, av.idAnnouncementVehicles desc limit 8`, { type: sequelize.QueryTypes.SELECT })
        .then((result) => {
            return res.json({ success: true, result: result })
        }).catch((err) => {
            return res.status(400).json(err)
        })

【问题讨论】：

在您的查询中没有别名为 md 的表 - 因此未知列 md.description
您也没有别名为 bn 的表。而且，大概您知道SELECT DISTINCT 适用于结果集的所有列。
@P.Salmon 我如何使用别名？
@O.Jones 我能做些什么来解决这个问题？
您在整个查询中使用表别名，因此您对此的评论令人惊讶。如果您真的不知道别名是什么，那么您应该在别处研究该主题。您的查询无法修复，因为我们不知道它应该做什么，并且从非工作代码进行逆向工程是一场无聊的游戏。

标签： mysql node.js mariadb sequelize.js mysql-workbench

【解决方案1】：

我解决了这个查询的问题

sequelize.query(`SELECT
        distinct av.idAnnouncementVehicles as id, 
        av.price, 
        CONCAT("https://autoparanaiba.s3-sa-east-1.amazonaws.com/" , SUBSTRING_INDEX(ap.image, '.', 1) , '-355x180.' , SUBSTRING_INDEX(ap.image, '.', -1)) as  image,
        am.worked_hours, 
        am.year AS ano,
    md.description AS modelo, 
        bn.description AS marca, 
        fu.description AS combustivel,
        pw.power AS potência,
        tc.description AS tração
        FROM announcement_vehicles AS av

        left join announcemment_photos AS ap on ap.annoucements_id = av.idAnnouncementVehicles
        left join persons AS pe on pe.id = av.personId

        left join agricultural_machine AS am on am.agricultural_machine_id = av.itemId 
        left join model_agricultural_machine as md on md.id = am.model_agricultural_machine_id
        left join brand_agricultural_machine as bn on bn.id = am.brand_agricultural_machine_id

        left join tractor as trc on trc.tractor_id = am.agricultural_machine_id
        left join power as pw on pw.id = trc.power_id


        left join itens AS it on it.id = av.itemId
        left join traction AS tc on tc.id = trc.traction_id

        left join fuel AS fu on fu.id = trc.fuel_id 

        where it.type=${type} and (av.plans_id IS NOT NULL or pe.type = 'jud')
        group by av.idAnnouncementVehicles order by av.plans_id, av.idAnnouncementVehicles desc limit 8`, { type: sequelize.QueryTypes.SELECT })
        .then((result) => {
            return res.json({ success: true, result: result })
        }).catch((err) => {
            return res.status(400).json(err)
        })

【讨论】：