node.js - 同时使用 'Where' 和 'Include' 语句拉取数据

Question

我已经设法让我的包含语句与我的外键一起工作，但是当我尝试向 findAll 语句添加“where”语句时，我收到以下错误。我已经多次检查我的外键和模型，我看不出它们有任何问题。
下面的 API 中的语法是否正确？键中的数据很重要，所以我不想花更多时间在解决方法上。我已经看到其他示例在其中一个包含键中包含 where 语句，但我想要提取的字段位于主表上，所以我认为这不是相同的情况。

Unhandled rejection Error: Include unexpected. Element has to be either a Model, an Association or an object.

API

const sequelize = require("pg");
const { Sequelize, Op, Model, DataTypes } = require("sequelize");

exports.schoolDiveLogApproval = (req, res) => {

    try {
        const schoolID = req.params.schoolID;

        diveLog.findAll({
            include: [{
                model: diveType, as: 'diveTypeID_FK2',
            }, {
                model: diveSchool, as: 'diveSchoolID_FK',
            }, {
                model: current, as: 'currentID_FK',
            }, {
                model: visibility, as: 'visibilityID_FK',
            }, {
                model: user, as: 'userID_FK1',
            }, {
                model: diveSpot, as: 'diveSpotID_FK2'
            }],
            where: {
                [Op.and]: [
                    {diveSchoolID: schoolID},
                    {diveVerifiedBySchool: false}
                ]
            },
        })

...........

更新
错误信息

{
    "message": "missing FROM-clause entry for table \"diveLog\""
}

带有原始 SQL 的 Controller API

exports.schoolDiveLogApproval = async (req, res) => {

    try {
        const schoolID = req.params.schoolID;
        const { QueryTypes } = require('sequelize');
        await diveLog.sequelize.query(
            'SELECT * '+
            'FROM "diveLogs" '+
            'LEFT OUTER JOIN "diveTypes" AS "diveTypeID_FK2" ON "diveLogs"."diveTypeID" = "diveTypeID_FK2"."diveTypeID" ' +
            'LEFT OUTER JOIN "diveSchools" AS "diveSchoolID_FK" ON "diveLog"."diveSchoolID" = "diveSchoolID_FK"."diveSchoolID" ' +
            'LEFT OUTER JOIN "currents" AS "currentID_FK" ON "diveLog"."currentID" = "currentID_FK"."currentID" ' +
            'LEFT OUTER JOIN "visibilities" AS "visibilityID_FK" ON "diveLog"."visibilityID" = "visibilityID_FK"."visibilityID" ' +
            'LEFT OUTER JOIN "userLogins" AS "userID_FK" ON "diveLog"."userID" = "userID_FK1"."userID" ' +
            'LEFT OUTER JOIN "diveSpots" AS "diveSpotID_FK2" ON "diveLog"."diveSpotID" = "diveSpotFK2"."diveSpotID" ' +
            'WHERE "diveLogs"."diveSchoolID" = ? ' +
            'AND "diveLogs"."diveVerifiedBySchool" = ?',
            {
                replacements: [schoolID, false],
                type: QueryTypes.SELECT
            }
        )

根据 pgAdmin 数据库，所有表名和列名看起来都是正确的，所以除了语法之外，我看不出问题出在哪里。
我已经偏离了在 IDE 终端中执行 SQL 的方式，因此它应该在理论上执行。这更有可能与我传递 id 的方式有关吗？

Answer 1

好吧，我真的在这个 Sequelize 上尝试了几分钟，但我真的建议你学习和使用原始 SQL，它比这更容易和通用。
在尝试设置和理解这个 sqlize 半小时后，看到了很多像这样的奇怪问题:

选择不存在的列，如id，在表“user_infoS(？)”的名称中添加一个“s”，如果我理解你的问题，我放弃并想出了一个可能对你有帮助的查询正确。
如果您发送正确的架构，我可以进一步帮助您，但是，对不起，我不能花更多时间在这个 Sequelize 中，这件事太糟糕了。

const { QueryTypes } = require('sequelize');
await sequelize.query(
  'SELECT *'+ 
  'FROM diveLog l'+
  'JOIN diveType t ON l.diveTypeId = t.id'+
  'JOIN diveSchool s ON l.diveSchoolId = s.id'+
  'JOIN current c ON l.currentId = c.id'+
  'JOIN visibility v ON l.visibilityId = v.id'+
  'JOIN user u ON l.userId = u.id'+
  'JOIN diveSpot sp ON l.diveSpotId = sp.id'+
  'WHERE diveSchoolID = ?'+
    'AND diveVerifiedBySchool = ?',
  {
    replacements: [schoolID, false],
    type: QueryTypes.SELECT
  }
);