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sequelize.js - sequelize 不包括所有 child ,如果任何一个匹配

转载 作者:行者123 更新时间:2023-12-05 04:38:43 25 4
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我有三个背靠背的关联表。这意味着 item_level_1 有很多 item_level_2,item_level_2 有很多 item_level_3。我使用搜索查询来查找名称中包含搜索文本的任何父项或子项。这意味着如果我键入 abc,那么我需要返回所有父项或子项的完整详细信息(父项和子项)。但在我的例子中,如果 item_level_3 的名称中有 abc,它会返回父级详细信息,但它只会返回 item_level_3 中带有 abc 的特定子项。我需要返回同一父级的 item_level_3 内的所有子级。

我在 AWS 中通过节点使用 MySQL 数据库

我检查了https://sequelize.org/master/manual/eager-loading.html#complex-where-clauses-at-the-top-level并尝试了不同的组合。但无济于事。我可能会错过一些东西。但是我找不到它。

exports.searchItems = (body) => {
return new Promise((resolve, reject) => {
let searchText = body.searchText.toLowerCase();
let limit = body.limit;
let offset = body.offset;

db.item_level_1.findAndCountAll({
where: {
[Sequelize.Op.or]: [
Sequelize.where(Sequelize.fn('lower', Sequelize.col("item_level_1.name")), Sequelize.Op.like, '%' + searchText + '%'),
Sequelize.where(Sequelize.fn('lower', Sequelize.col("item_level_2.name")), Sequelize.Op.like, '%' + searchText + '%'),
Sequelize.where(Sequelize.fn('lower', Sequelize.col("item_level_2.item_level_3.name")), Sequelize.Op.like, '%' + searchText + '%'),
],

[Sequelize.Op.and]: [
Sequelize.where(Sequelize.col("item_level_1.status"), Sequelize.Op.eq, body.status)
]
},
offset: offset,
limit: limit,
distinct: true,
subQuery: false,
attributes: ['id', 'name'],
include: [
{
model: db.item_level_2,
as: 'item_level_2',
where: {
status: body.status
},
attributes: ['id', 'name'],
required: true,
include: [{
model: db.item_level_3,
as: 'item_level_3',
where: {
status: body.status
},
required: false,
attributes: ['id', 'name']
}]
}
]
}).then(result => {
resolve({ [KEY_STATUS]: 1, [KEY_MESSAGE]: "items listed successfully", [KEY_DATA]: result.rows, [KEY_TOTAL_COUNT]: result.count });
}).catch(error => {
reject({ [KEY_STATUS]: 0, [KEY_MESSAGE]: "items list failed", [KEY_ERROR]: error });
});
})
}

预期结果

{
"status": 1,
"message": "Rent items listed successfully",
"data": [
{
"id": 21,
"name": "this is test parent one",
"item_level_2": [
{
"id": 39,
"name": "this is second test parent one",
"item_level_3": {
"id": 9,
"name": "this is the child description with abc"
}
},
{
"id": 40,
"name": "this is second test parent two",
"item_level_3": {
"id": 6,
"name": "this is the child description with def"
}
},
{
"id": 41,
"name": "this is second test parent three",
"item_level_3": {
"id": 70,
"name": "this is the child description with ghi"
}
}
]
}
],
"totalCount": 1
}

实际结果

{
"status": 1,
"message": "Rent items listed successfully",
"data": [
{
"id": 21,
"name": "this is test parent one",
"item_level_2": [
{
"id": 39,
"name": "this is second test parent one",
"item_level_3": {
"id": 9,
"name": "this is the child description with abc"
}
}
]
}
],
"totalCount": 1
}

item_level_1 模型

module.exports = (sequelize, DataTypes) => {
const item_level_1 = sequelize.define("item_level_1", {
id: { type: INTEGER, primaryKey: true, autoIncrement: true },
name: { type: STRING },
status: { type: BOOLEAN, defaultValue: 0 }
}, {
timestamps: false,
freezeTableName: true,
})
item_level_1.associate = function (models) {
item_level_1.hasMany(models.item_level_2, { as: 'item_level_2' });
};
return item_level_1;

}

item_level_2 模型

module.exports = (sequelize, DataTypes) => {
const item_level_2 = sequelize.define("item_level_2", {
id: { type: INTEGER, primaryKey: true, autoIncrement: true },
name: { type: STRING },
status: { type: BOOLEAN, defaultValue: 0 },
itemLevel2Id: { type: INTEGER },
itemLevel1Id: { type: INTEGER }
}, {
timestamps: false,
freezeTableName: true,
})
item_level_2.associate = function (models) {
item_level_2.belongsTo(models.item_level_3, { as: 'item_level_3', foreignKey: 'itemLevel2Id' });
};
return item_level_2;

}

item_level_2 型号

module.exports = (sequelize, DataTypes) => {
const item_level_3 = sequelize.define("item_level_3", {
id: { type: INTEGER, primaryKey: true, autoIncrement: true },
name: { type: STRING },
status: { type: BOOLEAN, defaultValue: 0 }
}, {
timestamps: false,
freezeTableName: true,
})
return item_level_3;

}

最佳答案

这是一个复杂的场景,需要一些解决方法。另外,我还没有测试所有场景,很抱歉它可能适用于示例案例,但不能满足您的所有需求。不过,我希望这会为您提供一些指导。

基于这里写的SQL,https://dba.stackexchange.com/a/140006 ,您可以在 item_level_2 之间创建 2 个 JOIN和 item_level_3 ,1 个用于过滤,1 个用于获取所有相关记录。

item_level_2.hasMany(item_level_3, { as: 'item_level_3' });
// This extra association will be used only for filtering.
item_level_2.hasMany(item_level_3, { as: 'filter' });

然后,

db.item_level_1.findAndCountAll({
where: {
[Sequelize.Op.or]: [
Sequelize.where(Sequelize.fn('lower', Sequelize.col("item_level_1.name")), Sequelize.Op.like, '%' + searchText + '%'),
Sequelize.where(Sequelize.fn('lower', Sequelize.col("item_level_2.name")), Sequelize.Op.like, '%' + searchText + '%'),
// Use the filter association to filter data.
Sequelize.where(Sequelize.fn('lower', Sequelize.col("item_level_2.filter.name")), Sequelize.Op.like, '%' + searchText + '%'),
],
...
include: [
{
model: db.item_level_2,
as: 'item_level_2',
where: {
status: body.status
},
attributes: ['id', 'name'],
required: true,
include: [
{
model: db.item_level_3,
as: 'item_level_3',
where: {
status: body.status
},
required: false,
attributes: ['id', 'name'] // This should fetch all associated data.
},
{
model: db.item_level_3,
as: 'filter',
where: {
status: body.status
},
required: false,
attributes: [] // Do not fetch any data from this association. This is only for filtering.
}
]
}
]
}
})

这涵盖了 1 个项目与 item_level_3 匹配的场景并且有多个项目与相同的 item_level_2 相关联.如果有多个 item_level_2,这将不起作用与 item_level_1 有关和 item_level_2 中的 1 个与 searchText 匹配.

我还没有测试过,但是,也许你可以为 item_level_1 做类似的事情如果您需要,也可以。

============================================= ==

更新:

如果item_level_2之间有关联和 item_level_3belongsTo ,上述解决方案将不起作用。

您需要 WHERE EXISTS查询 item_level_3 .(省略了错误的解决方案。)

============================================= ==

更新2:

使用内联 IN查询 item_level_3文本匹配。

在进行内联查询之前,确保转义将进入 Sequelize.literal 的动态内容稍后。

Important Note: Since sequelize.literal inserts arbitrary content without escaping to the query, it deserves very special attention since it may be a source of (major) security vulnerabilities. It should not be used on user-generated content.

引用:https://sequelize.org/master/manual/sub-queries.html

const escapedSearchText = sequelize.escape(`%${searchText}%`);

首先设置内联查询选项以提取 item_level_1 searchText 所在的 ID出现在任何 child 中(item_level_3)。为此,我只查询 item_level_2item_level_3表格和使用 GROUPHAVING .

const inQueryOptions = {
attributes: ['itemLevel1Id'], // This attribute name and the one in group could be different for your table.
include: [{
attributes: [],
model: db.item_level_3,
as: 'item_level_3',
where: {
name: {
[Sequelize.Op.like]: escapedSearchText
}
}
}],
group: 'itemLevel1Id',
having: Sequelize.literal('COUNT(*) > 0')
};

item_level_1 分组的 ID 和过滤 HAVING , 这将返回所有 item_level_1其任何子项位于 item_level_3 处的 ID有 searchText .

这仍然只搜索 item_level_3的名字。

接下来,将选项转换为内联查询。

const Model = require("sequelize/lib/model");
// This is required when the inline query has `include` options, this 1 line make sure to serialize the query correctly.
Model._validateIncludedElements.bind(db.item_level_2)(inQueryOptions);

// Then, pass the query options to queryGenerator.
// slice(0, -1) is to remove the last ";" as I will use this query inline of the main query.
const inQuery = db.sequelize.getQueryInterface().queryGenerator.selectQuery('item_level_2', inQueryOptions, db.item_level_2).slice(0, -1);

生成的inQuery看起来像这样。

SELECT `item_level_2`.`itemLevel1Id` 
FROM `item_level_2` AS `item_level_2`
INNER JOIN `item_level_3` AS `item_level_3`
ON `item_level_2`.`itemLevel3Id` = `item_level_3`.`id`
AND `item_level_3`.`name` LIKE '%def%'
GROUP BY `itemLevel1Id`
HAVING COUNT(*) > 0

最后,将这个生成的查询插入到主查询中。

db.item_level_1.findAndCountAll({
subQuery: false,
distinct: true,
where: {
[Op.or]: [
Sequelize.where(Sequelize.fn('lower', Sequelize.col("item_level_1.name")), Sequelize.Op.like, '%' + searchText + '%'),
Sequelize.where(Sequelize.fn('lower', Sequelize.col("item_level_2.name")), Sequelize.Op.like, '%' + searchText + '%'),
{
id: {
// This is where I am inserting the inline query.
[Op.in]: Sequelize.literal(`(${inQuery})`)
}
}
]
},
attributes: ['id', 'name'],
include: [{
attributes: ['id', 'name'],
model: db.item_level_2,
as: 'item_level_2',
required: true,
include: [{
attributes: ['id', 'name'],
model: db.item_level_3,
as: 'item_level_3',
required: false,
}]
}]
});

关于sequelize.js - sequelize 不包括所有 child ,如果任何一个匹配,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/70541443/

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