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mysql - Sequelize 与 mysql 的关系未执行正确的查询

转载 作者:行者123 更新时间:2023-11-29 17:20:18 32 4
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我正在尝试在sequelize中与mysql进行关联,但没有得到预期的结果

这是我的用户模型:

const bcrypt = require('bcrypt');

module.exports = (sequelize, DataTypes) => {
const user = sequelize.define('user', {
displayName: {
type: DataTypes.STRING,
allowNull: false,
unique: true,
},
email: {
type: DataTypes.STRING,
allowNull: false,
unique: true,
},
active: DataTypes.BOOLEAN,
description: DataTypes.STRING,
password: {
type: DataTypes.STRING,
allowNull: false
},
gender: DataTypes.CHAR(1),
presentationSentence: DataTypes.STRING,
hairId: DataTypes.INTEGER,
createdAt: DataTypes.DATE,
updatedAt: DataTypes.DATE,
}, {
hooks: {
beforeCreate: user => {
const salt = bcrypt.genSaltSync();

user.password = bcrypt.hashSync(user.password, salt);
}
},
isValidPassword: (password, salt) => {
return bcrypt.compareSync(password, salt);
}
});

user.associate = models => {
user.hasOne(models.hair, {
foreignKey: 'id',
sourceKey: 'hairId'
});

user.hasMany(models.photo, {
foreignKey: 'userId'
});
};

return user;
};

我有一个头发模型:

module.exports = (sequelize, DataTypes) => {
const hair = sequelize.define('hair', {
type: DataTypes.ENUM('Liso', 'Crespo', 'Ondulado', 'Cacheado', 'Calvo')
});

return hair;
};

问题是,当我执行查询时,它会查找带有用户 ID 的 hair 模型,而不是 hairId 字段;

SELECT `user`.`id`, `user`.`displayName`, `user`.`email`, `user`.`active`, `user`.`description`, `user`.`password`, `user`.`gender`, `user`.`presentationSentence`, `user`.`hairId`, `user`.`createdAt`, `user`.`updatedAt`, `hair`.`id` AS `hair.id`, `hair`.`type` AS `hair.type`, `hair`.`createdAt` AS `hair.createdAt`, `hair`.`updatedAt` AS `hair.updatedAt`, `photos`.`id` AS `photos.id`, `photos`.`url` AS `photos.url`, `photos`.`active` AS `photos.active`, `photos`.`avatar` AS `photos.avatar`, `photos`.`userId` AS `photos.userId`, `photos`.`createdAt` AS `photos.createdAt`, `photos`.`updatedAt` AS `photos.updatedAt` FROM `users` AS `user` LEFT OUTER JOIN `hairs` AS `hair` ON `user`.`id` = `hair`.`id` LEFT OUTER JOIN `photos` AS `photos` ON `user`.`id` = `photos`.`userId`;

所以,它总是给我带来 id 1 的头发字段

有谁知道为什么会这样吗?

我花了很多时间来解决这个问题,但无法找到解决方案

最佳答案

你的头发模型定义错误,给它一个自增主键和一个枚举值。

module.exports = (sequelize, DataTypes) => {
const hair = sequelize.define('hair', {
hair_id: {
type: DataTypes.BIGINT.UNSIGNED,
autoIncrement: true,
primaryKey: true,
},
email: {
type: DataTypes.STRING,
references: {
model: 'Users',
key: 'email',
}
}
hair_type: {
type: DataTypes.ENUM,
values: [
'Liso',
'Crespo',
'Ondulado',
'Cacheado',
'Calvo'
],
defaultValue: 'Liso',
notNull: true
}
});

return hair;
};

在协会中,

user.associate = models => {
user.hasOne(models.hair, {
foreignKey: 'email',
sourceKey: 'email'
});

user.hasMany(models.photo, {
foreignKey: 'userId'
});
};

关于mysql - Sequelize 与 mysql 的关系未执行正确的查询,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51296483/

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