尝试将“赞”与帖子关联时出现此错误。
未处理的拒绝SequelizeDatabaseError:列中的空值 “ userId”违反了非空约束
现在,以下代码获取正确的帖子ID和用户ID,我做了console log
。我可能做错了什么?
routes / posts.js
router.post('/:userId/like/:postId', (req, res)=> {
models.Post.findOne({
where:{
id: req.params.postId
}
})
.then( (like) => {
if(like){
models.Likes.create({
where:{
userId: req.params.userId,
postId: req.params.postId
},
like:true
}).then( (result) => {
res.status(200).send({
message: 'You have like this post',
like: result
})
})
}
}).catch( (err) => {
res.status(401).send({
message: "Something went wrong",
err: err
})
})
})
这是喜欢迁移
up: (queryInterface, Sequelize) => {
return queryInterface.createTable('Likes', {
id: {
allowNull: false,
autoIncrement: true,
primaryKey: true,
type: Sequelize.INTEGER
},
like: {
type: Sequelize.BOOLEAN
},
userId: {
allowNull: false,
type: Sequelize.INTEGER,
references: {
model: 'Users',
key: 'id'
}
},
createdAt: {
allowNull: false,
type: Sequelize.DATE
},
updatedAt: {
allowNull: false,
type: Sequelize.DATE
}
});
},
down: (queryInterface, Sequelize) => {
return queryInterface.dropTable('Likes');
}
};
帖子迁移
'use strict';
module.exports = {
up: (queryInterface, Sequelize) => {
return queryInterface.createTable('Posts', {
id: {
allowNull: false,
autoIncrement: true,
primaryKey: true,
type: Sequelize.INTEGER
},
title: {
type: Sequelize.STRING
},
post_content: {
type: Sequelize.STRING
},
createdAt: {
allowNull: false,
type: Sequelize.DATE
},
updatedAt: {
allowNull: false,
type: Sequelize.DATE
},
userId: {
type: Sequelize.INTEGER,
references: {
model: 'Users',
key: 'id'
}
},
username: {
type: Sequelize.STRING
},
});
},
down: (queryInterface, Sequelize) => {
return queryInterface.dropTable('Posts');
}
};
类似模型
'use strict';
module.exports = function(sequelize, DataTypes) {
const Like = sequelize.define('Likes', {
like:{
type:DataTypes.BOOLEAN,
allowNull:true
}
}, {});
Like.associate = function(models) {
Like.belongsTo(models.User, {
onDelete: "CASCADE",
sourceKey: 'userId'
})
Like.belongsTo(models.Post, {
onDelete: "CASCADE",
sourceKey: 'likeId'
})
}
return Like;
}
Post.model
module.exports = (sequelize, DataTypes) => {
const Post = sequelize.define('Post', {
title: DataTypes.STRING,
post_content: DataTypes.STRING,
username: DataTypes.STRING
}, {});
Post.associate = function(models) {
Post.belongsTo(models.User, { foreignKey: 'userId', targetKey: 'id' });
Post.hasMany(models.Likes, { foreignKey: 'postId', sourceKey: 'id' });
};
return Post;
};
额外
add_postId_to_likes
'use strict';
module.exports = {
up: function (queryInterface, Sequelize) {
return queryInterface.addColumn(
'Likes',
'postId',
{
type: Sequelize.INTEGER,
allowNull: true,
references: {
model: 'Posts',
key: 'id',
}
}
)
},
down: function (queryInterface, Sequelize) {
return queryInterface.removeColumn(
'Likes',
'postId'
)
}
};
答案 0 :(得分:0)
在解析器中的create调用中,您没有提供必要的值,没有where子句,但实际上没有为所需的userId提供它的值。.看起来您的模型中唯一的值是您要设置的布尔值< / p>
答案 1 :(得分:0)
我知道了。
我只是用body
代替了params
的{{1}}。
postId
将我喜欢的模型更改为此,我使用的是sourceKey而不是外键
router.post('/like', (req, res)=> {
models.Likes.create({
postId: req.body.postId,
userId: req.user.id,
like:true
}).then( (result) => {
res.status(200).send({
message: 'You have like this post',
like: result
});
}).catch( (err) => {
res.status(401).send({
message: "Something went wrong",
err: err
})
})
})
所以现在我可以喜欢一个帖子,它将把module.exports = function(sequelize, DataTypes) {
const Like = sequelize.define('Likes', {
like:{
type:DataTypes.BOOLEAN,
allowNull:true
},
// userId: {
// type: sequelize.INTEGER,
// references: {
// model: 'Users',
// key: 'id'
// }
// },
}, {});
Like.associate = function(models) {
Like.belongsTo(models.User, {
onDelete: "CASCADE",
foreignKey: 'userId'
})
Like.belongsTo(models.Post, {
onDelete: "CASCADE",
foreignKey: 'likeId'
})
}
return Like;
}
和postId
附加在likes表上。
像这样