Aar*_*tya 6 mysql database node.js typeorm
这是我的用户实体:
@PrimaryGeneratedColumn()
userId: number;
@Column({type:"varchar", length:"300"})
userName: string;
@OneToOne(type => UserProfile, {cascadeAll:true})
@JoinColumn()
userProfile: UserProfile;
@OneToOne(type => UserCredential, {cascadeAll:true, eager:true})
@JoinColumn()
userCredential: UserCredential;
@OneToOne(type => BusinessUnit, {cascadeAll:true})
@JoinColumn()
businessUnit: BusinessUnit;
@ManyToMany(type => ProductCategory)
@JoinTable()
productCategory: ProductCategory[];
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这是我要更新的新数据:
User {
userName: 'qweret@gmail.com',
userProfile:
UserProfile {
firstName: 'dcds',
lastName: 'Faiz',
mobile: '42423423',
addressLine1: 'Delhi',
addressLine2: 'Delhi',
city: '-',
country: '-',
zipCode: '234243',
homeTelephone: '-',
dayOfBirth: 0,
monthOfBirth: 0,
yearOfBirth: 0 },
userCredential: UserCredential { credential: 'abcd@123' } }
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我正在通过其userId搜索用户。
return await getManager()
.createQueryBuilder(User, "user")
.where("user.userId = :id", {id})
.getOne();
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上面的查询给了我结果:
User {
createdDate: 2018-03-29T06:45:16.322Z,
updatedDate: 2018-04-01T06:28:24.171Z,
userId: 1,
userName: 'qweret@gmail.com' }
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我想更新我的用户表及其相关表
manager.save(user)
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将在表格中插入新行,而不是更新现有行。有没有一种方法可以更新整个用户表及其相关表,而无需手动更新每一列?我不想执行这样的任务:
let user = await userRepositiory.findOneById(1);
user.userName = "Me, my friends and polar bears";
await userRepository.save(user);
let userProfile = await userProfileRepository.findOneById(1);
userProfile.firstName = "";
userProfile.lastName = "";
....// etc
await userRepository.save(userProfile);
// and so on update other tables.
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小智 5
试试这个方法:
await this. userRepository.update(
user.id ,
user
);
const updatedUser = await this.repository.findOne(user.id);
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