typeorm: saving an entity with many-to-many relationship ids
0
Is it possible to save an entity with many-to-many relation ids?
suppose I have following Project Entity with many-to-many relationship to userGroups table.
@Entity()
export class Project extends BaseEntity {
@Column({ type: 'varchar', length: 255 })
name: string
@RelationId((project: Project) => project.userGroups)
userGroupIds: number
@ManyToMany(type => UserGroup, userGroup => userGroup.projects)
@JoinTable()
userGroups: UserGroup
}
Since ids of the userGroups table are mapped to userGroupIds property of the Project class via @RelationId decorator, I thought I could save a new Project entity with userGroupIds like this:
let prj = new Project()
prj.name = 'foo'
prj.userGroupIds = [1, 2, 3]
prj.save()
but the above code only creates a project record... (no record is created on project - userGroups many-to-many relation table)
typescript typeorm
asked Nov 22 '18 at 8:31
sorasora
212311
add a comment |
0
Is it possible to save an entity with many-to-many relation ids?
suppose I have following Project Entity with many-to-many relationship to userGroups table.
@Entity()
export class Project extends BaseEntity {
@Column({ type: 'varchar', length: 255 })
name: string
@RelationId((project: Project) => project.userGroups)
userGroupIds: number
@ManyToMany(type => UserGroup, userGroup => userGroup.projects)
@JoinTable()
userGroups: UserGroup
}
Since ids of the userGroups table are mapped to userGroupIds property of the Project class via @RelationId decorator, I thought I could save a new Project entity with userGroupIds like this:
let prj = new Project()
prj.name = 'foo'
prj.userGroupIds = [1, 2, 3]
prj.save()
but the above code only creates a project record... (no record is created on project - userGroups many-to-many relation table)
typescript typeorm
asked Nov 22 '18 at 8:31
sorasora
212311
1
I think you should create the usergroup first, and add the created usergroup to project.
– Steve Ruben
Nov 22 '18 at 8:35
Thank you for your comment. You mean if i have 3 existing records(id=1,2,3) on user groups table, we need to create 3 instances of UserGroup entities from those records and add them to project entity?
– sora
Nov 22 '18 at 8:42
1
yes, you can do like that.
– Steve Ruben
Nov 22 '18 at 8:49
I see. Thanks! Would be better if we could simply pass ids tho...
– sora
Nov 22 '18 at 8:53
nope, because you don't always know the id, in test yes but in production you don't know
– Steve Ruben
Nov 22 '18 at 8:57
add a comment |
0
0
0
Is it possible to save an entity with many-to-many relation ids?
suppose I have following Project Entity with many-to-many relationship to userGroups table.
@Entity()
export class Project extends BaseEntity {
@Column({ type: 'varchar', length: 255 })
name: string
@RelationId((project: Project) => project.userGroups)
userGroupIds: number
@ManyToMany(type => UserGroup, userGroup => userGroup.projects)
@JoinTable()
userGroups: UserGroup
}
Since ids of the userGroups table are mapped to userGroupIds property of the Project class via @RelationId decorator, I thought I could save a new Project entity with userGroupIds like this:
let prj = new Project()
prj.name = 'foo'
prj.userGroupIds = [1, 2, 3]
prj.save()
but the above code only creates a project record... (no record is created on project - userGroups many-to-many relation table)
typescript typeorm
asked Nov 22 '18 at 8:31
sorasora
212311
Is it possible to save an entity with many-to-many relation ids?
suppose I have following Project Entity with many-to-many relationship to userGroups table.
@Entity()
export class Project extends BaseEntity {
@Column({ type: 'varchar', length: 255 })
name: string
@RelationId((project: Project) => project.userGroups)
userGroupIds: number
@ManyToMany(type => UserGroup, userGroup => userGroup.projects)
@JoinTable()
userGroups: UserGroup
}
Since ids of the userGroups table are mapped to userGroupIds property of the Project class via @RelationId decorator, I thought I could save a new Project entity with userGroupIds like this:
let prj = new Project()
prj.name = 'foo'
prj.userGroupIds = [1, 2, 3]
prj.save()
but the above code only creates a project record... (no record is created on project - userGroups many-to-many relation table)
typescript typeorm
typescript typeorm
asked Nov 22 '18 at 8:31
sorasora
212311
asked Nov 22 '18 at 8:31
sorasora
212311
asked Nov 22 '18 at 8:31
sorasora
212311
asked Nov 22 '18 at 8:31
sorasora
212311
asked Nov 22 '18 at 8:31
sorasora
212311
212311
1
I think you should create the usergroup first, and add the created usergroup to project.
– Steve Ruben
Nov 22 '18 at 8:35
Thank you for your comment. You mean if i have 3 existing records(id=1,2,3) on user groups table, we need to create 3 instances of UserGroup entities from those records and add them to project entity?
– sora
Nov 22 '18 at 8:42
1
yes, you can do like that.
– Steve Ruben
Nov 22 '18 at 8:49
I see. Thanks! Would be better if we could simply pass ids tho...
– sora
Nov 22 '18 at 8:53
nope, because you don't always know the id, in test yes but in production you don't know
– Steve Ruben
Nov 22 '18 at 8:57
add a comment |
1
I think you should create the usergroup first, and add the created usergroup to project.
– Steve Ruben
Nov 22 '18 at 8:35
Thank you for your comment. You mean if i have 3 existing records(id=1,2,3) on user groups table, we need to create 3 instances of UserGroup entities from those records and add them to project entity?
– sora
Nov 22 '18 at 8:42
1
yes, you can do like that.
– Steve Ruben
Nov 22 '18 at 8:49
I see. Thanks! Would be better if we could simply pass ids tho...
– sora
Nov 22 '18 at 8:53
nope, because you don't always know the id, in test yes but in production you don't know
– Steve Ruben
Nov 22 '18 at 8:57
1
1
I think you should create the usergroup first, and add the created usergroup to project.
– Steve Ruben
Nov 22 '18 at 8:35
I think you should create the usergroup first, and add the created usergroup to project.
– Steve Ruben
Nov 22 '18 at 8:35
Thank you for your comment. You mean if i have 3 existing records(id=1,2,3) on user groups table, we need to create 3 instances of UserGroup entities from those records and add them to project entity?
– sora
Nov 22 '18 at 8:42
Thank you for your comment. You mean if i have 3 existing records(id=1,2,3) on user groups table, we need to create 3 instances of UserGroup entities from those records and add them to project entity?
– sora
Nov 22 '18 at 8:42
1
1
yes, you can do like that.
– Steve Ruben
Nov 22 '18 at 8:49
yes, you can do like that.
– Steve Ruben
Nov 22 '18 at 8:49
I see. Thanks! Would be better if we could simply pass ids tho...
– sora
Nov 22 '18 at 8:53
I see. Thanks! Would be better if we could simply pass ids tho...
– sora
Nov 22 '18 at 8:53
nope, because you don't always know the id, in test yes but in production you don't know
– Steve Ruben
Nov 22 '18 at 8:57
nope, because you don't always know the id, in test yes but in production you don't know
– Steve Ruben
Nov 22 '18 at 8:57
add a comment |
1 Answer
1
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oldest
votes
1
suppose we
@Entity()
export class Project extends BaseEntity {
@Column({ type: 'varchar', length: 255 })
name: string
@RelationId((project: Project) => project.userGroups)
userGroupIds: number
@ManyToMany(type => UserGroup, userGroup => userGroup.projects)
@JoinTable()
userGroups: UserGroup
}
to persist, i can
let prj = new Project()
prj.name = 'foo'
userGroup a = // create a user group here
prj.userGroupIds.add(a)
prj.save()
after persist, you will see a record in the table project - userGroups
|project| |projet-usergroup| |usergroup|
--------- ------------------ -----------
|id | |idproj|idusergrp| |idusergrp|
|1 | |1 | 1 | | 1 |
after all you will have something like that
answered Nov 22 '18 at 8:56
Steve RubenSteve Ruben
322410
add a comment |
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suppose we
@Entity()
export class Project extends BaseEntity {
@Column({ type: 'varchar', length: 255 })
name: string
@RelationId((project: Project) => project.userGroups)
userGroupIds: number
@ManyToMany(type => UserGroup, userGroup => userGroup.projects)
@JoinTable()
userGroups: UserGroup
}
to persist, i can
let prj = new Project()
prj.name = 'foo'
userGroup a = // create a user group here
prj.userGroupIds.add(a)
prj.save()
after persist, you will see a record in the table project - userGroups
|project| |projet-usergroup| |usergroup|
--------- ------------------ -----------
|id | |idproj|idusergrp| |idusergrp|
|1 | |1 | 1 | | 1 |
after all you will have something like that
answered Nov 22 '18 at 8:56
Steve RubenSteve Ruben
322410
add a comment |
1
suppose we
@Entity()
export class Project extends BaseEntity {
@Column({ type: 'varchar', length: 255 })
name: string
@RelationId((project: Project) => project.userGroups)
userGroupIds: number
@ManyToMany(type => UserGroup, userGroup => userGroup.projects)
@JoinTable()
userGroups: UserGroup
}
to persist, i can
let prj = new Project()
prj.name = 'foo'
userGroup a = // create a user group here
prj.userGroupIds.add(a)
prj.save()
after persist, you will see a record in the table project - userGroups
|project| |projet-usergroup| |usergroup|
--------- ------------------ -----------
|id | |idproj|idusergrp| |idusergrp|
|1 | |1 | 1 | | 1 |
after all you will have something like that
answered Nov 22 '18 at 8:56
Steve RubenSteve Ruben
322410
add a comment |
1
1
1
suppose we
@Entity()
export class Project extends BaseEntity {
@Column({ type: 'varchar', length: 255 })
name: string
@RelationId((project: Project) => project.userGroups)
userGroupIds: number
@ManyToMany(type => UserGroup, userGroup => userGroup.projects)
@JoinTable()
userGroups: UserGroup
}
to persist, i can
let prj = new Project()
prj.name = 'foo'
userGroup a = // create a user group here
prj.userGroupIds.add(a)
prj.save()
after persist, you will see a record in the table project - userGroups
|project| |projet-usergroup| |usergroup|
--------- ------------------ -----------
|id | |idproj|idusergrp| |idusergrp|
|1 | |1 | 1 | | 1 |
after all you will have something like that
answered Nov 22 '18 at 8:56
Steve RubenSteve Ruben
322410
suppose we
@Entity()
export class Project extends BaseEntity {
@Column({ type: 'varchar', length: 255 })
name: string
@RelationId((project: Project) => project.userGroups)
userGroupIds: number
@ManyToMany(type => UserGroup, userGroup => userGroup.projects)
@JoinTable()
userGroups: UserGroup
}
to persist, i can
let prj = new Project()
prj.name = 'foo'
userGroup a = // create a user group here
prj.userGroupIds.add(a)
prj.save()
after persist, you will see a record in the table project - userGroups
|project| |projet-usergroup| |usergroup|
--------- ------------------ -----------
|id | |idproj|idusergrp| |idusergrp|
|1 | |1 | 1 | | 1 |
after all you will have something like that
answered Nov 22 '18 at 8:56
Steve RubenSteve Ruben
322410
answered Nov 22 '18 at 8:56
Steve RubenSteve Ruben
322410
answered Nov 22 '18 at 8:56
Steve RubenSteve Ruben
322410
answered Nov 22 '18 at 8:56
Steve RubenSteve Ruben
322410
322410
add a comment |
add a comment |
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1
I think you should create the usergroup first, and add the created usergroup to project.
– Steve Ruben
Nov 22 '18 at 8:35
Thank you for your comment. You mean if i have 3 existing records(id=1,2,3) on user groups table, we need to create 3 instances of UserGroup entities from those records and add them to project entity?
– sora
Nov 22 '18 at 8:42
1
yes, you can do like that.
– Steve Ruben
Nov 22 '18 at 8:49
I see. Thanks! Would be better if we could simply pass ids tho...
– sora
Nov 22 '18 at 8:53
nope, because you don't always know the id, in test yes but in production you don't know
– Steve Ruben
Nov 22 '18 at 8:57