How do you do UUID in Golangs Gorm?

Question

I have the following model...

type User struct {
    ID        string  `sql:"type:uuid;primary_key;default:uuid_generate_v4()"`
    FirstName string `form:"first_name" json:"first_name,omitempty"`
    LastName  string `form:"last_name" json:"last_name,omitempty"`
    Password  string `form:"password" json:"password" bindind:"required"`
    Email     string `gorm:"type:varchar(110);unique_index" form:"email" json:"email,omitempty" binding:"required"`
    Location  string `form:"location" json:"location,omitempty"`
    Avatar    string `form:"avatar" json:"avatar,omitempty"`
    BgImg     string `form:"bg_img" json:"bg_img,omitempty"`
    CreatedAt time.Time
    UpdatedAt time.Time
    DeletedAt time.Time
}

I've tried several different ways, but this way throws (pq: relation "users" does not exist). I have no related models, it's literally just that one model.

I've tried using...

func (user *User) BeforeCreate(scope *gorm.Scope) error {
    scope.SetColumn("ID", uuid.NewV4())
    return nil
}

Along with a uuid lib, but had no luck with that either.

That seems that your users table is not exist on the database. Have you automigrate it? — Apin, Apr 08 '16 at 01:15

Eric · Answer 1 · 2022-09-22T09:07:41.083

For postgresql, here is what I did:

go get github.com/google/uuid

Use uuid.UUID (from "github.com/google/uuid"), as type,
e.g

ID       uuid.UUID `gorm:"type:uuid;default:uuid_generate_v4()"`

Add uuid-ossp extension for postgres database,
e.g
```
CREATE EXTENSION IF NOT EXISTS "uuid-ossp";
```
Then, when you call DB's Create() method, the uuid is generated automatically.

Update: pg14+ `gen_random_uuid()`

(as mentioned in Doron Segal's comment)

pg 14 has built-in function gen_random_uuid() to generate uuid v4, e.g:

create table:

create table uuid_test (uid text default gen_random_uuid());

insert a row:

insert into uuid_test(uid) values (DEFAULT);

Then uid column is generated automatically.

Similiar, in go you can use the function as defaul value I think, e.g:

ID uuid.UUID gorm:"type:uuid;default:gen_random_uuid()"

BTW, the gen_random_uuid() function only support uuid v4 now, to use other versions, you still need uuid-ossp extension.

You don't need to use any extension if you're using Postgresql version > 14. just use gen_random_uuid() — Doron Segal, Sep 19 '22 at 06:22

score 13 · Answer 2 · answered Apr 10 '16 at 17:36

13

Turns out I was trying to store the UUID as the wrong type, I was doing...

func (user *User) BeforeCreate(scope *gorm.Scope) error {
    scope.SetColumn("ID", uuid.NewV4())
    return nil
}

When it needed to be...

func (user *User) BeforeCreate(scope *gorm.Scope) error {
    scope.SetColumn("ID", uuid.NewV4().String())
    return nil
}

answered Apr 10 '16 at 17:36

Ewan Valentine

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5

You should use`return scope.SetColumn("ID", uuid.NewV4().String())` instead of ignoring the result for `SetColumn` – Artmann Feb 10 '17 at 14:23

tim-montague · Answer 3 · 2023-05-06T02:11:39.137

7

This was my solution for Gorm v1.21

go get gorm.io/gorm
go get gorm.io/driver/postgres
go get github.com/google/uuid

import (
  "gorm.io/gorm"
  "github.com/google/uuid"
)

type User struct {
  Id string `gorm:"primaryKey"`
}

// Note: Gorm will fail if the function signature
//  does not include `*gorm.DB` and `error`

func (user *User) BeforeCreate(tx *gorm.DB) (err error) {
  // UUID version 4
  user.Id = uuid.NewString()
  return
}

Notes:

For the Google UUID package, the methods uuid.New() and uuid.NewString() use UUID version 4. This is not clearly stated in the documentation (http://pkg.go.dev/github.com/google/uuid), but by looking into the source code, you can see that these are wrappers around uuid.NewRandom() which is stated as being UUID version 4.
While some recommend the Satori UUID package (https://github.com/satori/go.uuid), benchmarks show that it has 3.3x lower performance than the Google UUID package (https://gist.github.com/mattes/69a4ab7027b9e8ee952b5843e7ca6955)

edited May 06 '23 at 02:11

answered Jul 13 '21 at 23:35

tim-montague

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It should be `uuid.New()` which works. Please update your answer. – HalfWebDev Aug 28 '21 at 08:28
@HalfWebDev To my understanding `uuid.New()` returns a 16-byte array, and not a string. If you have a convincing argument as to why using `uuid.New()` is better than `uuid.NewString()`, then I'm happy to update the example code. As I see it, `uuid.NewString()` is better because strings are readable, and well supported across different databases. – tim-montague Aug 29 '21 at 10:09
Umm, ok. From what I tried `NewString()` is not a method at all on `uuid` – HalfWebDev Sep 02 '21 at 18:14
@HalfWebDev `NewString` was added in v1.2.0 (https://pkg.go.dev/github.com/google/uuid#NewString). Does updating the package with `go get -u github.com/google/uuid` include the method? – tim-montague Sep 02 '21 at 20:59
Right, I thought I had latest package since I just worked on it. I am on `1.0.0` as it turns out. Will check for latest release – HalfWebDev Sep 03 '21 at 05:56

Peter Mghendi · Answer 4 · 2021-06-17T12:02:01.360

For this you will need gorm and go.uuid

go get github.com/jinzhu/gorm

go get github.com/satori/go.uuid

Try creating your own model base model in place of gorm.Model like so:

type Base struct {
 ID         string     `sql:"type:uuid;primary_key;default:uuid_generate_v4()"`
 CreatedAt  time.Time  `json:"created_at"`
 UpdatedAt  time.Time  `json:"updated_at"`
 DeletedAt  *time.Time `sql:"index" json:"deleted_at"`
}

You would then populate this field using a method called before creation of any record, like so:

func (base *Base) BeforeCreate(scope *gorm.Scope) error {
 id, err := uuid.NewV4()
 if err != nil {
   return err
 }
 return scope.SetColumn("ID", uuid.String())
}

Therefore, for your particular case, you would have:

type User struct {
    Base
    FirstName string `form:"first_name" json:"first_name,omitempty"`
    LastName  string `form:"last_name" json:"last_name,omitempty"`
    Password  string `form:"password" json:"password" bindind:"required"`
    Email     string `gorm:"type:varchar(110);unique_index" form:"email" json:"email,omitempty" binding:"required"`
    Location  string `form:"location" json:"location,omitempty"`
    Avatar    string `form:"avatar" json:"avatar,omitempty"`
    BgImg     string `form:"bg_img" json:"bg_img,omitempty"`
}

More details on this can be found here

Wouldn't `default:uuid_generate_v4()` automatically generate an UUID for us? — Carlos Martinez, Oct 12 '19 at 03:31
This certainly returns an error `uuid, err := uuid.NewV4().String()` — stelios, Nov 27 '20 at 09:36
Put the .String() in this line `return scope.SetColumn("ID", uuid.String())` — Thor, Dec 03 '20 at 11:33
This worked at the time of writing. I haven't followed up on it, nor I actively use Go in my current capacity, but I will make the requested changes. — Peter Mghendi, Dec 04 '20 at 12:15

score 1 · Answer 5 · answered Apr 08 '16 at 01:51

The error (pq: relation "users" does not exist) usually means that, the table users does not exists in the database. It has nothing to do with the relationship between two models.

So basically, You first need to create the table in the database (Or auto migrate the database As per @Apin suggest). And try to re-run the same code.

Justin G · Answer 6 · 2021-10-12T06:39:55.223

None of these worked for me using gorm v1.21. Here was my solution. Note that I'm using the satori/go.uuid library for generating UUID, but code with google's library is near identical.

type UUIDBaseModel struct {
    ID        uuid.UUID       `gorm:"primary_key" json:"id"`
    CreatedAt time.Time  `json:"created_at"`
    UpdatedAt time.Time  `json:"updated_at"`
    DeletedAt *time.Time `sql:"index" json:"deleted_at"`
}

func (base *UUIDBaseModel) BeforeCreate(tx *gorm.DB) error {
    uuid := uuid.NewV4().String()
    tx.Statement.SetColumn("ID", uuid)
    return nil
}

score 0 · Answer 7 · edited Aug 10 '22 at 18:06

0

Now I used Gorm 2.0 and this worked:

go get github.com/satori/go.uuid

type Tablename struct {
 ID         string     `sql:"type:uuid;primary_key;default:uuid_generate_v4()"`
 }

edited Aug 10 '22 at 18:06

mustaccio

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answered Aug 05 '22 at 03:24

dev-nnb

1
1

score 0 · Answer 8 · answered May 26 '23 at 14:52

With MySQL 8.0 and gorm (gorm.io/driver/mysql v1.5.0 and gorm.io/gorm v1.25.1):

type ModelUUID struct {
    ID        string         `gorm:"type:binary(16);column:id;primaryKey;default:(UUID_TO_BIN(UUID(), 1));" json:"id" validate:"required"`
    CreatedAt time.Time      `json:"createdAt"`
    UpdatedAt time.Time      `json:"-"`
    DeletedAt gorm.DeletedAt `json:"-" gorm:"index" swaggerignore:"true"`
}

How do you do UUID in Golangs Gorm?

8 Answers8

Update: pg14+ `gen_random_uuid()`

Linked

How do you do UUID in Golangs Gorm?

8 Answers8

Update: pg14+ gen_random_uuid()

Linked

Update: pg14+ `gen_random_uuid()`