Question

是否有一种方法告诉gorm仅选择具有hasOne关系（AttractionsData）的行（Attractions）？与在Laravel中使用“ :: has（'relativeName）”相同吗？以下代码将返回包含具有空关系的景点的集合，而不是跳过它们

type Attractions struct {
    ID              uint              `gorm:"primary_key"`
    ...
    Active                int       `json:"-"`
    AttractionsData AttractionsData `gorm:"foreignkey:AttractionID"`
}
type AttractionsData struct {
    ID                    uint      `gorm:"primary_key"`
    AttractionID          uint      `json:"-"`
    Lang                  string    `json:"lang"`
    Title                 string    `json:"title"`
    Active                int       `json:"-"`
}
    ...
func someFunc(c *gin.Context){
    ...
    db := config.DBConnect()
    defer db.Close()    
    var atts []Attractions
    db.Preload("AttractionsData", "lang = ? AND active = ?", lang, 1).
            Where("country_id = ?", countryID).
            Where("category_id = ?", categoryID).
            Where("active = ?", 1).
            Find(&atts) 
    ...

Answer 1

我只会先获取您想要的ID：

ids := []int{}
db.Model(&AttractionData{}).
   Where("lang = ? AND active = ?", lang, 1).
   Pluck("attraction_id", &ids)
db.Preload("AttractionsData").
   Where("id IN (?)", ids).
   Where("category_id = ?", categoryID).
   Where("active = ?", 1).
   Find(&atts)

Answer 2

此代码似乎对我有用。使用LEFT JOIN和WHERE可以过滤掉不需要的景点。尽管希望得到一个更优雅的解决方案。

...
db.Joins("LEFT JOIN attractions_data ON attractions_data.attraction_id = attractions.id").
    Preload("AttractionsData", "lang = ? AND active = ?", lang, 1).
    Where("attractions.country_id = ?", countryID).
    Where("attractions.category_id = ?", categoryID).
    Where("attractions.active = ?", 1).
    Where("attractions_data.active = ? AND attractions_data.lang = ?", 1,lang).
    Find(&atts)

Golang gorm具有（例如在Laravel中）等效的方法？

2 个答案: