Grails 使用findAllBy方法访问对象属性列表时出现的问题

下面是获取由params生成并从控制器传递到服务中的标记列表的代码。我正在尝试筛选餐馆列表，以仅包含具有匹配标记属性的餐馆。我不熟悉编程和grails，所以请原谅我犯的任何愚蠢的错误：

    class Eatery {
        String name
        String phone
        Date lastVisited
        List<Tag> tags
        static hasOne = [location: Location]
        static hasMany = [tags: Tag];
        static constraints = {
            name nullable: false, blank: false
            phone nullable: true, blank: true
            location nullable: true, blank: true
            tags nullable: true, blank: true
            lastVisited nullable: true, blank: true
        }
        public String toString() {
            return name
        }
}

以下是服务方法：

Eatery getRandomEatery(List<Tag> tagList) {
    List<Eatery> eateryList = Eatery.findAllByTags(tagList)
    Random random = new Random()
    Integer n = eateryList.size()
    Integer selection = Math.abs(random.nextInt(n))
    Eatery eatery = eateryList.get(selection)
    return eatery
}

下面是错误：

 Class
org.h2.jdbc.JdbcSQLException
Message
Parameter "#1" is not set; SQL statement: select this_.id as id2_0_, this_.version as version2_0_, this_.last_visited as last3_2_0_, this_.name as name2_0_, this_.phone as phone2_0_ from eatery this_ where this_.id=? [90012-164]
Around line 16 of grails-app/services/grubspot/RandomizerService.groovy
13://14://        }15:        List<Eatery> eateryList = Eatery.findAllByTags(tagList)16:        Random random = new Random()17:        Integer n = eateryList.size()18:        Integer selection = Math.abs(random.nextInt(n))19:        Eatery eatery = eateryList.get(selection)

改变

到

---

我想你可以用一个类似于

List<Eatery> eateryList = Eatery.withCriteria{
    tags {
        "in" "id", tagList.id
    }
}

还没有测试过，但应该给你一个想法

List<Eatery eateryList = Eatery.executeQuery("""
select e
from Eatery e left join e.tags as t
where t in (:tagList)""", [tagList: tagList])

List<Eatery> eateryList = Eatery.withCriteria{
    tags {
        "in" "id", tagList.id
    }
}