我正在使用Ebean构建Play2应用程序。我创建了一个服务类,其中包含通过 id 列表获取场地的方法:
public static List<Venue> getVenuesForIds(List<Long> list){
ArrayList<Venue> venues = new ArrayList<Venue>();
String sql = "select c.id, c.name from Company c where in (:ids)";
List<SqlRow> sqlRows =
Ebean.createSqlQuery(sql).setParameter("ids", list).findList();
for(SqlRow row : sqlRows) {
venues.add(new Venue(row.getLong("id"), row.getString("name")));
}
return venues;
}
但我得到:
[PersistenceException: Query threw SQLException:You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'in (201639091,201637666)' at line 1 Query was: select c.id, c.name from Company c where in (?,?) ]
我已经通读了 http://www.avaje.org/ebean/introquery.html 但可能错过了正确的语法。我想在原始 sql 中执行此操作。我错过了什么?
您的请求似乎不正确。
怎么样:
"select c.id, c.name from Company c where c.id in (:ids)";
你不需要
执行这种"复杂"的查询,如果你在Venue
模型中使用通用Finder<I,T>
(一次),那就足够了:
@Entity
@Table(name = "Company")
public class Venue extends Model {
@Id
public Long id;
public String name;
// other fields
public static Finder<Long, Venue> find
= new Finder<Long, Venue>(Long.class, Venue.class);
}
因此,您可以使用...方法中的单行代码:
public static List<Venue> getVenuesForIds(List<Long> ids){
return Venue.find.select("id,name").where().idIn(ids).findList();
}
或类似的表示:
public static List<Venue> getVenuesForIds(List<Long> ids){
return Venue.find.select("id,name").where().in("id",ids).findList();
}