我的设置标准如下:
std::unique_ptr
我认为它将返回二维列表。如上所述here。根据我的理解,第一个列表是所选项目的列表,第二个列表是项目列的列表。我的问题是
如何获取所有选定的项目?例如:
def e = Equipment.createCriteria()
def equipmentInstanceList = e.list {
createAlias "rentals", "r", org.hibernate.sql.JoinType.LEFT_OUTER_JOIN
projections {
property('name')
property('r.status')
property('r.dateRented')
property('r.dateReturned')
}
}
如何访问第二个列表中的特定列?例如:
def list = equipmentInstanceList.[all] //supposed to get all the selected items
def a = equipmentInstanceList[1] //will only get the 1st item in 1st list
修改
此查询与上述条件的结果相同。
def a = equipmentInstanceList[1].status //supposed to get the status column in 2nd list
我仍在尝试获取def equipmentForRent = Equipment.executeQuery("SELECT e.name, r.status, r.dateRented, r.dateReturned FROM ers.Equipment e LEFT JOIN e.rentals r")
中的列。到目前为止我在尝试什么。
equipmentForRent
仍然出现此错误:
each.equipmentForRent { e -> println e.dateReturned }
为什么它与此查询不同:
Exception evaluating property 'dateRented' for java.util.Arrays$ArrayList, Reason: groovy.lang.MissingPropertyException: No such property: dateRented for class: java.lang.String
任何?
答案 0 :(得分:1)
您在一个问题中提出了两个问题: 很快:
如何获取所有选定的项目?例如:
如何访问第二个列表中的特定列?例如:
def a = equipmentInstanceList[1].status //supposed to get the status column in 2nd list
获得第二名:
def a = equipmentInstanceList[2].status // this now points to 2nd element status
但这些都不是安全的,可能导致问题。 迭代具有索引的元素:
equipmentInstanceList?.eachWithIndex{ e,i->
println "-- $i is index $e is element"
}
您的查询已尝试在HQL中执行sql查询
def equipmentForRent = Equipment.executeQuery("SELECT e.name, r.status, r.dateRented, r.dateReturned FROM ers.Equipment e LEFT JOIN e.rentals r")
尝试从表格中选择新的地图(项目),如图所示
def equipmentForRent = Equipment.executeQuery("SELECT new map(e.name as name, r.status as status, r.dateRented as dateRented, r.dateReturned as dateReturned) FROM Equipment e LEFT JOIN e.rentals r")