Question

我是grails的新手......

我想为我的jquery自动完成

创建一个Controller

def arrSong = Song.executeQuery("select id, title as value, concat(artist, ' - ', title) as label from ${Song.name} where title like concat('%', :paramTitle, '%')", [paramTitle:params.term?.toString()])
render arrSong as JSON

使用此代码我得到这个JSON：

[[1,"Mr. Brightside","The Killers - Mr. Brightside"]]

我的期望是：

[{"id":1,"value":"Mr. Brightside","label":"The Killers - Mr. Brightside"}]

任何人都可以提供帮助吗？

Answer 1

executeQuery你写的方式会返回一个未命名的结果列表，必须使用索引来检索。

例如：

[[1,"Mr. Brightside","The Killers - Mr. Brightside"]]

arrSong.each{println it[0]} 
//Prints 1, similarly it[1] for "Mr. Brightside" and so on

您需要的是结果集的map表示，如下所示，您应该可以毫不费力地将地图渲染为JSON。

def query = """
             select new map(id as id, 
                            title as value, 
                            concat(artist, ' - ', title) as label) 
             from ${Song.name} 
             where title like concat('%', :paramTitle, '%')
            """

def arrSong = Song.executeQuery(query, [paramTitle: params.term?.toString()])

render arrSong as JSON

应该返回

[{"id":1,"value":"Mr. Brightside","label":"The Killers - Mr. Brightside"}]

Grails检索自定义查询到Mapped List

1 个答案: