“在"条款中的异常? [英] "In" clause in anorm?

问题描述

在 anorm 中使用in"子句似乎并不容易:

It seems no easy way to use "in" clause in anorm:

val ids = List("111", "222", "333")
val users = SQL("select * from users where id in ({ids})").on('ids-> ???).as(parser *)

如何替换???部分?

我试过了:

on('ids -> ids)
on('ids -> ids.mkString("'","','","'"))
on('ids -> ids.mkString("','")

但没有一个有效.

我在讨论中看到完全相同的问题:https://groups.google.com/d/topic/play-framework/qls6dhhdayc/discussion，作者有一个复杂的解决方案:

I see in the discussion the exactly same problem: https://groups.google.com/d/topic/play-framework/qls6dhhdayc/discussion, the author has a complex solution:

val params = List(1, 2, 3) 

val paramsList = for ( i <- 0 until params.size ) yield ("userId" + i) 

// ---> results in List("userId0", "userId1", "userId2") 

User.find("id in ({%s})"

    // produces "id in ({userId0},{userId1},{userId2})"
    .format(paramsList.mkString("},{")) 

    // produces Map("userId0" -> 1, "userId1" -> 2, ...) 
    .on(paramsList.zip(params))
    .list() 

这太复杂了.

有没有更简单的方法?还是应该让游戏变得更容易?

Is there any easier way? Or should play provide something to make it easier?

推荐答案

Anorm 现在从 2.3 开始支持这样的情况(以及更多):"使用多值参数"

Anorm now supports such a case (and more) since 2.3: "Using multi-value parameter"

回到它给出的初始示例:

Back to initial example it gives:

val ids = Seq("111", "222", "333")
val users = SQL("select * from users where id in ({ids})").on('ids-> ids).as(parser *)

这篇关于“在"条款中的异常?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！