带有Anorm和Scala Play框架的动态SQL参数 [英] Dynamic SQL Parameters with Anorm and Scala Play Framework

问题描述

是否可以为主动脉的"on"方法动态创建一个列表?

Is it possible to dynamically create a list for anorm's "on" method?

我有一个带有可选输入的表单，当前我检查每个Option并使用定义的Options创建一个列表，并试图将其传递给主动脉.目前，我收到此编译错误

I have a form with optional inputs and currently I check each Option and create a list with the defined Options and am trying to pass this through to anorm. Currently I get this compilation error

type mismatch; found : List[java.io.Serializable] required: (Any, anorm.ParameterValue[_])

我不确定如何创建此列表. 当前代码:

I'm not sure how I would go about creating this list. Current code :

val onList = List(
        'school_id = input.school,
        if(input.rooms isDefined)       ('rooms -> input.rooms) else "None" ,
        if(input.bathrooms isDefined)   ('bathrooms -> input.bathrooms) else "None" ,
        if(input.houseType isDefined)   ('houseType -> input.houseType) else "None" ,
        if(input.priceLow isDefined)    ('priceLow -> input.priceLow) else "None" ,
        if(input.priceHigh isDefined)   ('priceHigh -> input.priceHigh) else "None" ,
        if(input.utilities isDefined)   ('utilities -> input.utilities) else "None" 
).filter(_!="None")
SQL("SELECT * FROM Houses WHERE " + whereString).on(onList).as(sqlToHouse *)

我尝试这样做是因为最初我以为会和

I've tried doing this because initially I thought it would be the same as

.on('rooms -> input.rooms, 'bathroom -> input.bathrooms... etc)

代码现在为:

val onList = Seq(
        ('school_id -> input.school),
        if(input.rooms isDefined)       ('rooms -> input.rooms.get)         else None ,
        if(input.bathrooms isDefined)   ('bathrooms -> input.bathrooms.get) else None ,
        if(input.houseType isDefined)   ('houseType -> input.houseType.get) else None ,
        if(input.priceLow isDefined)    ('priceLow -> input.priceLow.get)   else None ,
        if(input.priceHigh isDefined)   ('priceHigh -> input.priceHigh.get) else None ,
        if(input.utilities isDefined)   ('utilities -> input.utilities.get) else None 
).filter(_!=None).asInstanceOf[Seq[(Any,anorm.ParameterValue[_])]]

使用SQL命令:

SQL("SELECT * FROM Houses WHERE " + whereString).on(onList:_*).as(sqlToHouse *)

现在获取异常

[ClassCastException: java.lang.Integer cannot be cast to anorm.ParameterValue]

推荐答案

重要的是，您必须创建类型为ParameterValue的值. 通常使用toParameterValue()函数来完成此操作.

The important thing is that you have to create values of type ParameterValue. This is normally done using the toParameterValue() function.

一种方法是创建要展平的选项序列:

One way would be to create a sequence of Options that you flatten:

val onList = Seq(
  Some('school_id -> input.school),
  input.rooms.map('rooms -> _),
  input.bathrooms.map('bathrooms -> _)
).flatten

然后可以将此序列映射为正确的值:

This sequence can then be mapped to correct values:

SQL(
  "SELECT * FROM Houses WHERE " + whereString
).on(
  onList.map(v => v._1 -> toParameterValue(v._2)): _*
)

---

可以这样简化:

---

This can be simplified like this:

val onList = Seq(
  Some('school_id -> input.school),
  input.rooms.map('rooms -> _),
  input.bathrooms.map('bathrooms -> _)
).flatMap(_.map(v => v._1 -> toParameterValue(v._2)))

SQL(
  "SELECT * FROM Houses WHERE " + whereString
).on(
  onList: _*
)

---

或者也许最简单的解决方案是这样:

---

Or maybe the simplest solution would be this:

val onList = Seq(
  Some('school_id -> toParameterValue(input.school)),
  input.rooms.map('rooms -> toParameterValue(_)),
  input.bathrooms.map('bathrooms -> toParameterValue(_))
).flatten

SQL(
  "SELECT * FROM Houses WHERE " + whereString
).on(
  onList: _*
)

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