以形式捕获持久关系 [英] Capturing Persistent Relations in a Form

问题描述

我在Persistent中定义了一对多关系，但无法弄清楚如何创建一个可以将其中一个外键作为输入的表单。简化我的用例，如下所示：

  Person 
 name String 
 
 Car 
 personId PersonId 
 name Text 
 type Text 
  

现在当我尝试要生成一个Form for Car，personId的字段类型应该是什么？我试过类似的东西，但得到一个错误：

  entryForm :: Maybe Car  - > Form Car 
 entryForm car = renderDivs $ Car 
< $> areq personIdFieldPersonNothing 
 *< *> areq textFieldCar Name（carName< $> car）
< areq textFieldType（carType< $> car）
  

当我运行上述时，错误：不在范围内：`personIdField'。


我试过 intField ，它说：

 无法匹配预期的类型`KeyBackend 
 persistent-1.2.1：Database.Persist.Sql.Types.SqlBackend Person'
与实际类型`文本'
预期类型：字段
 m0 
（KeyBackend 
 persistent-1.2.1：Database.Persist.Sql.Types.SqlBackend Person）
实际类型：字段m0文本
在'areq'的第一个参数中，即'intField'
在'（< $>）'的第二个参数中，即
 `areq intFieldPersonNothing 
  

理想情况下，我想填充一个人名下拉列表（if没有太多），或者有太多的时候有自由形式的文本字段（例如，自动完成）。关于如何将外键作为用户输入的建议？

更新：

我试过使用selectField，但不确定是否我我正确地做这件事。我仍然遇到错误。首先，我创建了一个where语句来获取personId：

 其中
 personId = do 
 person<  - runDB $ selectFirst [] [Asc PersonName] 
 
的个案人Just（Entity pId p） - >返回pId 
  - 无 - > ??? 
  

然后我将第一个areq改为

 < $> areq（selectField [（First，personId）]）Person NameNothing 
  

谢谢！ / p>

解决方案

我能弄清楚如何正确使用selectField。这是我最终做的：

 其中
 people = do 
实体<  -  runDB $ selectList [] [Asc PersonName] 
 optionsPairs $ map（\ p  - >（personName $ entityVal p，entityKey p））entities 
  

表单字段变为：

 < $> areq（selectField people）Person NameNothing 
  

我还没有想出自由表单条目但这是一个好的开始。

I have defined a one-to-many relationship in Persistent but could not figure out how to create a form that can take one of the foreign keys as input. Simplifying my use case to something like this:

Person
  name String

Car
  personId PersonId 
  name Text
  type Text

Now when I try to generate a Form for Car, what should be the field type for personId? I tried something like this but get an error:

entryForm :: Maybe Car -> Form Car
entryForm car = renderDivs $ Car
    <$> areq   personIdField   "Person" Nothing
    <*> areq   textField   "Car Name" ( carName <$> car)
    <*> areq   textField  "Type" ( carType <$> car)

When I run the above I get the error: Not in scope: `personIdField'.

I tried intField and it says:

Couldn't match expected type `KeyBackend
                                persistent-1.2.1:Database.Persist.Sql.Types.SqlBackend Person'
            with actual type `Text'
Expected type: Field
                 m0
                 (KeyBackend
                    persistent-1.2.1:Database.Persist.Sql.Types.SqlBackend Person)
  Actual type: Field m0 Text
In the first argument of `areq', namely `intField'
In the second argument of `(<$>)', namely
  `areq intField "Person" Nothing

Ideally I would like to populate a drop down of Person Names (if there are not too many) or have a free form text field (e.g., with autocomplete) when there are too many. Any suggestions on how I can get foreign key as an input from the user?

Update:

I tried using selectField as follows but not sure if I am doing this correctly. I still get an error. First I created a where statement to get personId:

 where
   personId = do  
       person <- runDB $ selectFirst [] [Asc PersonName] 
       case person of
           Just (Entity pId p) -> return pId
           -- Nothing             -> ???

and then I changed my first areq to

 <$> areq (selectField [("First", personId)]) "Person Name" Nothing

Thanks!

解决方案

I was able to figure out how to use selectField properly. This is what I ended up doing:

where
    people = do
        entities <- runDB $ selectList [] [Asc PersonName]
        optionsPairs $ map (\p -> (personName $ entityVal p, entityKey p)) entities

The form field became:

<$> areq   (selectField people) "Person Name" Nothing

I still have not figured out free form entry just yet but this is a good start.

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