如何提交包含Salat&amp ;;的其他模型列表的模型表单?玩框架?

时间:2014-08-19 14:26:07

标签: scala playframework playframework-2.2 salat

我有一个模特。它包含另一个模型的列表:

case class Account(
  _id: ObjectId = new ObjectId,
  name: String,
  campaigns: List[Campaign]
)

case class Campaign(
  _id: ObjectId = new ObjectId,
  name: String
)

我有一个用于显示和创建新帐户的表单和操作:

  val accountForm = Form(
    mapping(
      "id" -> ignored(new ObjectId),
      "name" -> nonEmptyText,
      "campaigns" -> list(
        mapping(
          "id" -> ignored(new ObjectId),
          "name" -> nonEmptyText
        )(Campaign.apply)(Campaign.unapply)
      )
    )(Account.apply)(Account.unapply)
  )

  def accounts = Action {
    Ok(views.html.accounts(AccountObject.all(), accountForm, CampaignObject.all()))
  }

  def newAccount = Action {
    implicit request =>    
    accountForm.bindFromRequest.fold(
      formWithErrors => BadRequest(views.html.accounts(AccountObject.all(), formWithErrors, CampaignObject.all())),
      account => {
        AccountObject.create(account)
        Redirect(routes.AccountController.accounts)
      }
    )
  }

最后,这是我对账户的看法:

@(accounts: List[models.mongodb.Account], account_form: Form[models.mongodb.Account], campaign_list: List[models.mongodb.Campaign])

@import helper._
@args(args: (Symbol, Any)*) = @{
    args
}
@main("Account List") {
    <h1>@accounts.size Account(s)</h1>
    <ul>
    @accounts.map { account =>
        <li>
            @account.name
        </li>
    }
    </ul>
    <h2>Add a New Account</h2>
    @form(routes.AccountController.newAccount()) {
        <fieldset>
            @inputText(account_form("name"), '_label -> "Account Name")
            @select(
                account_form("campaigns"),
                options(campaign_list.map(x => x.name):List[String]),
                args(
                    'class -> "chosen-select",
                    'multiple -> "multiple",
                    Symbol("data-placeholder") -> "Add campaigns",
                    'style -> "width:350px;"
                ): _*
            )
            <input type="submit" value="Create">
        </fieldset>
    }
}

问题是,当我提交此表单时,它会提交campaigns字段的字符串列表。当我发布表单提交时,这给了我400错误。

我想提交提交表单,其中包含广告系列列表而不是字符串,或者使用字符串列表提交表单,然后将字符串处理为控制器中的广告系列列表。哪种方式会更好,我该怎么做?谢谢!

1 个答案:

答案 0 :(得分:0)

我最终制作了一个临时表单来保存嵌套模型的字符串。然后我将这些字符串转换为控制器中的模型对象:

  val accountTempForm = Form(
    tuple(
      "id" -> ignored(new ObjectId),
      "name" -> nonEmptyText,
      "campaigns" -> list(text)
    )
  )

  def newAccount = Action {
    implicit request =>
      accountTempForm.bindFromRequest.fold(
        formWithErrors => {
          println("error")
          Redirect(routes.AccountController.accounts)
          //BadRequest(views.html.account.accounts(AccountObject.all(), formWithErrors, CampaignObject.all()))
        },
        account => {
          val campaign_string_list = account._3
          val campaign_list = campaign_string_list.map(x => CampaignObject.getOrCreateByName(x))
          val new_account = Account.apply(account._1, account._2, campaign_list)
          AccountObject.create(new_account)
          Redirect(routes.AccountController.accounts)
        }
      )
  }