如何捕捉scala playframework中的執行異常？

[APIError: You must provide an email address in order to create a ticket.]

有可能趕上這個錯誤，並張貼只有

"You must provide an email address in order to create a ticket."

這個錯誤應該在視圖頁面showup。

，讓錯誤代碼：

val question = uservoice.post("/api/v1/tickets.json", ticket).getJSONObject("ticket")

控制器：

def contactSave = withOptionUser { user => implicit request => 
    contactForm.bindFromRequest.fold(
     formWithErrors => BadRequest(html.anon.contact(user, formWithErrors)), 
     c => { 
      val uservoice = new com.uservoice.Client(SUBDOMAIN, API_KEY, API_SECRET) 

      val ticketMsg = Map("state" -> "open","subject" -> c._2, "message" -> c._3).toMap[String,Object].asJava 

      val ticket = Map("email" -> c._1, "ticket" -> ticketMsg).toMap[String,Object].asJava 

      Logger.debug(ticket.toString) 

      val question = uservoice.post("/api/v1/tickets.json", ticket).getJSONObject("ticket") 

      Logger.debug(question.toString) 

      Ok(views.html.anon.contactThanks(user)) 
     } 
    ) 
    }

HTML：

@main("Contact Us",user,"contact",stylesheet, scripts) { 
    @helper.form(routes.UservoiceController.contactSave) { 
<section class="contact"> 
    <div class="contactBox contentBox"> 
     <div class="leftColumn"> 
      <h1>Contact Us</h1> 
      <span>You can fill out this form for any general inquiries, comments, etc.</span> 
      <span>You can also find us on Facebook and Twitter!</span> 

      <div class="social"> 
       @form.globalError.map { error => 
       <span class="error" data-xpl="loginError"> 
        @error.message 
       </span>

來源

2014-02-11 Monnster

在發佈到該API之前，您不應該檢查有效的電子郵件地址嗎？盲目捕捉所有異常情況可能會掩蓋其他錯誤。 –

嗨肢體，代碼是使用第三方電子郵件發送，我無法找到他們如何驗證他們的電子郵件..所以我想在找出使用的驗證時趕上錯誤。 – Monnster

你可以使用try/catch塊，並創建一個表單

var form = contactForm.bindFromRequest() 
form.fold(
    formWithErrors => ... 
    c => { 
    ... 
    try { 
     val question = uservoice.post("/api/v1/tickets.json", ticket).getJSONObject("ticket") 
     Logger.debug(question.toString) 
     Ok(views.html.anon.contactThanks(user)) 
    } catch { 
     case e: Exception => 
     Logger.error("error ...", e) 
     val formWithError = form.withError("email", "You must provide an email address in order to create a ticket.") 
     BadRequest(html.anon.contact(user, formWithError)) 
    } 
    } 
)

來源

2014-02-11 13:24:50

您好@Yann西蒙我得到價值withError不是（String，String，String）的成員順便說一下，我在查看文件 @（user：Option [User]， form：Form [ （字符串，字符串，字符串）] ） – Monnster

是的，對不起，你必須先提取表單實例： VAL形式= contactForm.bindFromRequest form.fold（ formWithErrors => ...， C => ... 嘗試{ ... } catch { ... val formWithError = form.withError（...） ... } ） –

如何捕捉scala playframework中的執行異常？

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