8

在我的 Play 应用程序中，我有几个可以从不同页面触发的操作（例如“删除”实体）。触发操作后，我想将用户重定向回他们在执行操作之前所在的页面。在 Play 中是否有一个很好的模式可以用于此？

4

3 回答 3

5

您可以轻松地@request.getHeader("referer")在模板中使用，例如，如果您有一个取消按钮应该将您重定向到上一页，请使用：

<a  href="@request.getHeader("referer")">Cancel</a>

这样，您无需将任何额外信息传递给您的模板。（用 play 2.3.4 测试）

于 2014-12-29T21:47:06.407 回答
2

This is what I came up with in the end, although it isn't particularly elegant, and I'd be interested in better ways of doing it. I added a hidden input to my form with the current page URL:

@(implicit request: RequestHeader)

...

<form action="@routes.Controller.doStuff()" method="post">
  <input type="hidden" name="previousURL" value="@request.uri"/>
  ...
</form>

Then in my controller:

def doStuff() = Action { implicit request =>
  val previousURLOpt: Option[String] =
    for {
      requestMap <- request.body.asFormUrlEncoded
      values <- requestMap.get("previousURL")
      previousURL <- values.headOption
    } yield previousURL
   previousURLOpt match {
     case Some(previousURL) =>
       Redirect(new Call("GET", previousURL))
     case None =>
       Redirect(routes.Controller.somewhereElse)
   }
}
于 2013-07-02T08:15:25.983 回答
1

The easiest way I've found to do this, is from within your controller method, use this: 

String refererUrl = request().getHeader("referer");

So, you'd do something like: 

public static Result query(String queryStr, int page, int offset) {
    String refererUrl = request().getHeader("referer");
    Logger.info("refererUrl: " + refererUrl);
    if(queryStr.length() < 3) {
        flash(Application.FLASH_ERROR_KEY, "type a longer search than '" + queryStr.trim() + "'");
        return redirect(refererUrl);
    }
    return ok(listings.render(searchService.searchListings(queryStr)));
}

Keep in mind you need to do a redirect() and NOT a render() with a flash message.

于 2015-10-29T02:23:56.727 回答