Servlets 如何使用javax.portlet.ActionResponse.sendRedirect方法在java中发送Post请求？_Servlets_Portlet

Servlets 如何使用javax.portlet.ActionResponse.sendRedirect方法在java中发送Post请求？

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Servlets 如何使用javax.portlet.ActionResponse.sendRedirect方法在java中发送Post请求？,servlets,portlet,Servlets,Portlet,我想发布到URL，但下面让它成为GET，那么我如何发布 Object portletResponse = webAppAccess.getHttpServletRequest() .getAttribute(Constants.PORTLET_RESPONSE); if (portletResponse instanceof javax.portlet.ActionResponse) { javax.portlet.ActionRe

我想发布到URL，但下面让它成为GET，那么我如何发布

    Object portletResponse = webAppAccess.getHttpServletRequest()
            .getAttribute(Constants.PORTLET_RESPONSE);

   if (portletResponse instanceof javax.portlet.ActionResponse) {
            javax.portlet.ActionResponse actionResponse = (javax.portlet.ActionResponse) portletResponse;

            actionResponse.sendRedirect(URL);
        }

您是否尝试使用

RequestDispater

而不是response.sendRedirect

它保留原始请求，而不进行更改

所以，如果它是POST，它将保持POST。

我使用表单POST方法完成了这项工作，如下所示

 webAppAccess.processPage("importedPage");

已在模型中添加此导入页面：

<HTML>
<HEAD>
<title>New Page</title>
</HEAD>
<Body onload="">
<form id="FormID" method="POST" action="actionURL">
    <input type="hidden" name="id" id="ID" value="<%=webAppAccess.getVariables().getString("ID")%>"/>
    <noscript>
        <p>Your browser does not support JavaScript or it is disabled. 
            Please click the button below to process the request.
    </p>
        <input type="submit" value="Proceed " name ="submit"></input>
    </noscript>
    <script>
    document.getElementById('FormID').submit();
    </script>
</form>

根据定义，重定向总是GET，我想把它从WAS发送到tomcat

 @RequestMapping(value = {"/Details"}, method = RequestMethod.POST)
public String mthDetails(final Model p_model, @RequestParam(value = "id", required = false) final String p_ID){ 

    //code for further logic using ID
}