使用POST请求实现页面的跳转

项目情景:

当用户选择几个item之后,点击查看按钮之后, 页面跳转到展示items详情页面.

实现:

如果可以使用get请求, 直接在前端使用windows.loaction.href = "newUrl="xxx"&item=itemValue1&item=itemValue2"就行了

这样的问题是如果item比较多, 就有可能超出URL长度限制. 所以考虑用POST实现跳转.

server端:

@SlingServlet(resourceTypes = "/apps/xxx/components/pages/xxx-contentpage",
    extensions = "html",
    selectors = "selected",
    methods = {"GET", "POST"},
    metatype = true
)

public class MyServlet extends SlingAllMethodsServlet {
    @Override
    protected void doPost(SlingHttpServletRequest request, SlingHttpServletResponse response) throws ServletException, IOException {
        String newUrl= request.getParameter("newUrl");
        RequestDispatcher dispatcher = request.getRequestDispatcher(newUrl);
        GetRequest getRequest = new GetRequest(request);
        dispatcher.forward(getRequest, response);
    }

    @Override
    protected void doGet(SlingHttpServletRequest request, SlingHttpServletResponse response) throws ServletException, IOException {

　　　　doPost(request, response);

 } 

　　/** Wrapper class to always return GET for AEM to process the request/response as GET. */ 

private static class GetRequest extends SlingHttpServletRequestWrapper { 

public GetRequest(SlingHttpServletRequest wrappedRequest) { super(wrappedRequest); } 

@Override 

public String getMethod() { return "GET"; } 
} 

}

如果直接forward POST request,会报错:

500 error - operations.ModifyOperation Exception during response processing

所以必须把request转换成GET request.

前端:

如果使用ajax提交请求,返回的将是新页面的html代码. 必须用form提交请求.

提示: form里面的action URL 与 servlet 的newUrl 必须不一样 ,否则会发生无限循环错误.

参考: http://suryakand-shinde.blogspot.com/2016/07/aem-form-submission-handling-post.html

forward()和sendRedirect()的区别: https://examples.javacodegeeks.com/enterprise-java/servlet/java-servlet-sendredirect-example/