我需要通过使用以下模式在浏览器中输入URL来从数据库中获取对象:
https://host.com/<something>?action=<actionName>&object=<node|interface>&selection=<list>&tenants=<list>
“something”是检索参数的方法,参数包括:action,object,selection和tenants。
我如何在web.xml文件中建立模式以符合理解上述每个元素的方法?
<servlet>
<servlet-name>NewDynamicWebProject</servlet-name>
<servlet-class>com.test.package.NewDynamicWebProject</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>NewDynamicWebProject</servlet-name>
<url-pattern>/something/*</url-pattern>
</servlet-mapping>
我的课程:
@SuppressWarnings("serial")
public class NewDynamicWebProject extends HttpServlet {
@Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
something(req, resp);
}
protected void something(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
String actionName = req.getParameter("action");
PrintWriter out = resp.getWriter();
out.print("<div>" + actionName + "</div>");
}
}
如何确保我传递到我的网址的参数在我的“某事”方法的范围内有意义?
答案 0 :(得分:0)
我放弃了“某事”课,只做了以下几点:
@Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
String actionName = req.getParameter("action");
String objectType = req.getParameter("object");
String selectionList = req.getParameter("selection");
String tenantsList = req.getParameter("tenants");
PrintWriter out = resp.getWriter();
out.print("<div>" + actionName + " " + objectType + " " + selectionList + " " + tenantsList + "</div>");
}