Trying to access a Servlet from a button on HTML page
//Html Page
FORM method="GET" action="/StudentDBServlet">
yada yada
INPUT type="submit" value="Register" name="Register">
//My Servlet
#WebServlet(name="StudentDBServlet", urlPatterns={"/StudentDBServlet"})
public class StudentDBServlet extends HttpServlet {
The servlet is located in Package com.zzz.studentregistration
When I hit the "Register" Button this is the url create
http://localhost:8080/StudentDBServlet?FirstName
but it needs to be this to work properly
http://localhost:8080/com.zzz.studentregistration/StudentDBServlet?FirstName
How or where do I add the package name to the Servlet definition?
I tried adding to various parts if #WebServlet but no luck ???
Thanks
The servlet container couldn't care less about your servlets' package. Only the urlPatterns matter. Your code above should work just fine. It is not clear what (and why) do you want to achieve. You can simply write:
#WebServlet(urlPatterns={"/com.zzz.studentregistration/StudentDBServlet"})
But then the form has to point to this particular servlet:
<FORM method="GET" action="/com.zzz.studentregistration/StudentDBServlet">
Related
Can I call a servlet from JSP file without using a HTML form?
For example, to show results from database in a HTML table during page load.
You can use the doGet() method of the servlet to preprocess a request and forward the request to the JSP. Then just point the servlet URL instead of JSP URL in links and browser address bar.
E.g.
#WebServlet("/products")
public class ProductsServlet extends HttpServlet {
#EJB
private ProductService productService;
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
List<Product> products = productService.list();
request.setAttribute("products", products);
request.getRequestDispatcher("/WEB-INF/products.jsp").forward(request, response);
}
}
<%# taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c" %>
...
<table>
<c:forEach items="${products}" var="product">
<tr>
<td>${product.name}</td>
<td>${product.description}</td>
<td>${product.price}</td>
</tr>
</c:forEach>
</table>
Note that the JSP file is placed inside /WEB-INF folder to prevent users from accessing it directly without calling the servlet.
Also note that #WebServlet is only available since Servlet 3.0 (Tomcat 7, etc), see also #WebServlet annotation with Tomcat 7. If you can't upgrade, or when you for some reason need to use a web.xml which is not compatible with Servlet 3.0, then you'd need to manually register the servlet the old fashioned way in web.xml as below instead of using the annotation:
<servlet>
<servlet-name>productsServlet</servlet-name>
<servlet-class>com.example.ProductsServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>productsServlet</servlet-name>
<url-pattern>/products</url-pattern>
</servlet-mapping>
Once having properly registered the servlet by annotation or XML, now you can open it by http://localhost:8080/context/products where /context is the webapp's deployed context path and /products is the servlet's URL pattern. If you happen to have any HTML <form> inside it, then just let it POST to the current URL like so <form method="post"> and add a doPost() to the very same servlet to perform the postprocessing job. Continue the below links for more concrete examples on that.
See also
Our Servlets wiki page
doGet and doPost in Servlets
How to avoid Java code in JSP
Servlet returns "HTTP Status 404 The requested resource (/servlet) is not available"
You will need to use RequestDispatcher's Methods forward/include depending on your requirement to achieve same.
In JSP you need to use following tags:
jsp:include :
The element allows you
to include either a static or dynamic
file in a JSP file. The results of
including static and dynamic files are
quite different. If the file is
static, its content is included in the
calling JSP file. If the file is
dynamic, it acts on a request and
sends back a result that is included
in the JSP page. When the include
action is finished, the JSP container
continues processing the remainder of
the JSP file.
e.g.
<jsp:include page="/HandlerServlet" flush="true">
jsp:forward :
The element forwards the
request object containing the client
request information from one JSP file
to another file. The target file can
be an HTML file, another JSP file, or
a servlet, as long as it is in the
same application context as the
forwarding JSP file. The lines in the
source JSP file after the
element are not
processed.
e.g.
<jsp:forward page="/servlet/ServletCallingJsp" />
Check Advanced JSP Sample : JSP-Servlet Communication:
http://www.oracle.com/technology/sample_code/tech/java/jsps/ojsp/jspservlet.html
Sure you can, simply include it in your action in the form. But you have to write the correct doPost or doGet to handle the request!
If you want to call a particular servlet method than you also use Expression Language. For example, you can do something like:
Servlet
ForexTest forexObject = new ForexTest();
request.setAttribute("forex", forexObject);
JSP
<body bgcolor="#D2E9FF">
Current date : ${forex.rate}
</body>
I am new to servlets and i have developed a html page which has a submit button that triggers
my servlet.Everything is working fine .But now i want to use GET method as my html page is not posting anything.Hence i made the following changes:
1)In my page.html file, i replaced method="POST" with method="GET".
2)I changed doPost with doGet in my servlet.
But i'm getting error message that "GET not allowed here".Why is it so?
Here are the original files which work correctly(prior to making changes):
My page.html page:
<html>
<head>
<title>A simple revision of servlets</title>
</head>
<body>
<form method="POST" action="Idiot">
<input type="SUBMIT">
</form>
</body>
</html>
My Deployment Descriptor:
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
version="2.5">
<servlet>
<servlet-name>TangoCharlie</servlet-name>
<servlet-class>Revise</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>TangoCharlie</servlet-name>
<url-pattern>/Idiot</url-pattern>
</servlet-mapping>
</web-app>
And finally my servlet file named Revise.java
import javax.servlet.*;
import javax.servlet.http.*;
import java.io.*;
public class Revise extends HttpServlet
{
public void doPost(HttpServletRequest request,HttpServletResponse response) throws IOException,ServletException
{
response.setContentType("text/html");
PrintWriter out=response.getWriter();
out.println("<html><body><h3>Hello India</h3></body></html>");
out.println("Hello");
}
}
I don't want to add any input fields.My aim is just to run the servlet without a single button on html page that too called using "POST" method.
Just do the job in doGet() method (don't forget to properly rebuild/redeploy/restart the project after editing servlet code, otherwise you will still face a "HTTP 405: method not allowed" error) and invoke the URL of the servlet directly instead of the URL of the JSP in the browser's address bar.
So, the URL in browser address bar should be
http://example.com/contextname/Idiot
instead of
http://example.com/contextname/page.html
Unrelated to the concrete problem, emitting HTML in a servlet is bad design. It should be done by a JSP. You can use RequestDispatcher#forward() to forward the request to a JSP after finishing the doGet() business logic. Further, packageless classes are also a bad design. You should always put publicly reuseable Java classes in a package. Packageless servlets will only work on certain combinations of Tomcat + JVM versions.
See also:
Our servlets wiki page - contains concrete (and properly designed) Hello World examples
I creat a web application using eclipse and tomcat7 I had the following code in the html file and the java servlet class
in the html file:
<form action="UserAccessServlet" method = "get">
in the servlet class I had
#WebServlet ("/UserAccessServlet")
then I just made some small changes (new println statements) but it shows no effect I changed the server name with the following peice of code
html file: <form action="SQA_Servlet" method = "get">
java class: #WebServlet ("/SQA_Servlet")
but it seems that no reload take place and I got the following error:
HTTP Status 404 - /SQA_Learning/SQA_Servlet
--------------------------------------------------------------------------------
type Status report
message /SQA_Learning/SQA_Servlet
description The requested resource (/SQA_Learning/SQA_Servlet) is not available.
I tried clean the module, refresh, close the reopen the project with the same result
I replaced #WebServlet ("/SQA_Servlet") with #WebServlet(urlPatterns={"/SQA_Servlet"})
and still have no effect.. any suggestion.
The WebServlet name attribute cannot start with a /. Rather do,
#WebServlet("UserAccessServlet")
or leave it blank (if you want the WebServlet to use the name of your Servlet class name. Example:
#WebServlet
public class UserAccessServlet extends HttpServlet {
//Do stuff
}
I would recommend declaring your WebServlet annotations fully like in this example.
I'm not sure when and in what for conditions you are receiving this error. But if you are deploying to tomcat, the following might occur:
Assuming your webapp is called "my.webapp" resulting in my.webapp.war
assuming you have a Servlet "servlet1" which performs action1 => #WebServlet(urlPatterns = "/action1") (note the slash in front of action1)
Assuming you are calling this action with a html form:
<form action="/action1" method="GET"> this might not work because of the slash in front of action1
When it's there tomcat will redirect to localhost:8080/action1?..
while it should redirect to localhost:8080/my.project/action1?..
Solution alter the html so the form looks like:
<form action="action1" method="GET">, don't change the #WebServlet(urlPatterns = "/action1")
Hope this helps someone!
I am pretty new to servlets and web development in general.
So basically I have a servlet that queries a database and returns some values, like a name. What I want is to turn the name into a link that opens a details page for that name (which another servlet would handle). How can I send the name to the other servlet so it can query a database for the relevant details?
Maybe I'm taking the wrong approach?
Edit: I am using Tomcat 5.5
Pass it as request parameter.
Either add it to the query string of the URL of the link to the other servlet which is then available by request.getParameter("name") in the doGet() method.
link
Or add it as a hidden input field in a POST form which submits to the other servlet which is then available by request.getParameter("name") in the doPost() method.
<form action="otherservlet" method="post">
<input type="hidden" name="name" value="${name}" />
<input type="submit" />
</form>
See also:
Servlets info page - contains a Hello World
Not sure if I understand correctly, but you may look at javax.servlet.RequestDispatcher and forward the url to the second servlet.
The url could be created using the name:
http://myhost.mydomain/my.context/servlet2.do?name=John
I would create the URL either in the first servlet or in a client using a configurable template for the URL. This way both servlets are clearly separated - you can even have each one on different machine.
I have an app that may run at http://serverA/m/ or http://serverA/mobile/. I have a shared header with a search form that needs to go to http://serverA/installationName/search.
However, if I use <form action="/search"> it goes to the root of the server, not the tomcat application.
If I use <form action="search"> it goes to a path relative to the current page. (i.e http://serverA/m/someOtherPage/search
I've tried <c:url value="search"> and <c:url value="/search"> but neither of them seem to work.
In intelliJ, <c:url value="/search"> gives me "Cannot resolve controller URL /search" even though I have a controller defined with #RequestMapping("/search")
<form action="<c:url value="/search" />" />
Using <c:url> is the way. Ignore what the IDE tells you. They are not good at that. Just try to run it.
Bozho is right. I have used HTML BASE tag too:
<base href="${pageContext.request.scheme}://${pageContext.request.serverName}:${pageContext.request.serverPort}${pageContext.request.contextPath}/" />
If you can put this tag in a few places (ideally in only one JSP) you can get your code cleaner.
You can (apart from other responders hints) also use Spring JSP tag (spring:url) which is modeled after the JSTL c:url tag (see Bozhos reply). The tld reference:
http://static.springsource.org/spring/docs/3.0.x/spring-framework-reference/html/spring.tld.html#spring.tld.url
And the bottom of this mvc:resources block for an example use:
http://static.springsource.org/spring/docs/3.0.x/spring-framework-reference/html/mvc.html#mvc-static-resources
you will not be able to imbed the c:url tag directly in the attribute, if your form tag is a jsp tag (perhaps, <sf:form>).
In that situation I do the following:
<c:url var="someName" value="some uri value"/>
<sf:form path="${someName}" ...>