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關于easymock的測試

easymock的測試：
1、首先導入easymock.jar包。
2、 LoginServlet.java servlet類

package com.test.mock;
2

import java.io.IOException;
4

import javax.servlet.RequestDispatcher;
6

import javax.servlet.ServletContext;
7

import javax.servlet.ServletException;
8

import javax.servlet.http.HttpServlet;
9

import javax.servlet.http.HttpServletRequest;
10

import javax.servlet.http.HttpServletResponse;
11

public class LoginServlet extends HttpServlet {
13

/**
14

*
15

*/
16

private static final long serialVersionUID = 7020197334019200211L;
17

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
19

String username = request.getParameter("username");
20

String password = request.getParameter("password");
21

// check username & password:
22

if("admin".equals(username) && "123456".equals(password)) {
23

ServletContext context = getServletContext();
24

RequestDispatcher dispatcher = context.getNamedDispatcher("dispatcher");
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dispatcher.forward(request, response);
26

}
27

else {
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throw new RuntimeException("Login failed.");
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}
30

}
31

}
34

3、LoginServletTest.java

package com.test.mock;
2

import javax.servlet.http.*;
4

import org.easymock.*;
5

import junit.framework.TestCase;
6

public class LoginServletTest extends TestCase {
7

public void testLoginFailed() throws Exception {
8

MockControl mc = MockControl.createControl(HttpServletRequest.class);
9

HttpServletRequest request = (HttpServletRequest)mc.getMock();
10

request.getParameter("username");
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mc.setReturnValue("admin", 1);
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request.getParameter("password");
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mc.setReturnValue("1234", 1);
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mc.replay();
16

LoginServlet servlet = new LoginServlet();
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try {
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servlet.doPost(request, null);
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fail("Not caught exception!"); }
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catch(RuntimeException re) {
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assertEquals("Login failed.", re.getMessage());
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}
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mc.verify();
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}
25

}

測試結果：

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posted on 2007-09-16 17:33 apple0668 閱讀(425) 評論(0) 編輯收藏所屬分類: junit

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