Java 在 Windows 7 中使用命令提示符设置类路径
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Setting Classpath using command prompt in Windows 7
提问by user2779202
I am compiling my code through Windows 7 using command prompt -- here are details :
我正在使用命令提示符通过 Windows 7 编译我的代码 - 以下是详细信息：
I set the class path like this :
我这样设置类路径：
set classpath= %classpath%;C:\java-programes\Servlet-Programing-new1\TotalUsersOnline\lib\servlet-api\*.jar;C:\java-programes\Servlet-Programing-new1\TotalUsersOnline\lib\servlet\*.jar;
and then I tried to compile my file like :
然后我尝试编译我的文件，如：
javac -d ..\classe com\java\controller\LoginServlet.java
output:
输出：
com\java\controller\LoginServlet.java:7: package javax.servlet does not exist
import javax.servlet.RequestDispatcher;
^
com\java\controller\LoginServlet.java:8: package javax.servlet does not exist
import javax.servlet.ServletException;
^
com\java\controller\LoginServlet.java:9: package javax.servlet.http does not exist
import javax.servlet.http.HttpServlet;
^
com\java\controller\LoginServlet.java:10: package javax.servlet.http does not exist
import javax.servlet.http.HttpServletRequest;
^
com\java\controller\LoginServlet.java:11: package javax.servlet.http does not exist
import javax.servlet.http.HttpServletResponse;
^
com\java\controller\LoginServlet.java:13: cannot find symbol
symbol: class HttpServlet
public class LoginServlet extends HttpServlet{
^
com\java\controller\LoginServlet.java:21: cannot find symbol
symbol : class HttpServletRequest
location: class com.java.controller.LoginServlet
public void service(HttpServletRequest request, HttpServletResponse response)
^
com\java\controller\LoginServlet.java:21: cannot find symbol
symbol : class HttpServletResponse
location: class com.java.controller.LoginServlet
public void service(HttpServletRequest request, HttpServletResponse response)
^
com\java\controller\LoginServlet.java:22: cannot find symbol
symbol : class ServletException
location: class com.java.controller.LoginServlet
throws ServletException, IOException {
^
com\java\controller\LoginServlet.java:46: cannot find symbol
symbol : class RequestDispatcher
location: class com.java.controller.LoginServlet
RequestDispatcher dispatcher = request.getRequestDispatcher("/home.jsp");
^
com\java\controller\LoginServlet.java:20: method does not override or implement a method from a supertype
@Override
after that I tried like :
之后我尝试过：
javac -classpath C:\java-programes\Servlet-Programing-new1\TotalUsersOnline\lib\servlet-api\*.jar com\java\controller\LoginServlet.java
then the output I got is :
那么我得到的输出是：
javac: invalid flag: C:\java-programes\Servlet-Programing-new1\TotalUsersOnline\lib\servlet-api\servlet-api-2.5.jar
Usage: javac <options> <source files>
use -help for a list of possible options
Please Help on this As I am stuck on this point and I am not getting anything..how to go forward.I need help badly :(
请帮助解决这个问题因为我被困在这一点上，我没有得到任何东西..如何前进。我非常需要帮助:(
Thanks in Advance
提前致谢
回答by Gyanendra Dwivedi
Understanding class path wildcards
了解类路径通配符
Class path entries can contain the basename wildcard character , which is considered equivalent to specifying a list of all the files in the directory with the extension .jar or .JAR. For example, the class path entry foo/specifies all JAR files in the directory named foo. A classpath entry consisting simply of * expands to a list of all the jar files in the current directory.
A class path entry that contains * will not match class files. To match both classes and JAR files in a single directory foo, use either foo;foo/* or foo/*;foo. The order chosen determines whether the classes and resources in foo are loaded before JAR files in foo, or vice versa.
Subdirectories are not searched recursively. For example, foo/* looks for JAR files only in foo, not in foo/bar, foo/baz, etc.
类路径条目可以包含基本名称通配符，这被认为等同于指定目录中所有文件的列表，扩展名为 .jar 或 .JAR。例如，类路径条目 foo/指定名为 foo 的目录中的所有 JAR 文件。仅由 * 组成的类路径条目扩展为当前目录中所有 jar 文件的列表。
包含 * 的类路径条目将不匹配类文件。要匹配单个目录 foo 中的类和 JAR 文件，请使用 foo;foo/* 或 foo/*;foo。选择的顺序决定了 foo 中的类和资源是在 foo 中的 JAR 文件之前加载，反之亦然。
不递归搜索子目录。例如，foo/* 只在 foo 中查找 JAR 文件，而不在 foo/bar、foo/baz 等中查找。
Must read full detail here, it is awesome http://docs.oracle.com/javase/6/docs/technotes/tools/windows/classpath.html
必须在这里阅读完整的细节，它很棒http://docs.oracle.com/javase/6/docs/technotes/tools/windows/classpath.html
