如何从命令提示符编译servlet？ [英] How to compile servlets from command prompt?

问题描述

我想从命令提示符编译一个非常基本的 servlet ，但它总是不成功，编译器告诉我如下：

I'd like to compile a very basic servlet from command prompt, but it is always unsuccessful and the compiler tells me the following:

 error: package javax.servlet does not exist.

我搜索的解决方案，我发现我需要包括 servlet .jar 库到我的PATH。
我相信我。
我坚信这些库在我的计算机的位置是：

I googled for the solution and I found that I need to include the servlet.jar libraries into my PATH. I believe I did. I strongly believe that the location for those libraries in my computer is:

C:\apache-tomcat-7.0.23\lib\servlet-api.jar\ 

和结束部分）如下：

%JAVA_HOME%\bin;C:\apache-tomcat-7.0.23\lib\servlet-api.jar\

对我来说，它显然不是。任何人都可以告诉我可能是什么问题？

For me, it looks ok, but it is obviously not. Can anyone tell me what could be the problem?

推荐答案

classpath不是路径...你不需要它环境变量。
可以使用选项-cp或-classpath设置javac的类路径（还有其他几种方法）。
javac使用环境变量CLASSPATH来查找类，这可以是有用的，也可以是难以跟踪问题的源。

classpath not path ... and you don't need it as an evironment variable. You can set the classpath for javac with option -cp or -classpath (several other ways are also available). javac uses the environment variable CLASSPATH to look for classes, that can be useful and can also be a source for hard-to-track-down-problems.

编译使用库（即来自标准JavaSE外部的类）的java文件的示例如下：

An example to compile a java file that uses a library(that is classes from outside the standard JavaSE) would be:

javac -classpath C:\apache-tomcat-7.0.23\lib\servlet-api.jar MyTestServlet.java

如果您的环境变量CLASSPATH包含您需要的库，您可能需要：

if your environmental variable CLASSPATH contains libraries you need you might want to do:

javac -classpath %CLASSPATH%;C:\apache-tomcat-7.0.23\lib\servlet-api.jar MyTestServlet.java

（请注意，我没有访问Windows机器，因此没有测试上面语法的特殊部分）
（还要注意，在这个例子中C：​​\ apache-tomcat-7.0.23 \lib \servlet-api.jar是一个jar文件，而不是一个目录，它可能是从您的问题在您的机器上）
对于Windows操作系统上的命令行编译一个好主意是要正确设置环境变量JAVA_HOME，并在PATH中设置JDK的bin目录。

(please be aware that I don't have access to a windows machine, and therefore haven't tested the idiosyncratic parts of the syntax above) (also note that in this example "C:\apache-tomcat-7.0.23\lib\servlet-api.jar" is a jar file and not a directory which it might be from your question on your machine) For command line compiling on windows OS it is a good idea to have the environmental variable JAVA_HOME set correctly and the bin directory of the JDK in the PATH.

但我建议您编写compile-compile-execute-deploy通过/在IDE中进行servlet开发，然后通过命令行确定如何使用JDK。
Java Servlet不是独立的可执行类，但需要部署到例如要测试/使用的tomcat中。

I do however suggest getting to write-compile-execute-deploy via/in an IDE for servlet development before figuring out how to do it with just the JDK from a command line. Java Servlets are not stand-alone executable classes but needs to be deployed into for example tomcat to be tested/used.

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