为什么这个javac：File Not Found错误？ [英] Why this javac: File Not Found error?

问题描述

在CentOS 5 linux上使用一些基本的java应用程序，我将 classpath 设置为指向 home / pathToJava / bin 其中包含 javac 和 java

我在中有 .java 文件home / pathToFolderA / src

<当我运行 javac

时，

和 home / pathToFolderB / gen-java

和在 home / pathToFolderA / src 一切正常工作

但是当我在中运行 javac 时，home / pathToFolderB / gen-java c $ c> fileName.java 我得到一个文件未找到错误，特别是

  javac：file找不到：fileName.java 
用法：javac< options> <源文件> 
  

为什么会发生这种情况？

感谢所有帮助

解决方案

您不应将类路径设置为指向您的JDK bin目录 - 而是应该是 PATH 环境变量，它为类路径提供不同的目的。 （类路径定义了包含编译的Java .class代码的jar和目录的列表; PATH变量定义了一个路径列表，当shell在当前目录中找不到时，shell需要查找和定位要执行的程序，所以如果你类型例如 zip - 它将查看 PATH 中定义的所有目录，并确定 zip 程序位于 / usr / bin ）
其次，如果要从两个目录编译源代码， ：

• 所有源的路径（ home / pathToFolderA / src 和 home / pathToFolderB / gen-java ）


• 要生成编译的.class文件的路径


• 在类路径中指定您可能在源文件中使用的任何库

像这样编译：

  javac -d / home / pathToFolderWithResultsOfCompilation -classpath /path/to/some.jar:/path/到/ another.jar home / pathToFolderA / src / *。java home / pathToFolderB / gen-java / *。java 
  

$ b b

并运行您编译的程序：

  java -classpath /path/to/some.jar:/path/到/ another.jar：/ home / pathToFolderWithResultsOfCompilation full.name.of.your.Java 
  

Working with some basic java apps on CentOS 5 linux and I have my classpath set to point to home/pathToJava/bin which contains javac and java

and I have .java files in home/pathToFolderA/src

and home/pathToFolderB/gen-java

When I run javac and java in home/pathToFolderA/src everything works perfectly

But when I run javac from within home/pathToFolderB/gen-java on fileName.java I get a file not found error, specifically

javac: file Not found: fileName.java
Usage: javac <options> <source files>

Why could this be happening?

Thanks for all help

解决方案

You shouldn't set your classpath to point to your JDK bin directory -- instead it should be the PATH environment variable, which serves a different purpose to classpath. (The classpath defines a list of jars and directories containing compiled Java .class code; the PATH variable defines a list of paths where the shell needs to look and locate programs to execute when they are not found in the current directory -- so if you type for instance zip -- it would look in all the directories defined in PATH and figure out that zip program is located under /usr/bin) Secondly if you want to compile sources from both directory you need to specify:

• all the paths where the sources are (both home/pathToFolderA/src and home/pathToFolderB/gen-java)
• the path where the compiled .class files to be generated
• specify in the classpath any library you might use in your source files

To sum it up, it would be something like this to compile:

javac -d /home/pathToFolderWithResultsOfCompilation -classpath /path/to/some.jar:/path/to/another.jar home/pathToFolderA/src/*.java home/pathToFolderB/gen-java/*.java 

and to run your compiled programs:

java -classpath /path/to/some.jar:/path/to/another.jar:/home/pathToFolderWithResultsOfCompilation full.name.of.your.Java

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