为什么会出现 javac：找不到文件错误？
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 < code>java
中有 .java 文件
并且我在 home/pathToFolderA/src和 home/pathToFolderB/gen-java
当我跑步home/pathToFolderA/src 中的 javac 和 java 一切正常
但是当我从 中运行 on javac 时home/pathToFolderB/gen-javafileName.java 我收到文件未找到错误，具体来说
javac: file Not found: fileName.java
Usage: javac <options> <source files>
为什么会发生这种情况？
感谢所有帮助
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
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（5）
类路径用于查找类文件，而不是源文件。 （它也不用于查找
java和javac二进制文件；这些二进制文件可以在您的正常路径中找到。）您需要指定要显式编译的文件：显然，如果您'如果已经在
/home/pathToFolderA/src中，那么您可以只使用fileName.java，因为它被视为相对于您当前的目录。The classpath is used to find class files, not source files. (Nor is it used to find the
javaandjavacbinaries; those are found in your normal path.) You need to specify the files to compile explicitly:Obviously if you're already in
/home/pathToFolderA/srcthen you can just usefileName.javabecause that's treated as being relative to your current directory.您不应该将 classpath 设置为指向 JDK bin 目录 - 相反，它应该是 PATH 环境变量，它与类路径有不同的用途。 （类路径定义了包含已编译的 Java .class 代码的 jar 和目录的列表；PATH 变量定义了一个路径列表，当在当前目录中找不到程序时，shell 需要在这些路径中查找和定位要执行的程序 - 因此，如果您例如键入
zip——它将查找PATH中定义的所有目录，并找出zip程序位于/ 下usr/bin)其次，如果您想从两个目录编译源代码，则需要指定：
home/pathToFolderA/src和home/pathToFolderB/gen-java>)如下所示：
总而言之，编译：并运行已编译的程序
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 inPATHand figure out thatzipprogram is located under/usr/bin)Secondly if you want to compile sources from both directory you need to specify:
home/pathToFolderA/srcandhome/pathToFolderB/gen-java)To sum it up, it would be something like this to compile:
and to run your compiled programs:
这是错误的。类路径用于查找 *.class 文件，而不是特定于操作系统的可执行文件。 JDK 的
bin目录不属于类路径。请注意，类路径也不是用于查找 *.java 源文件。当您运行
javac时，您需要指定源文件的路径（如果它不在当前目录中）。That's wrong. The classpath is used to find
*.classfiles, not operating system specific executables. Thebindirectory of your JDK does not belong in the classpath. Note that the classpath is also not for finding*.javasource files.When you run
javacyou need to specify the path to the source file, if it isn't in the current directory.确保您的文件名不包含空格
例如：
通常当您通过复制过去重命名文件时会发生错误，这会导致名称和点之间出现空格（这是错误：
HelloWorld .java）。并确保您将目录更改为文件所在的同一文件夹
make sure that your file name contain no spaces
Eg:
usually the errors occur when you rename the file by copy past that will cause a space between the name and the dot (this is the mistake:
HelloWorld .java).and make sure you changed the directory to the same folder your file in
如果没有目录“gen-java”的列表和您正在键入的确切命令，我的猜测是您正在尝试编译一个不存在的文件。 Linux 区分大小写，所以也许这就是你的问题。或者该文件不存在。
Without a listing of the directory "gen-java" and the exact command you're typing,my guess would be that you're trying to compile a file that doesn't exist. Linux is case sensitive, so maybe that's your problem. Or the file doesn't exist.