什么是“找不到或加载主类”意思？ [英] What does "Could not find or load main class" mean?

问题描述

新的Java开发人员遇到的一个常见问题是，他们的程序无法运行并显示错误消息：无法找到或加载主类...

这是什么意思，是什么原因导致的，以及如何解决？

解决方案

h2> java< class-name> 命令语法

首先，您需要了解正确的方法使用 java （或 javaw ）命令启动程序。

正常的语法 ¹ 是这样的：

  java [< option> ...]< class-name> [< argument> ...] 
  

其中< option> 是一个命令行选项（以 - 字符开头），< class-name> 是一个完全限定的Java类名， < argument> 是传递给应用程序的任意命令行参数。

^{1 - executableJAR文件还有第二种语法}

完全限定的类名通常写成你在Java源代码中的样子;例如

  packagename.packagename2.packagename3.ClassName 
  

但是某些版本的 java 命令允许使用斜杠而不是句点;例如

  packagename / packagename2 / packagename3 / ClassName 
  

（令人困惑的）看起来像一个文件路径名，但不是一个。请注意，术语完全限定类名是标准的Java术语...不是我刚刚弥补的： - ）

java 命令应该类似的示例：

  Xmx100m com.acme.example.ListUsers fred joe bert 
  

上述将导致 java 命令执行以下操作：

1. 搜索 com.acme.example.ListUsers class。


2. 加载类。


3. 主要带签名+修饰符的方法 public static void main（String []）。


4. 调用该方法将命令行参数（fred，joe，bert）作为 String [] 传递。

Java无法找到类的原因

当您收到消息找不到或加载主类...，这意味着第一步失败。 java 命令无法找到该类。事实上，消息中的...将是 java 正在查找的完全限定类名。

那么为什么可能找不到类呢？基本上，有两个主要原因：

第一个可能的原因是你可能提供了错误的类名。考虑到上面的例子，这里有各种错误的方法来指定类名：

• 示例＃1 - 一个简单的类名：

    java ListUser 
    

  当在类如 com.acme中声明类时。示例，则必须在 java 命令中使用完整的类名，包括例如

    java com.acme.example.ListUser 
    




• 示例＃2 - 文件名或路径名，而不是类名称：

    java ListUser.class 
   java com / acme / example / ListUser.class 
    




• 示例＃3 - 包含不正确的类名：

    java com.acme.example.listuser 
    




• 示例＃4 - 错字

    java com.acme.example.mistuser 
    

第二个可能的原因是类名是正确的，但是 java 命令找不到类。要理解这一点，你需要理解classpath的概念。这在Oracle文档中有详细解释：

• c / c> c $

设置类路径。

• Java教程 - PATH和CLASSPATH

所以...如果您正确指定了类名，接下来要检查的是您正确指定了classpath：

1. 阅读上面链接的三个文件。 （是...阅读它们。一个Java程序员至少理解了Java类路径机制如何工作的基础是很重要的。）


2. 查看命令line和/或当您运行 java 命令时生效的CLASSPATH环境变量。


3. 如果类路径中有相对路径名，请检查它们是否正确解析...从当前的当您运行 java 命令时，它是有效的。


4. 检查错误消息中提到的类


$ b





1. 当你在classpath上放置一个目录时，它就是符合命名空间的根。类位于根目录下的目录结构中，通过将完全限定名映射到路径名。因此，例如，如果/ usr / local / acme / classes在类路径上，那么当JVM查找 com.acme.example.Foon ，它将查找具有此路径名的.class文件：

     / usr / local / acme / classes / acme / example / Foon.class 
     

   
   
   如果你输入/ usr / local / acme / classes /




2. 类路径需要包括所有的

   em>其他（非系统）类。 （系统类自动定位，您很少需要关心这一点。）








java -jar< jar file> 语法





用于可执行JAR文件的替代语法如下： p>



  java [< option> ...] -jar< jar-file-name> [< argument> ...] 
  

例如

  java -Xmx100m -jar /usr/local/acme-example/listuser.jar fred 
  

在这种情况下，入口点类的名称（即 com.acme.example.ListUser ）和类路径在JAR的MANIFEST中指定文件





IDEs





典型的Java IDE支持在IDE JVM中运行Java应用程序，一个子JVM。因为IDE使用自己的机制来构造运行时类路径，识别主类并创建 java ，所以这些通常免于此特定异常，命令行。





但是，如果你在IDE后面做事情打破事情，仍然可能发生此异常。例如，如果您以前在Eclipse中为Java应用程序设置了一个应用程序启动器，然后将包含main类的JAR文件移动到文件系统中的不同位置，而不告诉Eclipse ，Eclipse会不经意地启动带有不正确类路径的JVM。





简而言之，如果在IDE中遇到这个问题，请检查类似陈旧的IDE状态，或破坏的启动程序配置。





其他参考





< 常见问题（及其解决方案）

• 从Java文档：
  
  
  
  
  • Java命令 - 'java'和'javaw'
  
  
  • 类路径
  
  





A common problem that new Java developers experience is that their programs fail to run with the error message: Could not find or load main class ...

What does this mean, what causes it, and how should you fix it?

解决方案

The java <class-name> command syntax

First of all, you need to understand the correct way to launch a program using the java (or javaw) command.

The normal syntax¹ is this:

    java [ <option> ... ] <class-name> [<argument> ...]

where <option> is a command line option (starting with a "-" character), <class-name> is a fully qualified Java class name, and <argument> is an arbitrary command line argument that gets passed to your application.
^{1 - There is a second syntax for "executable" JAR files which I will describe at the bottom.}

The fully qualified classname is conventionally written as you would in Java source code; e.g.

    packagename.packagename2.packagename3.ClassName

However some versions of the java command allow you to use slashes instead of periods; e.g.

    packagename/packagename2/packagename3/ClassName

which (confusingly) looks like a file pathname, but isn't one. Note that the term fully qualified classname is standard Java terminology ... not something I just made up to confuse you :-)

Here is an example of what a java command should look like:

    java -Xmx100m com.acme.example.ListUsers fred joe bert

The above is going to cause the java command to do the following:

1. Search for the compiled version of the com.acme.example.ListUsers class.
2. Load the class.
3. Check that the class has a main method with signature + modifiers public static void main(String[]).
4. Call that method passing it the command line arguments ("fred", "joe", "bert") as a String[].

Reasons why Java cannot find the class

When you get the message "Could not find or load main class ...", that means that the first step has failed. The java command was not able to find the class. And indeed, the "..." in the message will be the fully qualified class name that java is looking for.

So why might it be unable to find the class? Basically, there are two main causes:

The first likely cause is that you may have provided the wrong class name. (Or ... the right class name, but in the wrong form.) Considering the example above, here a variety of wrong ways to specify the class name:

• Example #1 - a simple class name:

  java ListUser
  

  When the class is declared in a package such as com.acme.example, then you must use the full classname including the package name in the java command; e.g.

  java com.acme.example.ListUser
  

• Example #2 - a filename or pathname rather than a class name:

  java ListUser.class
  java com/acme/example/ListUser.class
  

• Example #3 - a class name with the casing incorrect:

  java com.acme.example.listuser
  

• Example #4 - a typo

  java com.acme.example.mistuser
  

The second likely cause is that the class name is correct, but that the java command cannot find the class. To understand this, you need to understand the concept of the "classpath". This is explained well by the Oracle documentation:

• The java command documentation
• Setting the Classpath.
• The Java Tutorial - PATH and CLASSPATH

So ... if you have specified the class name correctly, the next thing to check is that you have specified the classpath correctly:

1. Read the three documents linked above. (Yes ... READ them. It is important that a Java programmer understands at least the basics of how the Java classpath mechanisms works.)
2. Look at command line and / or the CLASSPATH environment variable that is in effect when you run the java command. Check that the directory names and JAR file names are correct.
3. If there are relative pathnames in the classpath, check that they resolve correctly ... from the current directory that is in effect when you run the java command.
4. Check that the class (mentioned in the error message) can be located on the effective classpath.

Additional Notes:

1. When you put a directory on the classpath, it notionally corresponds to the root of the qualified name space. Classes are located in the directory structure beneath that root, by mapping the fully qualified name to a pathname. So for example, if "/usr/local/acme/classes" is on the class path, then when the JVM looks for a class called com.acme.example.Foon, it will look for a ".class" file with this pathname:

       /usr/local/acme/classes/com/acme/example/Foon.class
   

   If you had put "/usr/local/acme/classes/com/acme/example" on the classpath, then the JVM wouldn't be able to find the class.

2. The classpath needs to include all of the other (non-system) classes that your application depends on. (The system classes are located automatically, and you rarely need to concern yourself with this.)

The java -jar <jar file> syntax

The alternative syntax used for "executable" JAR files is as follows:

    java [ <option> ... ] -jar <jar-file-name> [<argument> ...]

e.g.

    java -Xmx100m -jar /usr/local/acme-example/listuser.jar fred

In this case the name of the entry-point class (i.e. com.acme.example.ListUser) and the classpath are specified in the MANIFEST of the JAR file.

IDEs

A typical Java IDE has support for running Java applications in the IDE JVM itself or in a child JVM. These are generally immune from this particular exception, because the IDE uses its own mechanisms to construct the runtime classpath, identify the main class and create the java command line.

However it is still possible for this exception to occur, if you do things behind the back of the IDE to break things. For example, if you have previously set up an Application Launcher for your Java app in Eclipse, and you then moved the JAR file containing the "main" class to a different place in the file system without telling Eclipse, Eclipse would unwittingly launch the JVM with an incorrect classpath.

In short, if you get this problem in an IDE, check for things like stale IDE state, broken project references or broken launcher configurations.

Other References

• From the Oracle Java Tutorials - Common Problems (and Their Solutions)
• From the Java Documentation:
  • The Java command - 'java' and 'javaw'
  • The Classpath

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