java Class.forName 给出 ClassNotFound 异常
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Class.forName is giving ClassNotFound Exception
提问by Sashi Kant
I am using the following code ::
我正在使用以下代码::
String className = "SmsHelper"
Class c = Class.forName(className);
And I am getting the following
我得到以下信息
stackTrace ::
堆栈跟踪 ：：
inside main java.lang.ClassNotFoundException: SmsURLHelper
at java.net.URLClassLoader.run(Unknown Source)
at java.security.AccessController.doPrivileged(Native Method)
at java.net.URLClassLoader.findClass(Unknown Source)
at java.lang.ClassLoader.loadClass(Unknown Source)
at sun.misc.Launcher$AppClassLoader.loadClass(Unknown Source)
at java.lang.ClassLoader.loadClass(Unknown Source)
at java.lang.Class.forName0(Native Method)
at java.lang.Class.forName(Unknown Source)
at com.tcs.sms.actions.SmsUtil.invoke(SmsUtil.java:131)
Note : The SmsHelper class is present.
注意：存在 SmsHelper 类。
Please suggest if I have missed any thing.
如果我错过了什么，请提出建议。
回答by Mik378
Use the fully qualified name.
使用完全限定的名称。
Example:
例子：
Class.forName("com.yourownpackage.SmsHelper");
回答by Arun Kumar
Please write the Class name along with its package name(in which your class resides) as a String in the forNamemethod. Here is the sample to use:
请在forName方法中将类名及其包名（您的类所在的包名）写成一个字符串。这是要使用的示例：
String className = "com.testpackage.SmsHelper";
Class c = Class.forName(className);
回答by hmakholm left over Monica
class.forName()expects a fully qualified class name -- such ascom.tcs.foo.bar.SmsURLHelper.Why not just write
SmsHelper.class-- that will be checked at compile time, compiled to an appropriate call to class.forName().
class.forName()需要一个完全限定的类名——例如com.tcs.foo.bar.SmsURLHelper.为什么不直接写
SmsHelper.class——这将在编译时检查，编译为对 class.forName() 的适当调用。
