Java门路题目处理计划(国外英文资料).docVIP

Java路径问题解决方案(国外英文资料) Dir is specified, usually the invocation directory of the Java virtual machine. This means that you should not use a relative path when using classes in the java.io package. Otherwise, although it might be normal in the SE program, its a problem if you dont get it right in the EE program. The relative path of the classpath The relative path of the path relative to file: / D: / mywork/javaprj/MyTest/bin. In this case, bin is the classpath for the project. All the Java source files are compiled.class files are copied into this directory. Class 2 loading directory (that is, the load directory when running a class) Whether its a generic Java project or a web project, locate the first level of directory where you can see the package path InputStream is = readwrite.class.getclassloader ().getclassloader ().getclassloader ().getclassloader ().getclassloader () Among them, the path of the DeviceNO file is the project name \ SRC \ DeviceNO; The first level directory of the class ReadWrite is in the SRC directory. Similar to 1, the difference is that the method must begin with / InputStream is = readwrite.class.getresourceasstream ( DeviceNO ); Among them, the path of the DeviceNO file is the project name \ SRC \ DeviceNO; The first level directory of the class ReadWrite is in the SRC directory. Three. The web project root directory is obtained You can create a servlet and write the following statement in its init method The ServletContext sc = this.getservletcontext (); String temp = sc. GetRealPath (/); Get the output path of similar: D: \ Apache \ Tomcat6.0 \ webapps \ windpower \ (windpower project name), if is to call the s1 getRealPath ( ) is the output D: \ Apache \ Tomcat6.0 \ webapps \ windpower (note that in the last one less \) In httpServletRequest, the following statement can be used String cp = request.getsession ().getservletcontext ().getrealpath (/); The resulting output path is similar: D: \ Apache \ Tomcat6.0 \ webapps \ windpower \ Cl