Transfer from: http: //blog.csdn.net/yiluoak_47/article/details/7760385
Use a relative path java to read the file
1.java project environment, using examples java.io read the file using a relative path:
* directory structure:
DecisionTree
| src ___
| ___ com.decisiontree.SamplesReader.java
| Resource ___
| ___ train.txt, the Test. TXT
* SamplesReader.java:
String filepath = "Resource / train.txt"; // note the contents of filepath;
File File = new new File (filepath);
......
* we pay attention to the content filepath, java.io default to the current user directory ( "user.dir"), namely: the root Engineering
Catalog: the "D \ DecisionTree", therefore, at this time the relative path (path based user.dir path) is "resource / train.txt"
. In this way, JVM can get the full path based on "user.dir" and "resource / train.txt" (that is, absolute path
diameter) "D: \ DecisionTree \ resource \ train.txt", never found train.txt file.
* Note: the start of the free diagonals relative path "
filepath = "resource / train.txt";
instead = filepath "/ Resource / train.txt"; // error!
2, javaEE environment, an example of using Classloader read xml with a relative path:
* see previous article written by " xml file is read by a virtual path or a relative path, to avoid hard-coded. "
* As follows:
Java use a relative path to read xml file:
a, xml file storage location generally three:
1. Put the WEB-INF;
under 2.xml files in / WEB-INF / classes directory or the classpath jar package;
3. placed with its java parse a similar package, not necessarily the CLASSPATH;
two, corresponding to the relative path using two read methods:
method unverified :( a)
the discharge xml file in the WEB-INF directory, then the
program code:
the InputStream GetServletContext IS = () the getResourceAsStream ( "/WEB-INF/xmlfile.xml");.
method 2: xml files in / the WEB-INF / classes directory or classpath jar package, you can use a static ClassLoader
method getSystemResourceAsStream (String s) read;
program code:
S_xmlpath = String "COM / SPF / Web / EXT / Hotspot / hotspotxml / hotspot.xml";
the InputStream in = ClassLoader.getSystemResourceAsStream (s_xmlpath);
Method three: xml freely in the path of a packet:
String = s_xmlpath "COM / SPF /web/ext/hotspot/hotspotxml/hotspot.xml ";
ClassLoader HotspotXmlParser.class.getClassLoader classLoader = ();
the InputStream in = ClassLoader.getResourceAsStream (s_xmlpath);
