[英]Using wildcard when listing files in directory with java
Why does a wildcard not work in java code below? 为什么通配符在下面的Java代码中不起作用?
My request looks like http://localhost:8080/App/DataAccess?location=Dublin
我的请求看起来像
http://localhost:8080/App/DataAccess?location=Dublin
rob@work:~$ ls /usr/local/CustomAppResults/Dublin/*/.history
/usr/local/CustomAppResults/Dublin/team1/.history
/usr/local/CustomAppResults/Dublin/team2/.history
/usr/local/CustomAppResults/Dublin/team3/.history
Servlet code (DataAccess.java). Servlet代码(DataAccess.java)。
(DataAccess.java:27)
refers to the for loop .. (DataAccess.java:27)
引用for循环..
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
File[] files = finder("/usr/local/CustomAppResults/" +
request.getParameter("location") + "/*/");
for (int i = 0; i < files.length; i++){
System.out.println(files[i].getName());
}
}
private File[] finder(String dirName) {
File dir = new File(dirName);
return dir.listFiles(new FilenameFilter() {
public boolean accept(File dir, String filename) {
return filename.endsWith(".history");
}
});
}
Error: 错误:
The server encountered an internal error that prevented it
from fulfilling this request.
java.lang.NullPointerException
com.example.servlets.DataAccess.doGet(DataAccess.java:27)
javax.servlet.http.HttpServlet.service(HttpServlet.java:621)
javax.servlet.http.HttpServlet.service(HttpServlet.java:728)
The method public File[] listFiles(FilenameFilter filter)
方法
public File[] listFiles(FilenameFilter filter)
Returns null if this abstract pathname does not denote a directory, or if an I/O error occurs.
如果此抽象路径名不表示目录,或者发生I / O错误,则返回null。
( http://docs.oracle.com/javase/7/docs/api/java/io/File.html ) ( http://docs.oracle.com/javase/7/docs/api/java/io/File.html )
So, why do you get this situation? 那么,为什么会出现这种情况? You are trying to use a wildcard char (
*
) that is evaluated by your shell, but won't be evaluated in new File(path)
. 您正在尝试使用由外壳程序评估但不能在
new File(path)
评估的通配符char( *
new File(path)
。 The new File(path)
constructor only works for exact paths. new File(path)
构造函数仅适用于确切路径。
Things like DirectoryScanner
(Apache Ant) or FileUtils
(Apache commons-io) will solve your problem. DirectoryScanner
(Apache Ant)或FileUtils
(Apache commons-io)之类的东西可以解决您的问题。 See the comments above for further details on possible solutions, including the Java 7 NIO approach ( Files.newDirectoryStream( path, glob-pattern )
). 有关可能的解决方案的更多详细信息,请参见上面的注释,包括Java 7 NIO方法(
Files.newDirectoryStream( path, glob-pattern )
)。
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