If I try the above method with the filename as viewing.proprties and that file resides in
This is the error I get :
------- Error ----------
url = file:/C:/Program%20Files/viewing.properties java.io.FileNotFoundException: C:\Program%20Files\viewing.properties (The system cannot find the path specified) at java.io.FileInputStream.open(Native Method) at java.io.FileInputStream.<init>(FileInputStream.java:106) at java.io.FileInputStream.<init>(FileInputStream.java:66) at TestFile.getProperties(TestFile.java:79) at TestFile.main(TestFile.java:19)
** ANY IDEA How to resolve this
The code works fine - if there is no SPACE in the directory. Is there a way to get around this issue
SCEA, SCBCD, SCJP1.4, OOAD-UML, OCP 9i
Joined: Aug 10, 2005
Your error example shows %20 in the path to the file. Replace that with a space. How do you get the path?
See the URLDecoder class(?) for a method to remove it.