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Re: Open file in working directory
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Posted: May 14, 2007 11:30 PM
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> I created a JSP in a /webapps/TEST directory of my Tomcat
> installation and want to simply open/read a file that also
> exists in this same directory.
>
> My apps works correctly if I give the filename an absolute
> path:
>
> i.e.
>
> File configFile = new File("C:\\TEST\\test.cfg");
>
> however, I want to avoid using a hard-coded absolute path
> and just open the file from where my jsp also resides.
> I'm sure this is doable pretty easily, but can't find a
> a straight answer.
>
> Any help would be appreciated.
I am not getting exactly whats ur working dir. What u can do is,
1. Get the working directory.
2. Instead of hard-coding the entire path, use the relative path from there.
Lets say ur JSP is in c:\tomcat\webapps\test
And when u try to read the working directory, it gives c:\tomcat.
Then u can replace ur foll line:
<b>File configFile = new File("C:\\TEST\\test.cfg");</b>
with
<b>File configFile = new File("test\\test.cfg");</b>
Hope this works for u.
Regards,
Rajeev
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