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Home/ Questions/Q 756279
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T15:13:36+00:00 2026-05-14T15:13:36+00:00

I have a servlet that needs to write out files that have a user-configurable

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I have a servlet that needs to write out files that have a user-configurable name. I am trying to use URI encoding to properly escape special characters, but the JRE appears to automatically convert encoded forward slashes %2F into path separators.

Example:

File   dir = new File("C:\Documents and Setting\username\temp");
String fn  = "Top 1/2.pdf";
URI    uri = new URI( dir.toURI().toASCIIString() + URLEncoder.encoder( fn, "ASCII" ).toString() );
File   out = new File( uri );

System.out.println( dir.toURI().toASCIIString() );
System.out.println( URLEncoder.encode( fn, "ASCII" ).toString() );
System.out.println( uri.toASCIIString() );
System.out.println( output.toURI().toASCIIString() );

The output is:

file:/C:/Documents%20and%20Settings/username/temp/
Top+1%2F2.pdf   
file:/C:/Documents%20and%20Settings/username/temp/Top+1%2F2.pdf
file:/C:/Documents%20and%20Settings/username/temp/Top+1/2.pdf

After the new File object is instantiated, the %2F sequence is automatically converted to a forward slash and I end up with an incorrect path. Does anybody know the proper way to approach this issue?

The core of the problem seems to be that

uri.equals( new File(uri).toURI() ) == FALSE

when there is a %2F in the URI.

I’m planning to just use the URLEncoded string verbatim rather than trying to use the File(uri) constructor.

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  1. Editorial Team
    Editorial Team
    2026-05-14T15:13:36+00:00Added an answer on May 14, 2026 at 3:13 pm

    The new File(URI) constructs the file based on the path as obtained by URI#getPath() instead of -what you expected- URI#getRawPath(). This look like a feature “by design”.

    You have 2 options:

    1. Run URLEncoder#encode() on fn twice (note: encode(), not encoder()).
    2. Use new File(String) instead.
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