Re: Exact Name of File in Root-Part of IFS

Try stat() to get the fileid, then QlgGetPathFromFileID() to fetch the path
string. API Manual has a minimal example, just use stat() instead of
hardcoding the fileid.

Stu








On Mon, Dec 21, 2015 at 1:27 AM, <j.beckeringh@xxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:

Jan,

I ended up doing exactly what you describe (opendir / readdir), but I can
imagine that that is not feasible for directories of more than 50.000
files. I always found it strange that I can't find a stat-like api with
st_ino as input instead of filename.

Joep Beckeringh
Pantheon Automatisering

Jan Grove Vejlstrup <jgv@xxxxxxxx>

18-12-2015 16:44

Exact Name of File in Root-Part of IFS

Hello

The root part of the IFS isn't case-sensitive. This means, that I can't
have abc123.jpg and ABC123.jpg stored in the same directory. If I have
abc123.jpg in the IFS and I use the API access with the path-parameter
/directory/ABC123.jpg to check the existence of the file, the answer is
positive. But ABC123.jpg isn't stored, abc123.jpg is. Do you know of a
way to get the exact name of the file (in casu abc123.jpg) stored in the

IFS?

I know, I could use the APIs opendir and readdir to read through the
whole directory to get the exact names. But my directory has more than
50'000 files stored in it. So this wouldn't perform well, if I was
interested only in one file.

Thank you very much in advance.

Best regards

Jan

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