That nailed my problem. GLOB_ONLYDIR is a defined constant. It echos as 8192.
I inadvertently treated it as a string with quotes, which I didn't include in my
original post.
I had forgotten that most function flags are simply constants which can be
tested for and printed, etc.
Many thanks....
Jim Lucas wrote:
Al wrote:
I thought the "." had meaning since the manual user notes used it
Definitely, my php is version 4.4.4; it's on a virtual host.
I found the glob file for the GNU C Library and GLOB_ONLYDIR doesn't
seem to be defined correctly, the file appears to rather old.
echo GLOB_ONLYDIR;
does this work?
I used the numerical value "8196" and it works fine. I'm a little
reluctant to use the numerical value as I want my code to be generally
universal. Maybe the numerical value is safer anyhow. What do you
think?
I can alternatively just filter out the files in a second step.
Al.............
Jim Lucas wrote:
Al wrote:
I can't use flag "GLOB_ONLYDIR" to work on a Linux, php4.4.4. I
only want the dirs.
This works; sort of:
print_r(glob('../*',)); //It lists all the files AND directories.
Yet it seems to ignore the lack of ".*" and finds xxx.yyy. That's
OK, I can delete the files
remember this is linux you are talking about, not windows.
A '*' means anything, including a '.'
This fails:
print_r(glob('../*', GLOB_ONLYDIR));
well, double check your php version, because this should be built in
whit the version that you are running.
And I get: "Warning: glob() expects parameter 2 to be long, string
given in foo"
I can't find anything in the manual or when Googling.
Anyone can explain this for me?
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