directorys

[Robin]

How would I set chmod on a directory without making one and without using
exec or system...and how would I and what is a SUID bit for mkdir ("test",
0755)
is in the first number?

-Robin

 

[Simon Taylor]

How would I set chmod on a directory without making one and without using
exec or system...and how would I and what is a SUID bit for mkdir ("test",
0755)
is in the first number?

-Robin

The general solution is to use the chmod() function, see:

perldoc -f chmod

Here's a sample:

chmod 0600, $dir;

I'm sorry but I don't understand the second part of your question.

Regards,

Simon Taylor

 

[Gunnar Hjalmarsson]

How would I set chmod on a directory without making one and without
using exec or system...

You would do as the documentation for the chmod() function suggests.

 

[Tad McClellan]

How would I set chmod on a directory without making one and without using
exec or system...

perldoc -f chmod


Is it possible that you didn't even look for a like-named function?

 

[Robin]

the book I have said it works for files, and I was assuming it didn't work
for directories...

thanks a lot, appreciate it,

-Robin

 

[Robin]

another thing, is it just
$nameofdir = "somedirectory/"
-e "$nameofdir"

(not that this code runs)
or likewise filetest operator to test on a directory?

Thanks,
-Robin

 

[Ben Morrow]

another thing, is it just
$nameofdir = "somedirectory/"
-e "$nameofdir"

(not that this code runs)
or likewise filetest operator to test on a directory?

A directory *is* a file. Anything that works on a file works on a
directory (with the exception of opening it for writing).

Ben

 

[Tad McClellan]

another thing, is it just
$nameofdir = "somedirectory/"
-e "$nameofdir"

No, it is: -e $nameofdir


perldoc -q vars

What's wrong with always quoting "$vars"?



[snip TOFU. Please learn the proper way of quoting followups]