RE: [PHP] umask() and chmod()

On 15 April 2004 16:26, David T-G wrote: > Hi, all -- > > When I move_uploaded_file() a file into place, I want to give it the > correct permissions; by default they are 600 (rw-/---/--). I already > have the umask set correctly for any given situation in > anticipation of > creating directories, and that works for creating files from scratch, > but chmod() needs a permissions setting rather than a umask. > > The challenge is in representing this as octal. With some > mucking around > I was able to print > > $u = decoct(umask()) ; > $m = 0777 ; > $r = decoct($m) - $u ; > print "The setting is $r\n" ; I've just re-read this properly, and realised you want to feed the end value to chmod() -- so all the guff about octal representations is a complete red herring. What you want to feed to the 2nd parameter of chmod() is just: 0777 ^ umask() Within the computer, the value is just an integer, and that's what umask() provides and what chmod() expects for its 2nd argument -- so no conversions required. Even 0777 is an integer -- it's just a different way of representing 511 which is more understandable to us poor humans in that context. Your original offering was, in fact, barking up the wrong tree -- what it was printing out, although it looked right, was the *decimal* value 664, which in octal would be 01230 and really not what you want. I guess what this boils down to is that integers are just integers when you manipulate them in your script -- octal, decimal, hex and any other representations are just convenient notations for making them more human readable. Cheers! Mike --------------------------------------------------------------------- Mike Ford, Electronic Information Services Adviser, Learning Support Services, Learning & Information Services, JG125, James Graham Building, Leeds Metropolitan University, Beckett Park, LEEDS, LS6 3QS, United Kingdom Email: m.ford@leedsmet.ac.uk Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211