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umask
Changes the current umask
(PHP 4, PHP 5)
Example 673. umask() example<?php Code Examples / Notes » umask18-feb-2006 04:31
Using (cmask - umask) is a wrong way to calculate the new mask: 0022 - 0700 = 0656 WRONG 0700 & ~0022 = 0700 CORRECT Correct php code: <?php $rmask = ($cmask & ~$umask); ?> maulwuff
umask takes away the given values from the standard mask 777. A graphical view shows this better: standard: rwxrwxrwx = 777 will get with umask 002: rwxrwxr-x = 775 or will get with umask 077: rwx------ = 700 and so on. andi
To play around with umasks and permissions use this little fragment: <? $umask = 0012; $perm = 0777; printf("umask: %04o perm: %04o result: %04o\n", $umask,$perm,$perm & (0777 - $umask)); ?> sam
The first comment perhaps didn't quite make clear what's on with your umask and the permissions. The permission passed to a command is first bitwise ANDed with the _INVERSE_ of the current umask, then applied to the file. For example, umask = 0011 and permission = 0775 The inverse of 0011 = 0766 0775 AND 0766 = 111.111.101 AND 111.110.110 = 111.110.100 = 0764 shad@tw
Notice that directory(s) and file(s) sometimes have different results. <?php umask(0670); //- set umask $handle = fopen('file', 'w'); //- 0006 mkdir("/path/dir"); //- 0107 ?> calculate the result: <?php $umask = 0670; umask($umask); //- if you are creating a new directory, $permission = 0777; //- if you are creating a new file, $permission = 0666. printf( "result: %04o", $permission & ( 0777 - $umask) ); ?> BTW, as the manual said, the form of umask() is "int umask ( [int mask] )", so if you want to print/echo any umask, don't forget to convert it from DEC (because it returns a "int") to OCT. <?php $umask = umask(); //- returns the current umask, which is a "int" $umask = decoct($umask); //- Now, $umask is a "string" echo $umask; ?> Don't forget that the argument(parameter) is a "int", too. <?php umask(777); //- WRONG! Even though you maybe use "umask 777" in some OS. umask(0777); //- OK ?> If there was any mistake, please correct my statement. trisk
I thought I would clarify the numbering scheme used here, as it confused me at first. On the UNIX console, the command: umask "blah" In this instance, the umask command forces "blah" to be an octal number, regardless of how many digits you use and regardless of any leading zeroes. In PHP, umask() does not default to octal as the console command does, it uses whatever numeric format you specify. For example: umask(213); This uses the decimal integer 213 and not the octal number 213 as you would expect when using the console command. In this case, it would set the umask to the octal number "325". To enter the number as octal, just add one or more zeroes to the left of the number: umask(0213); umask(07); umask(0044); etc. sean
"It is better to change the file permissions with chmod() after creating the file." The usual lacking of security knowledge within the PHP team rears its head once again. You *always* want to have the file created with the proper permission. Let me illustrate why: (a) you create new file with read permissions (b) an attacking script opens the file (c) you chmod the file to remove read permissions (d) you write sensitive data to the file Now, you might think that the changes of an attacking script getting to open the file before you chmod them are low. And you're right. But low changes are never low enough - you want zero chance. When creating a file that needs increased permissions, you always need to create the file with the proper permissions, and also create it with O_EXCL set. If you don't do an exclusive create, you end up with this scenario: (a) attacker creates the file, makes it writable to everyone (b) you open the file with restricted permissions, but since it already exists, the file is merely opened and the permissions left alone (c) you write sensitive data into the insecure file Detecting the latter scenario is possible, but it requires a bit of work. You have to check that the file's owner and group match the script's (that is, posix_geteuid(), not myuid()) and check the permissions - if any of those are incorrect, then the file is insecure - you can attempt to unlink() it and try again while logging a warning, of course. The only time when it is reasonable or safe to chmod() a file after creating it is when you want to grant extra permissions instead of removing them. For example, it is completely safe to set the umask to 0077 and then chmoding the files you create afterward. Doing truly secure programming in PHP is difficult as is, and advice like this in the documentation just makes things worse. Remember, kids, anything that applies to security in the C or UNIX worlds is 100% applicable to PHP. The best thing you can possibly do for yourself as a PHP programmer is to learn and understand secure C and UNIX programming techniques. |
Change Languagebasename chgrp chmod chown clearstatcache copy delete dirname disk_free_space disk_total_space diskfreespace fclose feof fflush fgetc fgetcsv fgets fgetss file_exists file_get_contents file_put_contents file fileatime filectime filegroup fileinode filemtime fileowner fileperms filesize filetype flock fnmatch fopen fpassthru fputcsv fputs fread fscanf fseek fstat ftell ftruncate fwrite glob is_dir is_executable is_file is_link is_readable is_uploaded_file is_writable is_writeable lchgrp lchown link linkinfo lstat mkdir move_uploaded_file parse_ini_file pathinfo pclose popen readfile readlink realpath rename rewind rmdir set_file_buffer stat symlink tempnam tmpfile touch umask unlink |