Sponsored Content
Special Forums Cybersecurity More than 8 characters for passwd Post 302459926 by Daniel Gate on Tuesday 5th of October 2010 03:38:05 PM
Old 10-05-2010
More than 8 characters for passwd

Currently, the length of passwd is set to 8, which is the default.
I am to increase the length to more than 8. Please advise of the supported
options and steps.
 

10 More Discussions You Might Find Interesting

1. UNIX for Advanced & Expert Users

passwd

Hi, Besides of the command "passwd" on Solaris and HP-UX and IRIX that allow users to change their passwords on the system. Is there anyother way a user can change his/her own password. Do any of these systems have a GUI interface to allow the user doing so? Thanks (3 Replies)
Discussion started by: vtran4270
3 Replies

2. UNIX for Advanced & Expert Users

passwd

I have to change more then 200 User at once the password (security-dday). The programm passwd will answers (new password + again) How can i do this in a script? thanks for answers (5 Replies)
Discussion started by: Erwin Stocker
5 Replies

3. Shell Programming and Scripting

How to replace characters with random characters

I've got a file (numbers.txt) filled with numbers and I want to replace each one of those numbers with a new random number between 0 and 9. This is my script so far: #!/bin/bash rand=$(($RANDOM % 9)) sed -i s//$rand/g numbers.txtThe problem that I have is that it replaces each number with just... (2 Replies)
Discussion started by: hellocatfood
2 Replies

4. Solaris

passwd cmd reenables passwd aging in shadow entry

Hi Folks, I have Solaris 10, latest release. We have passwd aging set in /etc/defalut/passwd. I have an account that passwd should never expire. Acheived by emptying associated users shadow file entries for passwd aging. When I reset the users passwd using passwd command, it re enables... (3 Replies)
Discussion started by: BG_JrAdmin
3 Replies

5. Shell Programming and Scripting

Replace special characters with Escape characters?

i need to replace the any special characters with escape characters like below. test!=123-> test\!\=123 !@#$%^&*()-= to be replaced by \!\@\#\$\%\^\&\*\(\)\-\= (8 Replies)
Discussion started by: laknar
8 Replies

6. UNIX for Dummies Questions & Answers

etc/passwd help

Hi! i've been searching a way to display the users who are in the /etc/passwd directory...using ls or grep or cat command should do it?i can't find a way to get this right..i mean none of the preview commands function sounded good to me to use... (9 Replies)
Discussion started by: strawhatluffy
9 Replies

7. AIX

When did AIX start using /etc/security/passwd instead of /etc/passwd to store encrypted passwords?

Does anyone know when AIX started using /etc/security/passwd instead of /etc/passwd to store encrypted passwords? (1 Reply)
Discussion started by: Anne Neville
1 Replies

8. Shell Programming and Scripting

sed replacing specific characters and control characters by escaping

sed -e "s// /g" old.txt > new.txt While I do know some control characters need to be escaped, can normal characters also be escaped and still work the same way? Basically I do not know all control characters that have a special meaning, for example, ?, ., % have a meaning and have to be escaped... (11 Replies)
Discussion started by: ijustneeda
11 Replies

9. Shell Programming and Scripting

Remove first 2 characters and last two characters of each line

here's what im trying to do. i have a file containing lines similar to this: data.txt: 1hsRmRsbHRiSFZNTTA1dlEyMWFkbU5wUW5CSlIyeDFTVU5SYjJOSFRuWmpia0ZuWXpKV2FHTnRU 1lKUnpWMldrZFZaMG95V25oYQpSelEyWTBka2QyRklhSHBrUjA1b1kwUkJkd3BOVXpWM1lVaG5k... (5 Replies)
Discussion started by: SkySmart
5 Replies

10. Shell Programming and Scripting

Outputting characters after a given string and reporting the characters in the row below --sed

I have this fastq file: @M04961:22:000000000-B5VGJ:1:1101:9280:7106 1:N:0:86 GGGGGGGGGGGGCATGAAAACATACAAACCGTCTTTCCAGAAATTGTTCCAAGTATCGGCAACAGCTTTATCAATACCATGAAAAATATCAACCACACCA +test-1 GGGGGGGGGGGGGGGGGCCGGGGGFF,EDFFGEDFG,@DGGCGGEGGG7DCGGGF68CGFFFGGGG@CGDGFFDFEFEFF:30CGAFFDFEFF8CAF;;8... (10 Replies)
Discussion started by: Xterra
10 Replies
mkpwdict(1M)						  System Administration Commands					      mkpwdict(1M)

NAME
mkpwdict - maintain password-strength checking database SYNOPSIS
/usr/sbin/mkpwdict [-s dict1,... ,dictN] [-d destination-path] DESCRIPTION
The mkpwdict command adds words to the dictionary-lookup database used by pam_authtok_check(5) and passwd(1). Files containing words to be added to the database can be specified on the command-line using the -s flag. These source files should have a single word per line, much like /usr/share/lib/dict/words. If -s is omitted, mkpwdict will use the value of DICTIONLIST specified in /etc/default/passwd (see passwd(1)). The database is created in the directory specified by the -d option. If this option is omitted, mkpwdict uses the value of DICTIONDBDIR specified in /etc/default/passwd (see passwd(1)). The default location is /var/passwd. OPTIONS
The following options are supported: -s Specifies a comma-separated list of files containing words to be added to the dictionary-lookup database. -d Specifies the target location of the dictionary-database. FILES
/etc/default/passwd See passwd(1). /var/passwd default destination directory ATTRIBUTES
See attributes(5) for descriptions of the following attributes: +-----------------------------+-----------------------------+ | ATTRIBUTE TYPE | ATTRIBUTE VALUE | +-----------------------------+-----------------------------+ |Availability |SUNWcsu | +-----------------------------+-----------------------------+ |Interface Stability |Evolving | +-----------------------------+-----------------------------+ SEE ALSO
passwd(1), attributes(5), pam_authtok_check(5) SunOS 5.10 1 Jun 2004 mkpwdict(1M)
All times are GMT -4. The time now is 05:43 PM.
Unix & Linux Forums Content Copyright 1993-2022. All Rights Reserved.
Privacy Policy