Shell Programming and Scripting

it does not work

> usermod -p $(openssl passwd "test123" ) raj
Illegal variable name

Sorry try usermod -p `openssl passwd "test123"` raj

this one is working but shows me permission denied i will ask admin to run and advise

---------- Post updated at 01:52 PM ---------- Previous update was at 01:38 PM ----------

Hi Chubler,

My collegues also do not have admin/root right to run openss1 and below line

Code:

printf "%s\n" "$PASS" "$PASS" | (passwd raj) > /home/raj/Password/out.log

as we need to put the old passwords......but we do have rights to make cron job....is there any way we can input the old password also it will be there in the file.txt e.g today is 18-Jun so old password will be of 17-Jun

It is not an administrator command... Also, it is openssl, the letter, not openss1, the number.

...will probably not work because plain passwd is smart enough to demand a terminal on most systems. If you have expect you might be able to kludge something in that language, since it spoofs a terminal.

Code:

usermod -p `openssl passwd "liver12#"` raj

shows me permission denied
UX: usermod: ERROR: Permission denied.


other way of doing it might be getting the old password from file.txt also and using it in program to change password, when i try to do it manually it ask for existing password and twice the new password

Does the following work from the command line?

Code:

$ printf "%s\n" "oldpass" "newpass" "newpass" | (passwd raj)

If not do you have passwordless ssh access to the server? This could help as we can use the ssh -t option to force a pseudo-tty?

Code:

$ ssh -t raj@localhost id

Yes below mentioned is working

Code:

 printf "%s\n" "oldpass" "newpass" "newpass" | (passwd rsaini)

Enter existing login password:

i am not able to fetch 17-June password from below command

Code:

grep -e"``date +"%d`-1-date +%b"`"