sed - matching pattern one but not pattern two

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# 1  
Old 05-07-2009
sed - matching pattern one but not pattern two


I have the following file:
# /etc/pam.d/common-password - password-related modules common to all services
# This file is included from other service-specific PAM config files,
# and should contain a list of modules that define the services to be
# used to change user passwords. The default is pam_unix2 in combination
# with pam_pwcheck.
# The "nullok" option allows users to change an empty password, else
# empty passwords are treated as locked accounts.
# To enable Blowfish or MD5 passwords, you should edit
# /etc/default/passwd.
# Alternate strength checking for passwords should be configured
# in /etc/security/pam_pwcheck.conf.
# pam_make can be used to rebuild NIS maps after password change.
password required retry=5 ask_oldauthok min=disabled,8,8,8,8 passphrase=0 random=0 force=everyone
password required nullok
password required nullok use_authtok
#password required /var/yp

What I would like to do is generate a sed command that will:
Add "use_first_pass" to a line if the line contains "" but the line does *not* contain "use_first_pass".
The remainer of the file should be left alone.

I can generate the logic with:
a regex of:
(.*pam_unix2\.so.*) (?!.*use_first_pass.*)

I am having trouble translating that into a sed expression.

# 2  
Old 05-07-2009
If awk is allowed:

awk '/ && !/use_first_pass/{$0=$0 FS "use_first_pass"}1' file

# 3  
Old 05-07-2009
if you can use Python
for line in open("file"):
    if "" in line and not "use_first_pass" in line:
        print "use_first_pass "  + line
        print line

or inplace version
import fileinput
for line in fileinput.FileInput("file",inplace=1):
    if "" in line and not "use_first_pass" in line:
        print "use_first_pass "  + line
        print line

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