How to print without authtime?

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11-06-2015

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How to print without authtime?

Can someone help me how to print list of "cn" without lastauthtime.
Note: Each cn is separated with blank line.

Code:

Input:

cn=jersey
authtime=2012

cn=mike
authtime=2011

cn=steve

cn=jersey
authtime=2012

cn=rob

cn=matt

Code:

OutPut:

cn=steve
cn=rob
cn=matt

---------- Post updated at 03:26 PM ---------- Previous update was at 03:24 PM ----------

I tried below But its printing in reverse way.

Code:

BEGIN {
        FS="\n"
        RS="\n\n"
        lastauthtime  = ""

}

/lastauthtime/ && NF==2 {
        sub(/user=/,"",$1)
        lastauthtime = lastauthtime " " $1
}

END {
        print "lastauthtime::",lastauthtime
}

---------- Post updated at 04:36 PM ---------- Previous update was at 03:26 PM ----------

Can anyone help me on this?

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11-06-2015

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How about

Code:

awk '/cn=/ {getline X; if (!X) print} ' file
cn=steve
cn=rob
cn=matt

?

---------- Post updated at 22:44 ---------- Previous update was at 22:43 ----------

Or even

Code:

awk '/cn=/ {getline X}  !X ' file
cn=steve
cn=rob
cn=matt

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11-06-2015

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Assuming that every segment in your file contains a cn=name line, this might be a little bit simpler (and more portable) to get the output you said you wanted:

Code:

awk 'NF==1' RS= file

Modifying your code so it prints the names of users (if cn=name identifies a user's name) that do not have an authtime=year entry (although naming the variable that contains that list lastauthtime is not at all descriptive), you could try:

Code:

awk '
BEGIN {
        FS="\n"
        RS=""
}

NF==1 {
        sub(/cn=/,"",$1)
        lastauthtime = lastauthtime " " $1
}

END {
        print "lastauthtime::" lastauthtime
}' file

which produces the output:

Code:

lastauthtime:: steve rob matt

Or, more simply (and, perhaps, with less confusing names):

Code:

awk '
BEGIN {
	FS = "="
        RS = ""
}

NF==2 {
        noauthtime = noauthtime " " $2
}

END {
        print "noauthtime::" noauthtime
}' file

which produces the output:

Code:

noauthtime:: steve rob matt

As always, if you want to use any of these scripts on a Solaris/SunOS system, change awk in these scripts to /usr/xpg4/bin/awk or nawk.

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11-06-2015

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Just for fun:

Code:

(cat file;echo "") | sed -n -e '/^cn=/h;/^authtime=/{n;b};/^$/{x;p}'

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11-07-2015

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Hi.

Using a new utility sift (here written as gogrep to avoid a name collision at our site):

Code:

#!/usr/bin/env bash

# @(#) s1	Demonstrate conditional matching, gogrep.
# For "sift" (aka "gogrep" here), see:
# https://sift-tool.org/

# Utility functions: print-as-echo, print-line-with-visual-space, debug.
# export PATH="/usr/local/bin:/usr/bin:/bin"
LC_ALL=C ; LANG=C ; export LC_ALL LANG
pe() { for _i;do printf "%s" "$_i";done; printf "\n"; }
pl() { pe;pe "-----" ;pe "$*"; }
db() { ( printf " db, ";for _i;do printf "%s" "$_i";done;printf "\n" ) >&2 ; }
db() { : ; }
C=$HOME/bin/context && [ -f $C ] && $C gogrep

FILE=${1-data1}

pl " Input data file $FILE:"
cat $FILE

pl " Results:"
gogrep --not-followed-within="1:authtime" "cn=" $FILE

exit 0

producing:

Code:

$ ./s1

Environment: LC_ALL = C, LANG = C
(Versions displayed with local utility "version")
OS, ker|rel, machine: Linux, 2.6.26-2-amd64, x86_64
Distribution        : Debian 5.0.8 (lenny, workstation) 
bash GNU bash 3.2.39
gogrep sift 0.4.1 (linux/amd64)

-----
 Input data file data1:
cn=jersey
authtime=2012

cn=mike
authtime=2011

cn=steve

cn=jersey
authtime=2012

cn=rob

cn=matt

-----
 Results:
cn=steve
cn=rob
cn=matt

In addition to flexibility, sift is impressively fast.

Best wishes ... cheers, drl

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