How to Use Sed to Find Strings Between Two Patterns – Text Processing with Sed

sedtext processing

I have file content like this:

aaa accounting exec ...
aaa accounting exec ...
aaa accounting commands ..
aaa accounting commands ..
aaa accounting commands ..
aaa accounting commands ..
aaa accounting commands ..
aaa accounting commands ..
aaa accounting network ..
aaa accounting connection ..
aaa accounting system ..
!
aaa accounting exec default
 action-type start-only
 group tacacs+
!
aaa accounting exec default stop-only group tacacs+

The output should be like this:

aaa accounting exec default ..
aaa accounting exec default
 action-type start-only
 group tacacs+
!
aaa accounting exec default ..

I have tried following sed command:

sed -n '/aaa accounting exec default/,/!/p' AboveFileContent.txt

But it's not produce what I want.

What would be the solution? I have tried using awk also but same result is coming. What would be the command to get the exact output?

Best Answer

I'd use awk for this:

awk '
    /aaa accounting exec default/ {print; exec=1; next} 
    exec {
        if (/^ /) {print; next} else if (/^!/) {print}
        exec=0
    }
' filename

Passing the pattern, use awk's -v option, and then the pattern match operator ~:

awk -v patt='aaa accounting exec default' '
    $0 ~ patt {print; exec=1; next} 
    exec {
        if (/^ /) {print; next} else if (/^!/) {print}
        exec=0
    }
' filename

Related Solutions

Ubuntu – How to find all patterns between two characters

First of all, your grep -Po '"\K[^"]*' file idea fails because grep sees both "One" and ". the second is here" as being inside quotes. Personally, I'd probably just do

$ grep -oP '"[^"]+"' file | tr -d '"'
One
Two 
 Three 
Four

But that is two commands. To do it with a single command, you could use one of:

Perl
```
$ perl -lne '@F=/"\s*([^"]+)\s*"/g; print for @F' file 
One
Two 
Three 
Four
```
Here, the @F array holds all matches of the regex (a quote, followed by as many non-" as possible until the next "). The print for @F just means "print each element of @F.

Perl

$ perl -F'"' -lne 'for($i=1;$i<=$#F;$i+=2){print $F[$i]}' file 
One
Two 
 Three 
Four

To remove leading/trailing spaces from each match, use this:

perl -F'"' -lne 'for($i=1;$i<=$#F;$i+=2){$F[$i]=~s/^\s*|\s$//; print $F[$i]}' file

Here, Perl is behaving like awk. The -a switch causes it to automatically split input lines into fields on the character given by -F. Since I have given it ", the fields are:

$ perl -F'"' -lne 'for($i=0;$i<=$#F;$i++){print "Field $i: $F[$i]"}' file 
Field 0: first matched is 
Field 1: One
Field 2: . the second is here
Field 3: Two 
Field 0: and here are in second line
Field 1:  Three 
Field 2: 
Field 3: Four
Field 4: .

Because we are looking for text between two consecutive field separators, we know we want every second field. So, for($i=1;$i<=$#F;$i+=2){print $F[$i]} will print the ones we care about.

The same idea but in awk:

$ awk -F'"' '{for(i=2;i<=NF;i+=2){print $(i)}}' file 
One
Two 
 Three 
Four

Ubuntu – How to replace text between two patterns on different lines

It should work for you:

sed -e '/Security-Start/{ N; s/Security-Start.*Security-End/REDACTED/ }'

/Security-Start/ search for "Security-Start"
If you found it: "N;" means append the next line.
and do the replacements/Security-Start.*Security-End/REDACTED/ at the final result.

For more than of two line use this one:

sed -n '1h; 1!H; ${ g; s/Security-Start.*Security-End/REDACTED/p }'

Read here

Best Answer

Related Solutions

Ubuntu – How to find all patterns between two characters

Ubuntu – How to replace text between two patterns on different lines

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