在sed / awk start表达式上方打印四行

Question

I have the following sed command that fetches back a whole chunk of a log that has an <interface> XML within it that's being printed into a txt file:

sed -n '/(StartingExpression)/{:start /<\/Interface>/!{N;b start};/(SomeValueInTheXML)"/p}' *.log > File.txt

The issue is that I have some dynamic values (time stamp) that's always 4 lines above the (StartingExpression) ... is there a way to print something like:

line of (StartingExpression) -4

?

I found this question: search (eg awk, grep, sed) for string, then look for X lines above and another string below but the solution isn't really clear :\\

Thanks for the help.

also, if anyone has a good source to learn sed, I'll be thankful if you post it :)

EDIT per request:

one
two
1/1/2015
line 1
12345
(StartingExpression)
<Interface>
   <A>
   Name=Andy
   </A>
</Interface>
three
four
1/1/2015
line 1
12345
(StartingExpression)
<Interface>
   <A>
   Name=John
   </A>
</Interface>
hello
world

I would like to print from 1/1/2015 (which is 3 lines above (StartingExpression) - this can be dynamic as its a date ) until </Interface>

EDIT: I forgot to mention, there can be multiple instances of these interfaces... How do you also ensure to print ONLY the one that has Name=Andy ?

file.txt:

1/1/2015
line 1
12345
(StartingExpression)
<Interface>
   <A>
   Name=Andy
   </A>
</Interface>

Answer 1

This awk idiom will print n lines preceding and including pattern.

$awk -v n="$plines" '{a[p]=$0; p=(p+1)%n} /pattern/{for(i=p;i<p+n;i++) print a[i%n]}' file

For example

$awk -v n=5 '{a[p]=$0; p=(p+1)%n} /20/{for(i=p;i<p+n;i++) print a[i%n]}' <(seq 10 99)

16
17
18
19
20

UPDATE for additional logic.

You can incorporate printing after pattern match easily as well. This script will print the lines before; and now after the match until the end of close tag.

$ awk -v n=4 '           {a[p]=$0; p=(p+1)%n} 
   /(StartingExpression)/{for(i=p;i<p+n-1;i++) print a[i%n];f=1} 
                        f{print} 
            /\/Interface/{exit}' file

1/1/2015
line 1
12345
(StartingExpression)
<Interface>
   <A>
   </A>
</Interface>

UPDATE: filter based on attribute value

It's easier at this point just pass through another script instead of rewriting it. Replace the exit with f=0, the first script will output all the matching records and filter in a second script the record of interest.

$ awk -v n=4 '           {a[p]=$0; p=(p+1)%n}
   /(StartingExpression)/{for(i=p;i<p+n-1;i++) print a[i%n];f=1}
                        f{print}
            /\/Interface/{f=0}' file
| awk 'BEGIN{ORS=RS="</Interface>\n"} 
 /Name=Andy/'


1/1/2015
line 1
12345
(StartingExpression)
<Interface>
   <A>
   Name=Andy
   </A>
</Interface>

Answer 2

When you have a grep version that supports -A and -B, and your file is made with fixed lines, you can use

grep -A2 -B6 "Name=Andy" request

Using your vi-knowledge can help for making the following command:

echo "/(StartingExpression)/-3,/<.Interface>/ p" | ed -s request

I used a . for the slash in , I did not want to escape the character. This will only work when Andy is the first name.

Making it working with Andy somewhere else becomes more tricky:

rm /tmp/andy
echo "/Name=Andy/
?(StartingExpression)
-3,/<.Interface>/ w /tmp/andy" | ed -s request 2>&1 >/dev/null
cat /tmp/andy
rm /tmp/andy