Another way to print between two regular expressions 2

[FedoEx]

Test file

$cat  testfile
a b c 
1 2  3 
begin  line with first regex
block     1  11
between   2  12 
regex     3  13
end  line with last regex
4 5 6 
blah blah

I need another method printing the block between two regular expressions
where the entire code is in “{ }” brackets.
I was thinking something like this code that prints the line containing the first regex.

 awk '{if(/begin/) print }' testfile

I can't use any statement outside the brackets like.

awk '/begin/,/end/{print $0}' testfile

Thanks.

 

[LKBrwnDBA]

Try this:

awk '/begin/{s=1} /end/{s=0} {if(s>1) print $0; s+=1}' testfile

----------------------------------------------------------------------------
The person who says it can't be done should not interrupt the person doing it. -- Chinese proverb

 

[FedoEx]

The entire code needs to be enclosed in the "{....}" brackets.
Something like this incorrect expression.

 awk '{ if(/begin/,/end/) print $0  }' testfile

Thanks

 

[LKBrwnDBA]

awk '{if($1 ~ /begin/)s=1;if($1 ~ /end/)s=0;{if(s>1) print $0; s+=1}}' testfile

----------------------------------------------------------------------------
The person who says it can't be done should not interrupt the person doing it. -- Chinese proverb

 

[FedoEx]

Thanks. This almost works.
After the end regex the counter keeps increasing so the lines one after the regex gets printed out.
That is the line "blah blah" and after.
I can figure the fix for that.
Thanks again.

 

[FedoEx]

I am still looking for alternative way.
My idea is to get NR for the two regular expressions and then print for the range of lines where they are.
The code

awk '{
if($1~/begin/)begnr=NR;if($1~/end/)endnr=NR;
if(NR>3&&NR<7 ) print $0
}
END{ print begnr,endnr } ' testfile

gives me the desired output and I see that begnr,endnr was evaluated to 3,7 respectively.
However if I do

awk '{
if($1~/begin/)begnr=NR;if($1~/end/)endnr=NR;
if(NR>begnr&&NR<endnr ) print $0
}' testfile

I don't get any output.
I also tried

... if($1~/begin/)begnr=int(NR);if($1~/end/)endnr=int(NR);...

Any idea what might be wrong?

 

[LKBrwnDBA]

OK, I see...try this:

awk '{if($1 ~ /begin/)s=1;if($1 ~ /end/)e=1;{if(s>1&&e==0) print $0; s+=1}}' testfile

----------------------------------------------------------------------------
The person who says it can't be done should not interrupt the person doing it. -- Chinese proverb

 

[FedoEx]

That one works.

 awk '{if($1 ~ /begin/)s=1;if($1 ~ /end/)s=-1000;{if(s>1) print $0; s+=1}}' testfile

where I set s=-1000 or some other big number which I know is always much larger than the number of lines between the two regex.
But still I can't really use it due to the specific of my actual problem.
What would really be the best is something without any counters. Like this.

 awk '{ if(/begin/,/end/) print $0  }' testfile

 

[Annihilannic]

FedoEx, why do you have such strange restrictions on your code, out of curiosity?

Try:

awk '{ if ($1 ~ /begin/) { s=1; next; } if ($1 ~ /end/) s=0; if (s) print; }' testfile

Annihilannic.

 

[FedoEx]

Thanks.
To answer your question I need to open new thread.I will probably do that in a couple of days since I am pretty much stuck with the actual code.