Exiting an AWK statement after printing a block of text - bash

Question

My problem is that I have a very large database (10GB) and I want to save as much time as possible searching through it. I have an awk statement that is searching through the database and depending on the pattern, writes the data into another file.
I have an input file that will be fed into my script as a Terminal argument variable. There are several lines of data within it that will be used as the pattern for the awk statement.
Within the database, all the lines that match the pattern are all sorted next to each other, so essentially, after printing, there is no need to search any further into the database cause everything has already been found. Once the awk finds the first pattern matching line, all the other pattern matching lines are sequentially after it.
This problem is hard to explain with just words, so I've created a few examples of what my files, code, and the database look and operate like.
The input file via Terminal looks like this:
group_1
group_2
group_3
...
The 10GB database looks like this:
group_1 DATA ...
group_1 DATA ...
group_1 DATA ...
group_2 DATA ...
group_2 DATA ...
group_2 DATA ...
group_2 DATA ...
group_3 DATA ...
group_3 DATA ...
group_3 DATA ...
group_3 DATA ...
...
The script code with the awk statement in question looks like this:
IFS=$'\n'
set -f
for var in $(cat < "$1")
do
awk -v seq="$var" '{if (match($1, seq)) {print $0}}' filepath/database > pattern_matched.file
done
a brief explanation of what this code is doing is that it takes in the Terminal argument variable, a filename in this case, and opens it up for the for loop to begin looping. the pattern group_1, for example, is placed in var and the search through the database begins. If the first column matches the pattern, it saves the line into the file pattern_matched.file file.
Currently, it searches through the entire 10GB worth of data and prints the data into the file as intended, but it wastes a lot of time. After printing the lines that match the pattern, I want to stop the awk from continuing the search through the database and move on to the next pattern from the input file. An example behavior for group_2 would be the awk checking the first 3 lines of the database and sees that none of the lines have the matching pattern. However, line 4 contains the pattern, so it prints the line and the subsequent pattern matching lines after it. When the awk reaches line 8, it exits the awk statement and the for loop can then iterate to the next pattern to be searched for, group_3.
awk '{print $0; exit}' filename
Something like this does not work since it only prints the first instance and breaks out, I want something that can print all the matches and as soon as it finds the next non-pattern match, it breaks out.
Thanks in advance.
UPDATE:
The current problem now is that the solution given below makes logical sense. If it enters the if-statement, It would print the line into the file and iterate to the next line. If the line did not match, it would enter the else-if statement and exit the awk. This makes a lot of sense to me, but for some reason, once the flag variable has been set to 1 by the if-statement for the first matched line, it enters the else-if statement. Since the else-if condition evaluates to true, it exits before even scanning the next line. I confirmed this behavior with print statements everywhere in the awk statement.
This is my code with print statements:
awk -v seq="$seqid" '{if(match($1, seq)) {print "matched" ; print $1 ; flag=1} else if (flag) {print "not matched" ; exit}}'
which outputs this:
weird behavior

Answer 1

Can't you just read in the input file (input_file) into awk:
$ cat input_file
group_1
group_3
Awk script:
$ awk 'NR==FNR{a[$0];next} $1 in a' input_file database
group_1 DATA ...
group_1 DATA ...
group_1 DATA ...
group_3 DATA ...
group_3 DATA ...
group_3 DATA ...
group_3 DATA ...

Answer 2

Your shell code:
for var in $(cat < "$1")
do
awk 'script' filepath/database > pattern_matched.file
done
is using an anti-pattern to read the input file stored in $1, see http://mywiki.wooledge.org/BashFAQ/001, and will overwrite pattern_matched.file on every iteration of the loop. You should, I suspect, have written it as:
while IFS= read -r var
do
awk 'script' filepath/database
done < "$1" > pattern_matched.file
Your awk code:
awk -v seq="$var" '{if (match($1, seq)) {print $0}}'
is using match() unnecessarily since you just want to do a regexp comparison and aren't using the variables that match() populates to help you isolate the matching string (RSTART/RLENGTH) and it's using a defult null condition and then puting the real condition in the action space and then hard-coding the default action of printing the current record. It's equivalent to just:
awk -v seq="$var" '$1 ~ seq'
but I'm not convinced you actually need a regexp comparison - given your example you should be doing a string comparison instead:
awk -v seq="$var" '$1 == seq'
Given your posted example may be misleading you'd just choose which of these is appropriate based on whether you want a regexp or string and partial or full match on $1:
awk -v seq="$var" '$1 == seq' # full string
awk -v seq="$var" 'index($1,seq)' # partial string
awk -v seq="$var" '$1 ~ ("^"seq"$")' # full regexp
awk -v seq="$var" '$1 ~ seq' # partial regexp
Let's say we go with that first full string match match, then to exit once the matching $1 has been processed would be:
awk -v seq="$var" '$1 == seq{print; f=1; next} f{exit}'
which would make your full code:
while IFS= read -r var
do
awk -v seq="$var" '$1 == seq{print; f=1; next} f{exit}' filepath/database
done < "$1" > pattern_matched.file
BUT I doubt if you need a shell loop at all and you could just do this instead:
awk 'NR==FNR{seqs[$1]; next} $1 in seqs' "$1" filepath/database > pattern_matched.file
or some other variant that just has awk (or maybe just join) read the input files once. You can make the above exit after all seqs[] have been processed by:
awk '
NR==FNR { seqs[$1]; numSeqs++; next }
$1 in seqs { print; if ($1 !== prev) numSeqs--; prev = $1; next }
numSeqs == -1 { exit }
' "$1" filepath/database > pattern_matched.file
or similar.

Answer 3

I think this should do the trick:
awk -v seq="$var" '{if (match($1, seq)) {print $0; found=1} else if (found) { exit }}'
Similar to David C. Rankin answer, but no need to pass the rd=0 argument to awk since in awk any uninitialized variable is initialized to zero when its first used.

Answer 4

As we do not really know what you intend to do with your program, I will just give you an awk solution:
awk -v seq="$var" '($1!=seq) { if(p) exit; next }($1==seq){p=1}p'
This uses the flag p to check if it already met the sequence seq. A simple if condition determines if it should exit awk or move to the next record. Exiting is done after seq is found, moving to the next record is done before.
However, since you place this in a loop, this will read the file over and over and over again. If you want to make a subselection, you could use the solution of James Brown