Print a line in stdout that matches an expression if the output contains another expression

Question

This might be a common/easy task but I couldn't figure out from examples on the web or manuals of awk/sed/grep.

So, here is the scenario:

• There is an internal command line tool that prints out a multi-line result for each line in an input file.
• I have an input file with 500K lines.
• In the output of the tool, there is always a line which is like "src: /some/directory"
• I want to extract this line if and only if there is the specific string "foo" in the same output.

The number of rows between these lines might be different, so this question is somewhat related but not exactly what I'm trying to do. Match multiple regular expressions from a single file using awk

How can I do this using awk, sed or grep? I can do it using Python but I don't want to because I want to learn awk/sed and this might be a good example.

Here is what I tried with grep:

tool -inputfile | if grep "foo"; then grep "src: " ; fi > result.txt

This doesn't produce the result I expected, probably because of something related to buffering.

Trying with awk:

tool -inputfile | awk '{for (i=1;i<NF;i++) {if(match($i, "foo")) print ??? }}' > result.txt

How do I print the line that contains "src: " in this script?

Example outputs of the tool:

Output 1:

src: /usr/bin 
param1: value1 value2 
param2: "foo" 
param3: "bar" "spam" 
param4: "eggs" "spam" "spam"

Output 2:

src: /dev/null
param1: value1 value2
param2: "ham" "spam" "eggs"

So for these 2 cases, I am trying to extract just the 1st one, ie: src: /usr/bin

Answer 1

If you know that src: occurs at the beginning of a line and that foo is enclosed in quotes and preceded by a space and that there must be a colon earlier in the line, use

awk 'BEGIN{a=0} /^$/{if(a==1) print b; a=0} /:.* "foo"/{a=1} /^src:/{b=$0} END{if(a==1) print b}'

We use the variable a to remember whether or not the pattern foo occurs in the input block, and the variable b to store the src: line. At the beginning, a is set to 0. Whenever we find an empty line (i.e., ^$), we check the value of a, conditionally print b, and reset a. If we encounter "foo" preceded by a colon earlier in the line, we set a to 1. If we encounter src: at the beginning of a line (^), we store it in b. At the end, we check once more whether a == 1, if so, we print b.

Answer 2

Easy awk

awk '/src/{a=$0}/foo/{b=1}b&&a{print a;exit}'

If src or foo can be somewhere else in a different format or whatever

awk '/^src/{a=$0}/"foo"/{b=1}b&&a{print a;exit}'

If foo always comes after src

awk '/^src/{a=$0}/"foo"/{print a;exit}'

If there are multiple src blocks in a file and you want to print each one that contains foo

awk '/^src/{a=$0;b=0}/"foo"/{b=1}b&&a{print a;a=0}'