Substituting shell variables into awk pattern look

问题:

NOTE: I am a noob at bash scripts and the awk command - please excuse any dumb mistakes I make.

I am unable to substitute shell variables into my awk pattern. I am trying to scan through a file, find the first occurence of a specific string in the file, and print each line that succeed it in order until it hits an empty string/line.

I don't know the string I am searching for in advance, and I would like to substitute in that variable.

When I run this with the string directly specified (e.g "< main>:"), it works perfectly. I've already searched on how awk patterns work, and how to substitute in variables. I've tried using the -v flag for awk, directly using the shell variable - nothing works.

funcName="<${2}>:"

awk=`awk -v FN="$funcName" '/FN/,/^$/' "$ofile"`

rfile=search.txt

echo -e "$awk" > "$rfile"

The error is just that nothing prints. I want to print all the lines between my desired string and the next empty line.

回答1:

Could you please try following, haven't tested it because no clear samples but should work.

funcName="<${2}>:"
awk_result=$(awk -v FN="$funcName" 'index($0,FN){found=1} found; /^$/{found=""}' "$ofile")
rfile=search.txt
echo -e "$awk_result" > "$rfile"

Things fixed in OP's attempt:

1. NEVER keep same name of a variable as a binary's name or on a keyword's name so changed awk variable name to awk_result.
2. Use of backticks is depreciated now, so always wrap your variable for having values in var=$(......your commands....) fixed it for awk_result variable.
3. Now let us talk about awk code fix, I have used index method which checks if value of variable FN is present in a line then make a FLAG(a variable TRUE) and make it false till line is empty as per OP's ask.