Bash Shell脚本为grep搜索的输出分配新变量

Question

EDIT 2: I've decided to re-write this in order to better portray my outcome.

I'm currently using this code to output a list of files within various directories:

for file in /directoryX/*.txt

do
     grep -rl "Annual Compensation" $file
done

The output shows all files that have a certain table I'm trying to extract in a layout like this:

txtfile1.txt
txtfile2.txt
txtfile3.txt

I have been using this awk command on each individual .txt file to extract the table and then send it to a .csv:

awk '/Annual Compensation/{f=1} f{print; if (/<\/TABLE>/) exit}' txtfile1.txt > txtfile1.csv

My goal is to find a command that will run my awk command against each file in the list all at once. Thank you to those that have provided suggestions already.

Answer 1

If I understand what you're asking, I think what you want to do is add a line after the grep, or instead of the grep, that says:

awk '/Annual Compensation/{f=1} f{print; if (/<\/TABLE>/) exit}' $file > ${file}_new.csv

When you say ${file}_new.csv , it expands the file variable, then adds the string "_new.csv" to it. That's what you're shooting for, right?

Answer 2

Modifying your code:

for file in /directoryX/*.txt
do
    files+=($(grep -rl "Annual Compensation" $file))
done

for f in "${files[@]}";do
    awk '/Annual Compensation/{f=1} f{print; if (/<\/TABLE>/) exit}' "$f" > "$f"_new.csv
done

Alternative code:

files+=($(grep -rl "Annual Compensation" /directoryX/*))
for f in "${files[@]}";do
awk '/Annual Compensation/{f=1} f{print; if (/<\/TABLE>/) exit}' "$f" > "$f"_new.csv

In both cases, the grep results and awk results are not verified by me - it is just a copy - paste of your code.