如何知道awk命令是否打印出来而不会损坏原始输出

Question

I'm trying to use AWK command, it's printing for me very well and what exact way I want

My question is, how can I know from script if the awk command I used has printed something without spoiling and change the way it prints ?

commands I used: gawk 'BEGIN{RS=ORS="\\n\\n" {s=tolower($0)} s~/word1|word2/' input_file.log

I tried:

status=gawk 'BEGIN{RS=ORS="\\n\\n" {s=tolower($0)} s~/word1|word2/' input_file.log

if [ -z $status]
then
    //status is empty or null, means nothing printed in awk command
    echo "nothing"
else
    //printed something in awk command
    echo $status

the issue is that echo $status prints in sequence all the lines, without the "new lines" between the lines

how can print the original prints from awk without spoiling it?

example: input file:

line 0 no words in here

line 1 starting
line 1 word1

line 2 no words here as well

line 3 starting
line 3 word2
line 3 end

line 4 nothing
line 5 nothing

command: gawk 'BEGIN{RS=ORS="\\n\\n" {s=tolower($0)} s~/word1|word2/' input_file.log

expected output:

line 1 starting
line 1 word1

line 3 starting
line 3 word2
line 3 end

if I use: stat=$(gawk 'BEGIN{RS=ORS="\\n\\n" {s=tolower($0)} s~/word1|word2/' input_file.log) echo $stat

I get as output:

line 1 starting line 1 word1 line 3 starting line 3 word2 line 3 end

Thanks in advance!

Answer 1

Not fully sure as you haven't shown any sample Input_file or expected output, so could you please try using echo "$status" .

EDIT: As you have edited your post now, so you should change your code to following and it should fly then.

status=$(awk 'BEGIN{RS=ORS="\n\n"} {s=tolower($0)} s~/word1|word2/' Input_file)
if [[ -z $status ]]
then
    echo "nothing"
else
    echo "$status"
fi

Answer 2

You can use exit code for checking whether awk has printed something or not

Correct your code

From

gawk 'BEGIN{RS=ORS="\n\n" {s=tolower($0)} s~/word1|word2/' input_file.log

TO

status=$(gawk 'BEGIN{RS=ORS="\n\n"}tolower($0)~/word1|word2/' input_file.log)

and (with quotes)

echo "$status"

This happens because when you quote a parameter (whether that parameter is passed to echo , test or some other command), the value of that parameter is sent as one value to the command. If you don't quote it, the shell does its normal magic of looking for whitespace to determine where each parameter starts and ends.

To correct existing code

#!/usr/bin/env bash

status=$(gawk 'BEGIN{RS=ORS="\n\n"}tolower($0)~/word1|word2/' input_file.log)
if [  -z "$status" ]; then
       echo "looks like nothing matched and so nothing printed"
else
       echo "awk matched regex and printed something"
fi

Here is code for checking awk has printed something or not using exit code :

gawk 'BEGIN{RS=ORS="\n\n"}f=(tolower($0)~/word1|word2/){e=1}f; END{exit !e}' input_file.log

# check exit code 
if [ "$?" -eq 0 ]; then
       echo "awk matched regex and printed something"
else
       echo "looks like nothing matched and so nothing printed"
fi

Test Results:

$ cat test.sh 
#!/usr/bin/env bash

gawk 'BEGIN{RS=ORS="\n\n"}f=(tolower($0)~/word1|word2/){e=1}f; END{exit !e}' "$1"
if [ "$?" -eq 0 ]; then
       echo "awk matched regex and printed something"
else
       echo "looks like nothing matched and so nothing printed"
fi

sample files for testing

$ echo 'word1' >file1

$ echo 'nothing' >file2

Contents of files

$ cat file1
word1

$ cat file2
nothing

Execution with first file

$ bash test.sh file1
word1

awk matched regex and printed something

Execution with second file

$ bash test.sh file2
looks like nothing matched and so nothing printed