How to grep specific a word in bash

Question

I want to catch the word that is after "on" in several lines at a file, and if the word appears already in the file, I want to skip it. I tried to do so:

#!/bin/bash
echo "" > missig_packages.txt
cat log_file.txt | grep depends > dependsLog.txt
function createListOfPackages {
    if grep "$1" missig_packages.txt; then
        continue
    else
        echo "$1" >> missig_packages.txt
    fi  
}
while read line; do
    package=`cat dependsLog.txt | cut -d" " -f5`
    createListOfPackages $package
done < dependsLog.txt

The file dependsLog.txt contains lines like this:

  libgcc1:amd64 depends on **gcc-4.9-base** (= 4.9.1-0ubuntu1); however:
  cinder-volume depends on **cinder-common** (= 1:2015.1.1-0ubuntu2~cloud2); 
  python-cryptography depends on **python-cffi**.
  python-pycadf depends on **python-netaddr**.

How can I grep the words between ** and ** (which are not themselves in the text)? Each line begins with "".

Answer 1

this is awk's jobs.

line with on

awk '$3 == "on" '

the word you are looking for

awk '$3 == "on" { print $4 ;}'

• $3 == "on" will get the line with "on" as third word
• { print $4 ;} will print fourth word

Answer 2

Your entire shell script fragment can be replaced with:

awk '/depends on/ { print $4}' log_file.txt | sed -e 's/\.$//' | sort -u > missing_packages.txt

The sed script strips the trailing . from package names where the input line doesn't have version information.

Answer 3

Try following command which will grep word after 'on' from dependLogs.txt and then will insert "" at start of line.

cat dependsLog.txt | grep -oP "(?<=on )[^ ]+" | sed 's/^/\"\"/' >> missig_packages.txt

To make sure lines do not get duplicated you can sort and uniq by following command.

cat dependsLog.txt | grep -oP "(?<=on )[^ ]+" | sed 's/^/\"\"/' | sort | uniq >> missig_packages.txt