Linux Bash: extracting text from file int variable

Question

I haven't found anything that clearly answers my question. Although very close, I think...

I have a file with a line:

# Skipsdata for serienummer 1158

I want to extract the 4 digit number at the end and put it into a variable, this number changes from file to file so I can't just search for "1158". But the "# Skipsdata for serienummer" always remains the same.

I believe that either grep , sed or awk may be the answer but I'm not 100 % clear on their usage.

Answer 1

You can use grep with the -o switch, which prints only the matched part instead of the whole line.

Print all numbers at the end of lines from file yourFile

grep -Po '\d+$' yourFile

Print all four digit numbers at the end of lines like described in your question:

grep -Po '^# Skipsdata for serienummer \K\d{4}$' yourFile

-P enables perl style regexes which support \\d and especially \\K .
\\d matches any digit (0-9). \\d{4} matches exactly four digits.
\\K lets grep forget the previously matched part, such that only the part afterwards is printed.

Answer 2

Using Awk as

numberRequired=$(awk '/# Skipsdata for serienummer/{print $NF}' file)
printf "%s\n" "$numberRequired"
1158

Answer 3

There are multiple ways to find your number. Assuming the input data is in a file called inputfile :

mynumber=$(sed -n 's/# Skipsdata for serienummer //p' <inputfile) will print only the number and ignore all the other lines;

mynumber=$(grep '^# Skipsdata for serienummer' inputfile | cut -d ' ' -f 5) will filter the relevant lines first, then only output the 5th field (the number)