sed 检索部分行

Question

I have lines of code that look like this:

hi:12345:234 (second line)

How do I write a line of code using the sed command that only prints out the 2nd item in the second line?

My current command looks like this:

sed -n '2p' file which gets the second line, but I don't know what regex to use to match only the 2nd item '12345' and combine with my current command

Answer 1

Could you please try following, written and tested with shown samples in GNU sed .

sed -n '2s/\([^:]*\):\([^:]*\).*/\2/p' Input_file

Explanation: Using -n option of sed will stop the printing for all the lines and printing will happen only for those lines where we are explicitly mentioning p option to print(later in code). Then mentioning 2s means perform substitution on 2nd line only. Then using regex and sed 's capability to store matched regex into a temp buffer by which values can be retrieved later by numbering 1,2...and so on . Regex is basically catching 1st part which comes before first occurrence of : and then 2nd part after first occurrence of : to till 2nd occurrence of : as per OP's request. So while doing substitution mentioning /2 will replace whole line with 2nd value stored in buffer as per request, then mentioning p will print that part only in 2nd line.

Answer 2

A couple of solutions:

echo "hi:12345:234" | sed -n '2s/.*:\([0-9]*\):.*/\1/p'
echo "hi:12345:234" | sed -n '2{s/^[^:]*://; s/:.*//p; q}'
echo "hi:12345:234" | awk -F':' 'FNR==2{print $2}'

All display 12345 .

sed -n '2s/.*:\([0-9]*\):.*/\1/p' only displays the captured value thanks to -n and p option/flag. It matches a whole string capturing digits between two colons, \1 only keeps the capture.

The sed -n '2{s/^[^:]*://;s/:.*//p;q}' removes all from start till first : , all from the second to end, and then quits ( q ) so if your file is big, it will be processed quicker.

awk -F':' 'FNR==2{print $2}' splits the second line with a colon and fetches the second item.