How to take first numbers and ignore the rest using sed

Question

I have this line of code. The sed is taking 10 and 5. How can I extract 10 and ignore 5?

$ NUMBER=$(echo "The food is ready in 10mins and transport is coming in 5mins." | sed 's/[^0-9]*//g') ; echo $NUMBER

Answer 1

You're just removing everything that isn't a digit, so 5 is left.

Instead, use a capture group to capture the first number, and use that in the replacement.

sed 's/^[^0-9]*\([0-9]*\).*/\1/'

In the regular expression:

• ^ matches the beginning of the line
• [^0-9]* matches all non-digits at the beginning of the line
• \( and \) surround a capture group
• [0-9]* matches digits. These are captured in the group
• .* matches the rest of the line.

In the replacement:

• \1 copies the part of the line that was matched by the capture group. These are the first set of digits on the line.