Check if a string matches a regex in Bash script

Question

One of the arguments that my script receives is a date in the following format: yyyymmdd.

I want to check if I get a valid date as an input.

How can I do this? I am trying to use a regex like: [0-9]\{\8}

Answer 1

You can use the test construct, [[ ]], along with the regular expression match operator, =~, to check if a string matches a regex pattern.

For your specific case, you can write:

[[ $date =~ ^[0-9]{8}$ ]] && echo "yes"

Or more a accurate test:

[[ $date =~ ^[0-9]{4}(0[1-9]|1[0-2])(0[1-9]|[1-2][0-9]|3[0-1])$ ]] && echo "yes"
#           |^^^^^^^^ ^^^^^^ ^^^^^^  ^^^^^^ ^^^^^^^^^^ ^^^^^^ |
#           |   |     ^^^^^^^^^^^^^  ^^^^^^^^^^^^^^^^^^^^^^^^ |
#           |   |          |                   |              |
#           |   |           \                  |              |
#           | --year--   --month--           --day--          |
#           |          either 01...09      either 01..09     end of line
# start of line            or 10,11,12         or 10..29
#                                              or 30, 31

That is, you can define a regex in Bash matching the format you want. This way you can do:

[[ $date =~ ^regex$ ]] && echo "matched" || echo "did not match"

where commands after && are executed if the test is successful, and commands after || are executed if the test is unsuccessful.

Note this is based on the solution by Aleks-Daniel Jakimenko in User input date format verification in bash.

---

In other shells you can use grep. If your shell is POSIX compliant, do

(echo "$date" | grep -Eq  ^regex$) && echo "matched" || echo "did not match"

In fish, which is not POSIX-compliant, you can do

echo "$date" | grep -Eq "^regex\$"; and echo "matched"; or echo "did not match"