2

I have the following lines in a bash script:

 while true; do
    DATE=date
    FORMAT="%Y%m%d"
    read -p "Enter start date (YYYYMMDD)  " STARTDATE
    if date=$(date -d "$STARTDATE" +'+%Y%m%d'); then
       start_date=`$DATE +$FORMAT -d $STARTDATE`
       echo $start_date
       break
    fi
    echo "Please use right format (YYYYMMDD)  "
 done

This works fine as long as the input are 8 numbers (e.g. it accepts "20210901" and rejects "20213131"). However, when the input is totally off (e.g. "a" or "nonsense"), it simply takes today's date. How could the code be modified for a stricter check of the format?

The modified code reads:

 valid=0
 while true; do
    DATE=date
    FORMAT="%Y%m%d"
    read -p "Enter start date (YYYYMMDD)  " initialdate
    if date=$(date -d "$initialdate" +'+%Y%m%d'); then
       start_date=`$DATE +$FORMAT -d $initialdate`
       if [[ "$start_date" =~ ^[[0123456789]]{8}$ ]]
       then
         valid=1 
         echo "valid"; 
         echo $start_date
         break
       else
         echo "Invalid format" 
       fi
    fi
    echo "Please use right format (YYYYMMDD)  "
 done
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1
  • Please don't edit the solution to your question. If you want to share what ultimately worked for you, you can add your own answer instead. Commented Sep 14, 2021 at 9:16

3 Answers 3

Reset to default
3

A common way to do input validation is to use an infinite loop that you break out of as soon as the input has been validated.

In this case, "validated" means that the input string contains only eight digits and that GNU date is able to use that date as an option argument to its -d option.

With bash, we could write this as

while true; do
    read -r -p 'Input date: '

    if [[ $REPLY =~ ^[[:digit:]]{8}$ ]] && date -d "$REPLY" >/dev/null 2>&1
    then
        break
    fi

    echo 'Invalid format, try again' >&2
done

thedate=$REPLY

This reads the date at the start of our loop, then tests it with a regular expression that will match if the string contains only eight digits. If that test passes, the string is also tested with GNU date. If that goes well, the loop is exited via a break statement.

In sh, we need to replace the regular expression match with a shell pattern match:

while true; do
    printf 'Input date: ' >&2
    read -r REPLY

    case $REPLY in
        *[![:digit:]]*)
            ;;
        ????????)
            date -d "$REPLY" >/dev/null 2>&1 && break
    esac

    echo 'Invalid format, try again' >&2
done

thedate=$REPLY

The test here first tests whether the string contains any other character than a digit. If that test fails, the next test is for exactly eight characters. If that test succeeds we have a string of eight digits and we test with GNU date and exit the loop if the date is ok.

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1
  • This seems to work seamlessly. Thanks. Commented Sep 13, 2021 at 20:04
3

Since you can reduce the task to an additional check demanding that the input consists of exactly 8 digits, and you are using Bash, you can harness the regular expression comparison operator:

if [[ "$startdate" =~ ^[0123456789]{8}$ ]]; then echo "valid"; else echo "invalid"; fi

This will check if the content of $startdate conforms to a regular expression consisting of exactly eight digits.

You could combine this as follows:

valid=0
while (( valid == 0 )); do
    valid=1
    IFS= read -r -p "Enter start date (YYYYMMDD): " startdate

    if [[ ! "$startdate" =~ ^[0123456789]{8}$ ]]; then valid=0;
    elif ! date -d "$startdate" +%Y%m%d > /dev/null 2>&1 ; then valid=0; fi

    if (( valid == 0 )); then printf "Invalid date entered\n"; fi
done

# < ... operations for valid date here ... >

As a general rule:

  • It is recommended to use shellcheck, also available as standalone program on many Linux distributions, to verify your shell scripts.
  • It is discouraged to use all-uppercase variable names unless you want to export them as environment variables, in order to avoid overwriting critical settings such as $PATH by accident.
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5
  • I tried the 1st suggestion and it was able to sort out invalid entries such as those with letters. However, it is taking "2021" as valid. It's a minor thing but I don't understand why it's not using {8}. Commented Sep 13, 2021 at 16:26
  • 1
    You're not specifying the delimiter for read. You'd likely want IFS= read... Commented Sep 13, 2021 at 16:56
  • @afernandezody Sorry, I noticed two very stupid type-copy errors. Corrected now. Commented Sep 14, 2021 at 7:05
  • @glennjackman Yes, the anchoring would be "very helpful" indeed ;). Commented Sep 14, 2021 at 7:10
  • @StéphaneChazelas Once again, we deplore that lack of an "ASCII-digit" character class ... :( Commented Sep 14, 2021 at 7:11
1

Or without the infinite loop, and assuming BSD's date(1) command:

REPLY=
valid=false

while ! $valid; do

    read -r -p 'Input date: '

    if ([[ $REPLY =~ ^[[:digit:]]{8}$ ]]) && (date -j +%Y%m%d ${REPLY}0000 >/dev/null 2>&1)
    then
        valid=true
    else
        echo 'Invalid format, try again' >&2
    fi

done

thedate=$REPLY

The ${REPLY}0000 syntax is necessary because BSD date wants a full YmdHM, but defaulting the HM to 0000 is sufficient.

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