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AgileSoul
Asked: 2022-04-22 21:06:06 +0800 CST 2022-04-22 21:06:06 +0800 CST

How do I validate dates with regular expressions in Bash?

  • 772

I leave my script to help me understand where the error is and to be able to correct it:

#!/bin/bash
fechaDeNacimientoRegex='^19|20[0-9]{2}[1-12]{2}[1-31]$'
#Validacion y entrada de la fecha
read -p "Por favor ingresa la fecha formato [yyyyMMdd]:
" fechaNacimiento
if [[ $fechaNacimiento =~ $fechaDeNacimientoRegex ]]; then
    echo "La fecha $fechaNacimiento es valida"
else  
    echo "La fecha $fechaNacimiento es invalida"
fi

The output when I enter a date between 1900 and 1999 works:

Por favor ingresa la fecha formato [yyyyMMdd]:
19921129
La fecha 19921129 es valida

But it doesn't work when I put 2000 or greater:

 Por favor ingresa la fecha formato [yyyyMMdd]:
 20110330
 La fecha 20110330 es invalida
regex

3 Answers

  1. If you want to validate a date, what better than to use datesame to validate it. If you pass it with date -d "$fecha", the exit code will tell you if it's good or not:

    if date -d "$fecha" &>/dev/null; then
    echo "fecha válida"
else
    echo "fecha NO válida"
fi

    Let's see it with a couple of examples:

    $ for f in 20211231 19001300; do if date -d "$f" &>/dev/null; then echo "fecha $f válida"; else echo "fecha $f NO válida"; fi; done
fecha 20211231 válida
fecha 19001300 NO válida
    • 6
  2. Best Answer

    As I said in a comment, the regular expression is wrong. You can't just say you want numbers between 1 and 12, you can say which characters to accept and in what order.

    I have a regular expression, rather longer, but which seems to work for what you want:

    ^(19[0-9]{2}|20[0-9]{2})(0[1-9]|10|11|12)(0[1-9]|1[0-9]|2[0-9]|3[0-1])$

    Part by part:

    • Year: (19[0-9]{2}|20[0-9]{2})This is similar to what you have, but in parentheses to capture it.
    • Month: (0[1-9]|10|11|12)means a 0 followed by a number between 1 and 9, or a 10, or an 11, or a 12. Between parentheses to capture the result.
    • Day: (0[1-9]|1[0-9]|2[0-9]|3[0-1])means a 0 followed by a number between 1 and 9, or a 1 followed by a number between 0 and 9, or a 2 followed by a number between 0 and 9, or a 3 followed by a 0 or 1. Enter parentheses to capture the result.

    Thanks to capturing the result, we can do:

    fechaDeNacimientoRegex='^(19[0-9]{2}|20[0-9]{2})(0[1-9]|10|11|12)(0[1-9]|1[0-9]|2[0-9]|3[0-1])$'
read -p "Por favor ingresa la fecha formato [yyyyMMdd]:
" fechaNacimiento
if [[ $fechaNacimiento =~ $fechaDeNacimientoRegex ]]; then
    anio=${BASH_REMATCH[1]}
    mes=${BASH_REMATCH[2]}
    dia=${BASH_REMATCH[3]}
    echo "Año: $anio"
    echo "Mes: $mes"
    echo "Día: $dia"
fi

    Now, the variables $anio, $mesand $diacontain the part that matches the regular expression, and you can do more validations, for example, that February does not accept more than 29 days, or indicate which number makes the date invalid:

    if [ "$mes" -gt "12" ]; then
    echo "El mes $mes es invalido"
elif [ "$dia" -gt "31" ]; then
    echo "El dia $dia es invalido"
else
    echo "La fecha $fechaNacimiento es valida"
fi
    • 5

  3. It sounds like that regular expression would be extremely verbose.

    I propose an alternative with a perl type regex and grep:

    $ grep -P '^(19|20)\d{2}\d{2}\d{2}$' <<< "19921129" && echo valida || echo 
invalida
19921129
valida

    Now with an invalid:

    $ grep -P '^(19|20)\d{2}\d{2}\d{2}$' <<< "1992119" && echo valida || echo invalida
invalida

    Used in your code, it would look something like this:

    #Validacion y entrada de la fecha

read -p "Por favor ingresa la fecha formato [yyyyMMdd]:" fechaNacimiento

if grep -qP '^(19|20)\d{2}\d{2}\d{2}$' <<< "$fechaNacimiento"; then
    echo "La fecha $fechaNacimiento es valida"
else  
    echo "La fecha $fechaNacimiento es invalida"
fi

    You could also do an analysis by number. That is, by the first four numbers that are between "1900" and "2099". And the following that belong to a regular expression.

    • 0