Accessing last x characters of a string in Bash

I found out that with ${string:0:3} one can access the first 3 characters of a string. Is there a equivalently easy method to access the last three characters?


  • Last three characters of string:

    ${string: -3}



    (mind the space between : and -3 in the first form).

    Please refer to the Shell Parameter Expansion in the reference manual:

    Expands to up to length characters of parameter starting at the character
    specified by offset. If length is omitted, expands to the substring of parameter
    starting at the character specified by offset. length and offset are arithmetic
    expressions (see Shell Arithmetic). This is referred to as Substring Expansion.
    If offset evaluates to a number less than zero, the value is used as an offset
    from the end of the value of parameter. If length evaluates to a number less than
    zero, and parameter is not ‘@’ and not an indexed or associative array, it is
    interpreted as an offset from the end of the value of parameter rather than a
    number of characters, and the expansion is the characters between the two
    offsets. If parameter is ‘@’, the result is length positional parameters
    beginning at offset. If parameter is an indexed array name subscripted by ‘@’ or
    ‘*’, the result is the length members of the array beginning with
    ${parameter[offset]}. A negative offset is taken relative to one greater than the
    maximum index of the specified array. Substring expansion applied to an
    associative array produces undefined results.
    Note that a negative offset must be separated from the colon by at least one
    space to avoid being confused with the ‘:-’ expansion. Substring indexing is
    zero-based unless the positional parameters are used, in which case the indexing
    starts at 1 by default. If offset is 0, and the positional parameters are used,
    $@ is prefixed to the list.

    Since this answer gets a few regular views, let me add a possibility to address John Rix's comment; as he mentions, if your string has length less than 3, ${string: -3} expands to the empty string. If, in this case, you want the expansion of string, you may use:


    This uses the ?: ternary if operator, that may be used in Shell Arithmetic; since as documented, the offset is an arithmetic expression, this is valid.

    Update for a POSIX-compliant solution

    The previous part gives the best option when using Bash. If you want to target POSIX shells, here's an option (that doesn't use pipes or external tools like cut):

    # New variable with 3 last characters removed
    # The new string is obtained by removing the prefix a from string

    One of the main things to observe here is the use of quoting for prefix inside the parameter expansion. This is mentioned in the POSIX ref (at the end of the section):

    The following four varieties of parameter expansion provide for substring processing. In each case, pattern matching notation (see Pattern Matching Notation), rather than regular expression notation, shall be used to evaluate the patterns. If parameter is '#', '*', or '@', the result of the expansion is unspecified. If parameter is unset and set -u is in effect, the expansion shall fail. Enclosing the full parameter expansion string in double-quotes shall not cause the following four varieties of pattern characters to be quoted, whereas quoting characters within the braces shall have this effect. In each variety, if word is omitted, the empty pattern shall be used.

    This is important if your string contains special characters. E.g. (in dash),

    $ string="hello*ext"
    $ prefix=${string%???}
    $ # Without quotes (WRONG)
    $ echo "${string#$prefix}"
    $ # With quotes (CORRECT)
    $ echo "${string#"$prefix"}"

    Of course, this is usable only when then number of characters is known in advance, as you have to hardcode the number of ? in the parameter expansion; but when it's the case, it's a good portable solution.