How to remove last n characters from a string in Bash?

How to remove last n characters from a string in Bash?

You can do like this (in bash v4 and higher):


v=some string.rtf


echo $v --> $v2

Note: macos uses bash 3.x by default

To remove four characters from the end of the string use ${var%????}.

To remove everything after and including the final . use ${var%.*}.

How to remove last n characters from a string in Bash?

First, its usually better to be explicit about your intent. So if you know the string ends in .rtf, and you want to remove that .rtf, you can just use var2=${var%.rtf}. One potentially-useful aspect of this approach is that if the string doesnt end in .rtf, it is not changed at all; var2 will contain an unmodified copy of var.

If you want to remove a filename suffix but dont know or care exactly what it is, you can use var2=${var%.*} to remove everything starting with the last .. Or, if you only want to keep everything up to but not including the first ., you can use var2=${var%%.*}. Those options have the same result if theres only one ., but if there might be more than one, you get to pick which end of the string to work from. On the other hand, if theres no . in the string at all, var2 will again be an unchanged copy of var.

If you really want to always remove a specific number of characters, here are some options.

You tagged this bash specifically, so well start with bash builtins. The one which has worked the longest is the same suffix-removal syntax I used above: to remove four characters, use var2=${var%????}. Or to remove four characters only if the first one is a dot, use var2=${var%.???}, which is like var2=${var%.*} but only removes the suffix if the part after the dot is exactly three characters. As you can see, to count characters this way, you need one question mark per unknown character removed, so this approach gets unwieldy for larger substring lengths.

An option in newer shell versions is substring extraction: var2=${var:0:${#var}-4}. Here you can put any number in place of the 4 to remove a different number of characters. The ${#var} is replaced by the length of the string, so this is actually asking to extract and keep (length – 4) characters starting with the first one (at index 0). With this approach, you lose the option to make the change only if the string matches a pattern; no matter what the actual value of the string is, the copy will include all but its last four characters.

You can leave the start index out; it defaults to 0, so you can shorten that to just var2=${var::${#var}-4}. In fact, newer versions of bash (specifically 4+, which means the one that ships with MacOS wont work) recognize negative lengths as the index of the character to stop at, counting back from the end of the string. So in those versions you can get rid of the string-length expression, too: var2=${var::-4}.

If youre not actually using bash but some other POSIX-type shell, the pattern-based suffix removal with % will still work – even in plain old dash, where the index-based substring extraction wont. Ksh and zsh do both support substring extraction, but require the explicit 0 start index; zsh also supports the negative end index, while ksh requires the length expression. Note that zsh, which indexes arrays starting at 1, nonetheless indexes strings starting at 0 if you use this bash-compatible syntax. But zsh also allows you to treat scalar parameters as if they were arrays of characters, in which case the substring syntax uses a 1-based count and places the start and (inclusive) end positions in brackets separated by commas: var2=$var[1,-5].

Instead of using built-in shell parameter expansion, you can of course run some utility program to modify the string and capture its output with command substitution. There are several commands that will work; one is var2=$(sed s/.{4}$// <<<$var).

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