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How does this code work? (sh.itjust.works)

submitted 2 months ago by zarkanian@sh.itjust.works to c/bash@lemmy.ml

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I'm studying bash, and I came across this Stackoverflow thread which contains this bit of code:

var="abcde"
echo ${var%d*}

The output is abc, but I can't figure out why. I understand that %d is used to indicate an integer number and * represents anything, but I can't figure out why those together would truncate var to only 3 characters.

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[–] beatnik86@lemmy.world 0 points 2 months ago

That's manipulation of the string held in var. Dollar sign and curly braces wrap the variable, the single percent sign say remove the shortest matching substring from the end of the variable contents, and the 'd*' is the substring to match, a lowercase d followed by any characters in this instance. var will still hold the entire original string, only the echo output is modified here. You can find more documentation here, https://tldp.org/LDP/abs/html/string-manipulation.html.