Menu
I have a set of valid characters [0-9a-z_] and a variable that is assigned one of these characters. What I want to do is to be able to increment that variable to the next in the set.
If need be I can handle the "special" cases where it would increment from '9' to 'a' and 'z' to '_', but I can't figure out how to increment letters.
#!/bin/bash
y=b
echo $y # this shows 'b'
y=$((y+1))
echo $y # this shows '1', but I want it to be 'c'
787
$1 means an input argument and -z means non-defined or empty. You're testing whether an input argument to the script was defined when running the script. Follow this answer to receive notifications.
In both cases, the -z flag is a parameter to the bash's "test" built-in (a built-in is a command that is built-into the shell, it is not an external command). The -z flag causes test to check whether a string is empty. Returns true if the string is empty, false if it contains something.
So as far as I can tell, %% doesn't have any special meaning in a bash function name. It would be just like using XX instead. This is despite the definition of a name in the manpage: name A word consisting only of alphanumeric characters and under- scores, and beginning with an alphabetic character or an under- score.
Bash is a widely used shell in Linux, and it supports the '+=' operator to concatenate two variables. As the example above shows, in Bash, we can easily use the += operator to concatenate string variables. Bash's += works pretty similar to compound operators in other programming languages, such as Java.
y=b
echo "$y" # this shows 'b'
y=$(echo "$y" | tr "0-9a-z" "1-9a-z_")
echo "$y" # this shows 'c'
Note that this does not handle the case where $y = "_" (not sure what you want then, and in any case it'll probably require separate handling), and if $y is more than one character long it'll "increment" all of them (i.e. "10" -> "21", "09g" -> "1ah", etc).
125
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
