问题
I was reviewing some of my old code and came across this syntax:
extractDir="${downloadFileName%.*}-tmp"
The only information I found searching refers to a list of commands, but this is just one variable. What does this curly-brace syntax mean in bash?
回答1:
In this context, it is a parameter substitution.
The ${variable%.*} notation means take the value of $variable, strip off the pattern .* from the tail of the value — mnemonic: percenT has a 't' at the Tail — and give the result. (By contrast, ${variable#xyz} means remove xyz from the head of the variable's value — mnemonic: a Hash has an 'h' at the Head.)
Given:
downloadFileName=abc.tar.gz
evaluating extractDir="${downloadFileName%.*}-tmp" yields the equivalent of:
extractDir="abc.tar-tmp"
The alternative notation with the double %:
extractDir="${downloadFileName%%.*}-tmp"
would yield the equivalent of:
extractDir="abc-tmp"
The %% means remove the longest possible tail; correspondingly, ## means remove the longest matching head.
回答2:
It indicates that parameter expansion will occur.
回答3:
It is used when expanding an environment variable adjacent to some text that is not the variable, so the shell does not include all of it in the variable name.
来源：https://stackoverflow.com/questions/9558986/what-does-the-curly-brace-syntax-var-mean