bash GREP：正则表达式中的变量
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GREP: variable in regular expression
提问by KeinReverb Leonardo Gabrielli
If I want to look whether a string is alphanumeric and shorter than a certain value, say 10, I would do like this (in BASH+GREP):
如果我想查看一个字符串是否是字母数字并且比某个值短，比如 10，我会这样做（在 BASH+GREP 中）：
if grep '^[0-9a-zA-Z]\{1,10\}$' <<<$1 ; then ...
if grep '^[0-9a-zA-Z]\{1,10\}$' <<<$1 ; then ...
(BTW: I'm checking for $1, i.e. the first argument)
（顺便说一句：我正在检查 $1，即第一个参数）
What if I want the value 10 to be written on a variable, e.g.
如果我想将值 10 写在变量上怎么办，例如
UUID_LEN=10
if grep '^[0-9a-zA-Z]\{1,$UUID_LEN\}$' <<< ; then ...
I tried all sort of escapes, braces and so on, but could not avoid the error message
我尝试了各种转义、大括号等，但无法避免错误消息
grep: Invalid content of \{\}
After googling and reading bash and grep tutorials I'm pretty convinced it can't be done. Am I wrong? Any way to go around this?
在谷歌搜索和阅读 bash 和 grep 教程后，我非常确信这是不可能的。我错了吗？有什么办法可以解决这个问题吗？
采纳答案by chepner
You need to use double quotes so that the shell expands the parameter before passing the resulting argument to grep:
您需要使用双引号，以便 shell 在将结果参数传递给之前扩展参数grep：
if grep "^[0-9a-zA-Z]\{1,$UUID_LEN\}$" <<< ; then ...
bashcan perform regular expression matching itself, without having to start another process to run grep:
bash可以自己执行正则表达式匹配，而不必启动另一个进程来运行grep：
if [[ =~ ^[0-9a-zA-Z]{1,$UUID_LEN}$ ]]; then
