Bash shell if 语句匹配正则表达式和花括号 {}
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Bash shell if statement matching regex with curly braces {}
提问by Colin
I've been writing a little bash script to aid less experienced Linux users with some commands. One thing seems to be escaping me and that's using curly braces when doing pattern matching for an if statement.
我一直在编写一个小 bash 脚本来帮助经验不足的 Linux 用户使用一些命令。一件事似乎在逃避我，那就是在为 if 语句进行模式匹配时使用花括号。
regex="[A-Za-z0-9]/{5/}"
if [[ =~ $regex ]]
then
num=
else
echo "Invalid entry"
exit 1
fi
This should capture anything A-Z, a-z or 0-9 that is exactly 5 characters, should it not?
这应该捕获正好是 5 个字符的任何 AZ、az 或 0-9，不是吗？
I've tried many times, many variations, many quotes, with and without escaping... Nothing seems to be working:
我已经尝试了很多次，很多变化，很多引用，有和没有转义......似乎没有任何工作：
+ regex='[A-Fa-f0-9]/{5/}'
+ [[ abcd1 =~ [A-Za-z0-9]/{5/} ]]
+ echo 'Invalid entry'
Any ideas what I'm missing?
任何想法我错过了什么？
GNU bash, version 3.2.39(1)-release
回答by Nico Rodsevich
Maybe the problem is here:
也许问题出在这里：
regex="[A-Za-z0-9]/{5/}"
Should be:
应该：
regex="[A-Za-z0-9]\{5\}"
(watch backslashes)
（注意反斜杠）
回答by murgatroid99
You should try
你应该试试
regex="[A-Za-z0-9]{5}"
Edit:Another change you could try is
编辑：您可以尝试的另一个更改是
if [[ "" =~ $regex ]]
Edit2:If you just want to determine if $2is 5 characters long, you could try something like
Edit2：如果您只想确定是否$2为 5 个字符长，您可以尝试类似
if [ ${#2} = 5]
回答by glenn Hymanman
Hmm, works for me in bash version "GNU bash, version 4.1.10(4)-release"
嗯，在 bash 版本“GNU bash，版本 4.1.10(4)-release”中对我有用
$ regexp="[[:alnum:]]{5}"
$ set -- "" abcde
$ [[ "" =~ $regexp ]] && echo y || echo n
y
$ set -- "" foo
$ [[ "" =~ $regexp ]] && echo y || echo n
n
Note that the right-hand side must notbe quoted:
请注意，右侧不得引用：
$ set -- "" abcde
$ [[ "" =~ "$regexp" ]] && echo y || echo n
n
You don't actually need a regular expression for this:
您实际上不需要为此使用正则表达式：
$ glob="[[:alnum:]][[:alnum:]][[:alnum:]][[:alnum:]][[:alnum:]]"
$ set -- "" abcde
$ [[ "" == $glob ]] && echo y || echo n
y
$ set -- "" foo
$ [[ "" == $glob ]] && echo y || echo n
n
