Bash-如何将非字母数字字符转换为“_”
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Bash- How to convert non-alphanumerical character to "_"
提问by kheraud
I am trying to store user input in a variable and clean that variable in order to keep only alphanumerical caract + some others (I mean [a-zA-Z0-9-_]).
我试图将用户输入存储在一个变量中并清理该变量，以便仅保留字母数字字符 + 其他一些字符（我的意思是 [a-zA-Z0-9-_]）。
I tried using this but it isn't exhaustive :
我尝试使用它，但它并不详尽：
SERVICE_NAME=$(echo $SERVICE_NAME | tr A-Z a-z | tr ' ' _ | tr \' _ | tr \" _)
Do you have some help for this?
你有什么帮助吗？
采纳答案by Steve Prentice
$ echo 'asd!@QCW@@D' | tr A-Z a-z | sed -e 's/[^a-zA-Z0-9\-]/_/g'
asd__qcw__d
I would use sedfor this and use the ^(not) operator in your set of valid characters and replace everything else with an underscore. The above shows the syntax with the output.
我会sed为此使用并^在您的一组有效字符中使用(not) 运算符，并用下划线替换其他所有内容。上面显示了输出的语法。
And, as a bonus, if you want to replace a run of invalid characters with one underscore, just add +to your regular expression (and use the -rswitch to sedto make it use extended regular expressions:
而且，作为奖励，如果您想用一个下划线替换一系列无效字符，只需添加+到您的正则表达式中（并使用-r开关sed使其使用扩展正则表达式：
$ echo 'asd!@QCW@@D' | tr A-Z a-z | sed -r 's/[^a-zA-Z0-9\-]+/_/g'
asd_qcw_d
回答by user unknown
Bash's string substitution is a fine thing: ${var//pat/rep}
Bash 的字符串替换是个好东西：${var//pat/rep}
val='Foo$%!*@BAR###baZ'
echo ${val//[^a-zA-Z_-]/_}
Foo_____BAR___baZ
A small explanation: The slash introduces a search/replace, a little like in sed (where it just delimits patterns). But you use a single slash for one replacement:
一个小解释：斜线引入了搜索/替换，有点像在 sed 中（它只是分隔模式）。但是您使用一个斜杠替换一个：
val='Foo$%!*@BAR###baZ'
echo ${val/[^a-zA-Z_-]/_}
Foo_%!*@BAR###baZ
Two slashes // mean replace all. Uncommon, but it has some logic, multiple slashes to mean multiple replace (please excuse my poor English).
两个斜线 // 表示全部替换。不常见，但它有一些逻辑，多个斜线表示多个替换（请原谅我的英语不好）。
And note how the $ is separated from the variable, but it is hard to modify a literal constant this way (which would be nice for testing). Modifying $1 isn't a no-brainer as well, afaik.
并注意 $ 是如何与变量分开的，但很难以这种方式修改文字常量（这对测试很有好处）。修改 1 美元也不是一件轻而易举的事，afaik。
回答by anubhava
I believe it can all be done in 1 single sed command like this:
我相信这一切都可以在 1 个 sed 命令中完成，如下所示：
echo 'Foo$%!*@BAR###baZ' | sed -e 's/[A-Z]/\L&/g' -e 's/[^a-z0-9\-]/_/g'
OUTPUT
输出
foo_____bar___baz
回答by jm666
perl way:
perl方式：
perl -ple 's/[^\w\-]/_/g'
pure bash way
纯粹的 bash 方式
a='foo-BAR_123,.:goo'
echo ${a//[^[:alnum:]-]/_}
produces:
产生：
foo-BAR_123___goo
