bash 在bash变量中转义点
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Escaping dots in bash variables
提问by Tim Bellis
I want to escape dots from an IP address in Unix shell scripts (bash or ksh) so that I can match the exact address in a grep command.
我想从 Unix shell 脚本（bash 或 ksh）中的 IP 地址转义点，以便我可以匹配 grep 命令中的确切地址。
echo $ip_addr | sed "s/\./\\./g"
works (outputs 1\.2\.3\.4), but
有效（输出 1\.2\.3\.4），但是
ip_addr_escaped=`echo $ip_addr | sed "s/\./\\./g"`
echo $ip_addr_escaped
Doesn't (outputs 1.2.3.4)
不（输出 1.2.3.4）
How can I correctly escape the address?
如何正确转义地址？
Edit:It looks like
编辑：看起来像
ip_addr_escaped=`echo $ip_addr | sed "s/\./\\\\./g"`
works, but that's clearly awful!
有效，但这显然很糟糕！
回答by chepner
bashparameter expansion supports pattern substitution, which will look (slightly) cleaner and doesn't require a call to sed:
bash参数扩展支持模式替换，它看起来（稍微）更干净并且不需要调用sed：
echo ${ip_addr//./\.}
回答by ruakh
Yeah, processing of backslashes is one of the strange quoting-related behaviors of backticks `...`. Rather than fighting with it, it's better to just use $(...), which the same except that its quoting rules are smarter and more intuitive. So:
是的，反斜杠的处理是反引号的一种奇怪的引用相关行为`...`。与其与之抗争，不如直接使用$(...)，除了它的引用规则更智能和更直观之外，它是相同的。所以：
ip_addr_escaped=$(echo $ip_addr | sed "s/\./\\./g")
echo $ip_addr_escaped
But if the above is really your exact code — you have a parameter named ip_addr, and you want to replace .with \.— then you can use Bash's built-in ${parameter/pattern/string} notation:
但是，如果以上确实是您的确切代码——您有一个名为 的参数ip_addr，并且您想替换.为\.——那么您可以使用 Bash 的内置符号：${parameter/pattern/string}
ip_addr_escaped=${ip_addr//./\.}
Or rather:
更确切地说：
grep "${ip_addr//./\.}" [FILE...]
回答by Jonas Berlin
Replace the double quotes with single quotes.
用单引号替换双引号。
