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sed - 从字符串中删除特定的下标

发布于 2024-12-13 11:05:52 字数 97 浏览 1 评论 0原文

请向我提供一个 sed oneliner，它提供以下输出： sdc3 sdc2 对于输入： sdc3[1] sdc2[0]

我的意思是从字符串中删除所有下标值..

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执笔绘流年 2024-12-20 11:05:52

sed 's/\[[^]]*\]//g'

读取：用文字“[”替换任何字符串，后跟零个或多个不是“]”的字符，然后用空字符串替换结束的“]”。

您需要 [^]] 位来防止贪婪匹配将“[1] sdc2[0]”视为示例字符串中的单个匹配项。

至于您的评论：

sed 's#\([^[ ]*\)\[[^]]*\]#/dev/\1#g'

我将分隔符从通常的“/”切换为“#”，只是为了避免转义您要求的 /dev/ 位（我不会说“为了清楚起见”）
\(...\) 位匹配一个子组，这里是 sdc2 或其他，所以我们可以在替换中引用它，
子组使用与我们丢弃索引的字符类类似的字符类： <代码>[^[ ] 表示除“[”（同样，为了避免贪婪地匹配索引）或空格（假设您的值按照您的帖子以空格分隔）之外的任何字符，
替换现在是文字“/dev/”后面跟着第一个（也是唯一的）子组匹配，
最后的 g 标志告诉它每行执行多个匹配，而不是在第一个匹配处停止

sed 's/\[[^]]*\]//g'

reads: substitute any string with literal "[" followed by zero or more characters that aren't a "]", and then the closing "]", with an empty string.

You need the [^]] bit to prevent greedy matching treating "[1] sdc2[0]" as a single match in your sample string.

As for your comment:

sed 's#\([^[ ]*\)\[[^]]*\]#/dev/\1#g'

I switch the seperator from the usual '/' to '#', just to avoid escaping the /dev/ bit you asked for (I won't say "for clarity")
the \(...\) bit matches a subgroup, here sdc2 or whatever, so we can refer to it in the replacement
the subgroup uses a similar character class to the one we used discarding the index: [^[ ] means any character except an "[" (again, to avoid greedily matching the index) or a space (assuming your values are space-delimited as per your post)
the replacement is now the literal "/dev/" followed by the first (and only) subgroup match
the g flag at the end tells it to perform multiple matches per line, instead of stopping at the first one

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