使用 sed 在分隔符之间打印文本 [英] Print text between delimiters using sed

问题描述

假设我有 op(abc)asdfasdf 并且我需要 sed 在括号之间打印 abc.什么对我有用?(注意:我只想要一行中第一对分隔符之间的文本，如果特定的输入行没有一对括号，则不需要.)

Suppose I have op(abc)asdfasdf and I need sed to print abc between the brackets. What would work for me? (Note: I only want the text between first pair of delimiters on a line, and nothing if a particular line of input does not have a pair of brackets.)

推荐答案

sed -n -e '/^[^(]*(\([^)]*\)).*/s//\1/p'

该模式查找以零个或多个非开括号字符列表开头的行，然后是开括号；然后开始记住零个或多个不是右括号的字符列表，然后是右括号，然后是任何字符.用您记住的列表替换输入并打印它.-n 表示默认不打印" - 任何没有括号的输入行都不会被打印.

The pattern looks for lines that start with a list of zero or more characters that are not open parentheses, then an open parenthesis; then start remembering a list of zero or more characters that are not close parentheses, then a close parenthesis, followed by anything. Replace the input with the list you remembered and print it. The -n means 'do not print by default' - any lines of input without the parentheses will not be printed.

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