我需要一种宏观/功能,从顶头开始,以寻找一种具体模式。 删除第1行的斜线。
举例来说,本说明:
"This is the first line
"This is the third line
if exists("did_something")
如果它发现这种模式,否则就会停止。
All help appreciated!
让我说一下你的模式是这封信。
a. 寻求一条没有评论的行文,但包含以下内容:
/^\%(s*")@!.*e
使用了负头号(@!),以确认该行的起步(<>>>/code>)后面是一开始评论(whitespace, 然后是一番引用和s*>),但随后是一(.*e)。 <代码/>%(>>>>>使所附模式成为其他操作者(如负光头)可以作为一个单位运作的一种原子。
采用<代码>:g
:g/^\%(s*")@!.*e/echo "found one"
查看现有线路是否匹配,使用<代码>match()。
:echo match( getline(line( . )), ^\%(s*")@!.*e ) >= 0 ? true : false
监管部门几乎总是一样,问题在于你想要与对应关系做些什么?
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