变量str
包含多行字符串。我想从中掏出一条线。我想grep字符串行kernel release 721
并将其分配给变量。有没有办法这样做?
irb(main):152:0> print str
*************************** Component ********************
c:\temp\agent\bin\CAR.exe: 721, patch 618, changelist 1633822, NTAMD64, opt
**********************************************************
--------------------
AGENT information
--------------------
kernel release 721
kernel make variant 721_REL
compiled on NT 6.1 7601 S x86 MS VC++ 14.00 for NTAMD64
compiled for 64 BIT
compilation mode Non-Unicode
compile time Mar 21 2016 21:07:50
patch number 12
latest change number 1659167
---------------------
supported environment
---------------------
operating system
Windows NT 5.0
Windows NT 5.1
Windows NT 5.2
Windows NT 6.0
Windows NT 6.1
答案 0 :(得分:5)
我认为您对该值感兴趣,即721
:
kernel_release = str[/^kernel release\s+(\w+)$/, 1]
#=> "721"
正则表达式匹配以kernel release
开头的行,后跟空格,以一个或多个单词字符结尾。后者被捕获,第二个参数1
指的是捕获组。
答案 1 :(得分:3)
string.lines.grep /pattern/
或
string.each_line.grep /pattern/
可能效率更高