从正则表达式获取子字符串-gettingasubstringfromaregularexpression

3  

Here is how to do it using sed

以下是使用sed的方法

grep SRC=\'.*\' | sed 's/SRC=.\(.*\)./\1/'

2  

Are the lines to match always in the format SRC='nnn' ? Then you could use

是否匹配的行始终采用SRC ='nnn'格式?然后你可以使用

grep SRC | cut -d"'" -f2

2  

You can use the -o option on grep to return only the part of the string that matches the regex:

您可以在grep上使用-o选项仅返回与正则表达式匹配的字符串部分:

echo "SRC='999'" | grep -o -E '[0-9]{3}'

1  

You can't do it with plain grep. As the man page on my box states: "grep - print lines matching a pattern" grep only prints lines, not part of lines.

你不能用普通的grep做到这一点。正如我的框上的手册页所述:“grep - 匹配模式的打印行”grep只打印行,而不是行的一部分。

I would recommend awk since it can do both the pattern matching and sub-line extracting:

我会推荐awk,因为它可以同时进行模式匹配和子行提取:

awk -F\' ' /SRC/ {print $2}'

1  

just sed will do

只是sed会做的

$ echo SRC='999' | sed '/SRC/s/SRC=//'
999

0  

Platform grep or general regular expression?

平台grep或一般正则表达式?

Regex

SRC\=\'(\d{3})\'

0  

why grep? How about..

为什么grep?怎么样..

substr("SRC='999'",6,3)

0  

depends on your platform / language.

取决于您的平台/语言。

in Ruby:

string = "SRC = '999'"

string.match(/([0-9]{3})/).to_s.to_i 

will return an integer

将返回一个整数