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Bash-查找正则表达式并用另一个字符串替换

[英]Bash - Find and replace regex with another string

I have the following string libVersion = '1.23.45.6' and I need to replace 1.23.45.6 with 1.23.45.7 . 我有以下字符串libVersion = '1.23.45.6' ,我需要将1.23.45.6替换为1.23.45.7

Obviously the version could be any number with similar format (it does not have to be 4 numbers). 显然,该版本可以是格式相似的任何数字(不一定是4个数字)。

I tried to use the following but doesn't work 我尝试使用以下内容,但不起作用

echo "libVersion = '1.23.45.6'" |sed "s/([0-9\\.]+)/1.23.45.7/g"

Basic sed, ie sed without any arguments uses BRE (Basic Regular Expression). 基本sed,即不带任何参数的sed使用BRE (基本正则表达式)。 In BRE , you have to escape + , to bring the power of regex + which repeats the previous token one or more times, likewise for the capturing groups \\(regex\\) BRE ,您必须转义+ ,以使regex +重复一次或多次重复先前的标记,对于捕获组\\(regex\\)

echo "libVersion = '1.23.45.6'" | sed "s/[0-9.]\+/1.23.45.7/"

You may also use a negated char class to replace all the chars exists within single quotes. 您也可以使用否定的char类来替换单引号中存在的所有char。

echo "libVersion = '1.23.45.6'" | sed "s/'[^']*'/'1.23.45.7'/"

Since the replacement should occur only one time, you don't need a g global modifier. 由于替换只发生一次,因此不需要g全局修饰符。

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