使用grep查找以$开头的模式

Question

I need to find a pattern that starts with a $ is followed by two numbers, a single character that is not a number, and anything else.

I know how to find a pattern starting in a dollar sign and followed by two numbers but I can't figure out how to check for one character that is not a number.

I also need to count how many lines have this pattern.

I have this so far:

grep -Ec '\$[0-9][0-9].....

I don't know what to do. Can someone please help? Any help would be much appreciated.

Answer 1

插入符号反转选择组，因此，如果[0-9]为“匹配任何数字”，则[^0-9]为“匹配任何非数字”。

Answer 2

You can possibly try this regex \\$[0-9][0-9][^0-9].*

\\$[0-9][0-9][^0-9].*

• \\$ matches the character $ literally
• [0-9] match a single character present in the list below. 0-9 a single character in the range between 0 and 9
• [0-9] match a single character present in the list below. 0-9 a single character in the range between 0 and 9
• [^0-9] match a single character not present in the list below. 0-9 a single character in the range between 0 and 9
• .* matches any character (except newline) Quantifier: * Between zero and unlimited times, as many times as possible, giving back as needed [greedy]

Answer 3

I would second @realspirituals answer, and if you need to count how many lines have this pattern, you can count how many lines grep ouputs by piping to wc -l . In order to both show the lines and count them in one fell swoop, pipe the output like so

grep "\$[0-9]{2}[^0-9].*" | tee >(wl -l)

where tee will split the output between wl and STDOUT . {2} will cause the prior [0-9] to match twice.