I have a large file and I want to print the lines that do not match a particular undesired pattern. The following does the exact opposite of what I want, namely it retains all the undesirable lines.
我有一个大文件，我想打印与特定不需要的模式不匹配的行。以下与我想要的完全相反，即它保留了所有不需要的行。
grep -e '\[0.0, 0.0\]' locscore.txt
How can I get the lines that DON'T have the above pattern? I tried
如何获得不具有上述模式的线条？我试过了
grep -e '^*(?!\[0.0, 0.0\])*$' locscore.txt
but it produces nothing.
但它什么都没产生。
1 个解决方案
#1
3
if you use grep, there is an option -v, it does what you need.
如果你使用grep，有一个选项-v，它可以满足你的需要。
from man page:
来自手册页：
-v, --invert-match
Invert the sense of matching, to select non-matching lines. (-v is specified by POSIX.)
#1
3
if you use grep, there is an option -v, it does what you need.
如果你使用grep，有一个选项-v，它可以满足你的需要。
from man page:
来自手册页：
-v, --invert-match
Invert the sense of matching, to select non-matching lines. (-v is specified by POSIX.)