使用grep处理文本文件

Question

Ok, so I have a text file formatted like this, it has several rows going down.

EXAMPLE1:EXAMPLE2:EXAMPLE3
EXAMPLE1:EXAMPLE2:EXAMPLE3
EXAMPLE1:EXAMPLE2:EXAMPLE3

I want to use grep to change it to look like this

EXAMPLE2:EXAMPLE3

Basically I want to take out the first part before the :

If someone could please tell me how to do this it would be greatly appreciated.

Answer 1

You can use the cut utility to filter out columns given a delimiter.

In this example, the delimiter is : and you want every column starting with the second:

$ cut -d: -f2- < input.txt

Answer 2

You can use -o flag to select the part of the input which matches a regex in grep .

Example

$ grep -o "[^:]*:[^:]*$" input
EXAMPLE2:EXAMPLE3
EXAMPLE2:EXAMPLE3
EXAMPLE2:EXAMPLE3

Answer 3

Another way to do this. Hope this helps.

awk -F":" '{printf "%s:%s\n" ,$1,$2}' file.txt

Output

EXAMPLE1:EXAMPLE2
EXAMPLE1:EXAMPLE2
EXAMPLE1:EXAMPLE2

Answer 4

Here it is:

cut -d':' -f2,3 filename

Output:

EXAMPLE2:EXAMPLE3
EXAMPLE2:EXAMPLE3
EXAMPLE2:EXAMPLE3