Doubt in grep command

I think the single quote doesn't matter in the scenario!!!

hmmm okay.. pls come to a conclusion and lemme know..

Here ^ stands for negation.The output shows all lines which contain characters other than abcd.Since all your lines in the file have characters other than abcd, all lines are printed.

This is the conclusion! And quotes doesn't matter!

okie:) thnx all of ya

[336560@oracleclient commands]$ cat 5.txt
asdf
1234
asdrg
awsrdq
kjhgkji


[336560@oracleclient commands]$ grep [^asd] 5.txt
1234


According to the conclusion made.,
The output should print all lines which contains characters other than a,s and d. So all lines must have been printed. But it isn't the case.

I thought it works this way. Prints all lines which has neither of the characters inside [ ].
If it works this way,last 2 lines should have been printed. But it isn't the case too

Someone explain.

Code:


cat > 5.txt
asdf
1234
asdrg
awsrdq
kjhgkji

[335818@oracleclient ~]$ grep [^asd] 5.txt
asdf
1234
asdrg
awsrdq
kjhgkji


This is the output I got.I don't understand why you get a different output!

ls | grep "a*"

is displaying all the files in the directory including those whose names do not contain a

What is the error?

How to use -l with grep command?

Syntax

Code:

$ grep -l 'orange' *

This ll print name of all the files having 'orange'.