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Grep only line starting with....

 
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# 1  
Old 07-24-2013
Grep only line starting with....
Hello,
I have a command that show some application information. Now, I have to grep there informations, like:

Code:
# showlog | grep 1266
1266.1369866124 :: 
1266.1304711286 :: 
41031.1161812668 :: 
41078.1301266480 :: 
41641.712662564 :: 
1266.333792515 :: 
41462.1512661988 :: 
1266.54932671 :: 
12666.54345932671 ::

What I need from grep is the line that start with 1266. (with dot) But that string is also contained into other rows and the grep reports wrong output.

The output I'm expecting is:
Code:
1266.1369866124 :: 
1266.1304711286 :: 
1266.333792515 :: 
1266.54932671 ::

Thanks who can help me.
Lucas
# 2  
Old 07-24-2013
Hello,


Could you please try the following and let me know if this helps.


1st way:
Code:
 
$ cat grep_starting | grep '^1266*'

1266.1369866124 ::
1266.1304711286 ::
1266.333792515 ::
1266.54932671 ::
$


2nd Day:

Code:
 
$ awk '/^1266/' grep_starting

1266.1369866124 ::
1266.1304711286 ::
1266.333792515 ::
1266.54932671 ::
$



Kindly try same and let me know if you have any queries.



Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
Lord Spectre
# 3  
Old 07-24-2013
Code:
showlog | grep "^1266\."

This User Gave Thanks to Yoda For This Post:
Lord Spectre
# 4  
Old 07-24-2013
Thanks, this one:
Code:
| grep '^1266*'*


Works perfect!!

EDIT: Also the one provided by Yoda works! Smilie

# 5  
Old 07-24-2013
Note that '^1266*' will return lines starting with 1266 followed by any character:
Code:
1266.1369866124 ::
1266.1304711286 ::
1266.333792515 ::
1266.54932671 ::
12666.54345932671 ::

You should use "^1266\." to get only thoses starting with 1266.

Period is a metacharacter, hence escaped to preserve its literal meaning.
This User Gave Thanks to Yoda For This Post:
RavinderSingh13
# 6  
Old 07-24-2013
Thanks Yoda for information, I was about to say the same.
This User Gave Thanks to RavinderSingh13 For This Post:
Yoda
 
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