UNIX for Dummies Questions & Answers

I want to grep based on first column and print all the columns. for eg.

Code:

root       12344       /sh
root       33389       /bash
oracle      87378      /tech/oracle
oracle      26367      /tmp/oracle

Now I want to grep based on user "root" in first column and print like below.

Code:

root       12344       /sh
root       33389       /bash

Please help me.

Code:

awk -v val='root' '$1 == val' myFile

Code:

cat file| grep root*

file

Code:

root 12344 /sh
root 33389 /bash
oracle 87378 /tech/oracle
oracle 26367 /tmp/oracle

output

Code:

root 12344 /sh
root 33389 /bash

You keep doing this, this is a useless use of cat. Almost no commands need cat's help to read an individual file.

Also, your regex is wrong. grep doesn't work that way. Your regex would match "root", "roott", "roottttttttt", and so forth. Which doesn't actually change anything since you don't have the regex anchored anywhere. You might as well have left the * out completely...

Code:

grep "^root" file

Please consider a little more carefully before giving solutions that are worse than ones already given.

the value 'root' can be in any field. I want to grep the value root only in field-1 and print all the fields based on that.

The expression I gave, "^root", anchors "root" to the beginning of the line.

The awk solution tests against the first token.

Slight modification:

Code:

grep -w "^root" file

OR

Code:

grep "^root " file