Hi guys


I want to view one field of a file and i just cannot figure out how to with the cut -d command.The file will be something like
now i want to view only the ip address "10.0.0.1" nothing else please help


t

How about this?

cut's default delimiter is the TAB, use -d to set the delimiter to SPACE and -f to select the field.

The -d option to cut is meant to define a delimiter different from the default one (TAB). If you want to use a blank space as delimiter you have to do something like
where the blank space after -d is escaped to let the shell interpret it correctly. In alternative you can embed it in single or double quotes:
Regarding the suggested awk command, two notes: piping the output from cat is not necessary, since awk accepts one or more file names as arguments. Also the definition of FS is not needed, because the blank space is the default field separator. Hence the command would be simply:

thanks guys that helped alot one more questino is there a way to view a certain line of a file. for example

or
or
or...

I think this is correct one.....

sed -n 7p file




.

Oh yeah, just a typo. I stand corrected. Thanks.

thanks guys that helped alot...thats what i love about linux we have a community of SKILLED people around the world who are willing to help and there is no question that cannot be answered

Hi,

You can run this command to be able to get the third field:

echo "ip address 10.0.0.1 hostname linux1 model dell" |sed s/' \{1,\}'/";"/g|cut -d";" -f3

sed s/' \{1,\}'/";"/g will drop all blank space between each field.