Extracting fields from an output 8-)

csaha · #1 (**permalink**) 01-19-2006

I am getting a variable as x=2006/01/18

now I have to extract each field from it.
Like x1=2006, x2=01 and x3=18.

Any idea how?

Thanks a lot for help.

Thanks
CSaha

vino · #2 (**permalink**) 01-19-2006

Code:

[/tmp]$ echo "2006/01/18" | { IFS="/" read a b c; echo $a $b $c ; }
2006 01 18

vino · #3 (**permalink**) 01-19-2006

On a deeper note, I think I know where you are heading to...

If that solution does not work, pick up some clues from this thread Breaking input with "read" command

csaha · #4 (**permalink**) 01-19-2006

Excellent Vino !!!

After posting I tried this and resolved

$ echo "2006/01/18" | awk -F"/" '{print $1}'
2006

$ echo "2006/01/18" | awk -F"/" '{print $2}'
01

$ echo "2006/01/18" | awk -F"/" '{print $3}'
18

See this also ...

matrixmadhan · #5 (**permalink**) 01-19-2006

here is one more in sed,

Code:

echo 2006/01/18 | sed 's/\(.*\)\/\(.*\)\/\(.*\)/year: \1 month: \2 day: \3/'

year: 2006 month: 01 day: 18

aigles · #6 (**permalink**) 01-20-2006

You can also use 'cut'

year=`echo 2006/01/18 | cut -d/ -f1`
month=`echo 2006/01/18 | cut -d/ -f2`
day=`echo 2006/01/18 | cut -d/ -f3`

But, i think tah the better solution is Vino post

echo 2006/01/18 | IFS=/ read year month day

Jean-Pierre.

mona · #7 (**permalink**) 01-19-2006

Code:

echo "2006/01/18" | awk -F"/" '{print "x1="$1",x2="$2 ",x3=" $3}'

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