Shell Programming and Scripting

I have a file where a line has the following form:

n0=7.00 !Central density [10**19 m-3]

and I want to extract the value 7.00. I used to do this with the order below, which finds the string "n0" and take the rest of the line parting from the separator "=", but the comment "Central density..." makes this solution not applicable to this case,

>n0=` grep "n0" myfile.dat | cut -d "=" -f2,3 `

Anyone knows how can I improve it in order to take only the number?. Thanks in advance

this will do i guess

Code:

echo "n0=7.00 !Central density [10**19 m-3]"|awk -F"[=!]" '/^n0/{print $2}'

Thanks for the reply, but actually n0 can take any value, it is not 7.00 necessarily. I've solved it concatenating a second cut command which defines another delimiter " ":

> n0=` grep "n0" $bsim | cut -d "=" -f2 | cut -d " " -f1`

Thanks again!

www:~# echo "n0 \u003d 7,00! Central densidade [10 ** 19 m-3]" | awk '{print $3}' |cut -f1 -d\
7,00
www:~#


Espero ter ajudado.

---------- Post updated at 12:45 PM ---------- Previous update was at 12:42 PM ----------

www:~# echo "n0=7.00 \!Central density [10**19 m-3]" | awk '{print $1}' |cut -f1 -d\! |cut -f2 -d\=
7.00
www:~#

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or...

echo 'n0=7.00 !Central density [10**19 m-3]' | awk -F '=|!' '{print $2}'

Or -

Code:

$
$ cat f1
first line
n0=7.00 !Central density [10**19 m-3]
third line
$
$ perl -nle '/^n0=(.*?) / && print $1' f1
7.00
$
$

tyler_durden