Find a string and extract a value from a file

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# 1  
Old 08-19-2009
Find a string and extract a value from a file

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
# 2  
Old 08-19-2009
this will do i guess
echo "n0=7.00 !Central density [10**19 m-3]"|awk -F"[=!]" '/^n0/{print $2}'

# 3  
Old 08-19-2009
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!
# 4  
Old 09-04-2009
voce ja conseguiu? se nao .. tai

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

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\=
# 5  
Old 09-04-2009

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# 6  
Old 09-04-2009

echo 'n0=7.00 !Central density [10**19 m-3]' | awk -F '=|!' '{print $2}'
# 7  
Old 09-04-2009
Or -

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

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