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How to replace a value in a file in perl?

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# 1  
Old 09-09-2010
How to replace a value in a file in perl?
Hi,

I have a file with the following contents and i want to replace that value in a file with some other value.

Code:
#file.txt

/local/Disk/data
n
192.168.55.98
52035
3
1
2
1
1
192.168.55.98

Here is my code.
Code:
open(FH,"file.txt");
@array=<FH>;
print "\n array[7]\n";


I want to replace the $array[7] (i.e 1) with 2 in a file.

The output should be like this:
Code:
/local/Disk/data
n
192.168.55.98
52035
3
1
2
2
1
192.168.55.98

How can i do it in perl?

Regards
Vanitha
# 2  
Old 09-09-2010
For infile replacement -

Code:
$
$
$ cat -n f10
     1  /local/Disk/data
     2  n
     3  192.168.55.98
     4  52035
     5  3
     6  1
     7  2
     8  1
     9  1
    10  192.168.55.98
$
$
$ perl -pi.bak -le '$.==8 && s/$_/2/' f10
$
$ cat -n f10
     1  /local/Disk/data
     2  n
     3  192.168.55.98
     4  52035
     5  3
     6  1
     7  2
     8  2
     9  1
    10  192.168.55.98
$
$
$ cat -n f10.bak
     1  /local/Disk/data
     2  n
     3  192.168.55.98
     4  52035
     5  3
     6  1
     7  2
     8  1
     9  1
    10  192.168.55.98
$
$
$

tyler_durden
# 3  
Old 09-10-2010
Quote:
Originally Posted by durden_tyler
For infile replacement -

Code:
$
$
$ cat -n f10
     1  /local/Disk/data
     2  n
     3  192.168.55.98
     4  52035
     5  3
     6  1
     7  2
     8  1
     9  1
    10  192.168.55.98
$
$
$ perl -pi.bak -le '$.==8 && s/$_/2/' f10
$
$ cat -n f10
     1  /local/Disk/data
     2  n
     3  192.168.55.98
     4  52035
     5  3
     6  1
     7  2
     8  2
     9  1
    10  192.168.55.98
$
$
$ cat -n f10.bak
     1  /local/Disk/data
     2  n
     3  192.168.55.98
     4  52035
     5  3
     6  1
     7  2
     8  1
     9  1
    10  192.168.55.98
$
$
$

tyler_durden
Hi,

Thanks for the reply but i am not able to execute i am getting syntax error is there any other way then perl one liners?

Code:
Can't find string terminator "'" anywhere before EOF at -e line 1.

Regards
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