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Finding the pattern and replacing the pattern inside the file

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# 8  
Old 07-25-2013
Assuming that you are using the variable after assigning it on the file...(else it will print a blank in place of &xxx string)

Code:
$ perl -pe '$hash{$1}=$2 if /\&(\w+)\=(\w+)/;s/\&(\w++)(?!\=)/$hash{$1}/g;' file
go to &abc=ddd;
 if file found &ccc=10;
 no the value name is ddd;
  and the age is 10;

# 9  
Old 07-25-2013
Hi rdrtx1,

It not working for the generic thing. i dont have same line there will be different lines but with the same format. if you have any idea share it.

---------- Post updated at 03:32 AM ---------- Previous update was at 03:31 AM ----------

Guys if you have idea help me out in this.
Last edited by saaisiva; 07-25-2013 at 05:47 AM..
# 10  
Old 07-25-2013
Have you tried the perl solution above..?
# 11  
Old 07-25-2013
Hi Rajamadhavan,

Code:
Nested quantifiers in regex; marked by <-- HERE in m/&(\w++ <-- HERE )(?!=)/ at -e line 1.

i am getting following error when i run your code
# 12  
Old 07-25-2013
Whats your perl version ? Possessive quantifiers (\w++) works on versions > v5.10. If you can upgrade to perl 5.12 or latest version, it will work...

---------- Post updated at 04:23 AM ---------- Previous update was at 04:14 AM ----------

Try this one..should work in older version (>= 5.8)
Code:
perl -pe '$hash{$1}=$2 if /\&(\w+)\=(\w+)/;s/\&((?>\w+))(?!\=)/$hash{$1}/g;' file

# 13  
Old 07-25-2013
Hi Rajamadhavan,

Ya now it works but the it should not replce the undeclared variable tp space. as i need those values.
# 14  
Old 07-25-2013
will get slightly complicated.
Code:
perl -pe '$hash{$1}=$2 if /\&(\w+)\=(\w+)/; m/\&((?>\w+))(?!\=)/g;s/\&((?>\w+))(?!\=)/$hash{$1}/g if(defined $hash{$1});' file

Please note the above solutions is under the assumption that you have one assignment and/or one substitution per line.

if you have something like below, we need a bigger script.
Code:
&abc value is &bcd
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