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find and Replace String in Perl - Regexp


 
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# 1  
Old 10-30-2015
find and Replace String in Perl - Regexp

Trying to find and replace one string with another string in a file

Code:
#!/usr/bin/perl

	$csd_table_path = "/file.ntab";
	$find_str = '--bundle_type=021';
	$repl_str = '--bundle_type=021 --target=/dev/disk1s2';
	if( system("/usr/bin/perl -p -i -e 's/$find_str/$repl_str/' $csd_table_path") != 0 ) {
		print "\n\n\nERROR: adding new target to <$csd_table_path>!\n\n\n";
		exit 1;
	}
	print "Msg: file modified \n";

exit 0;

however getting errors ...

Bareword found where operator expected at -e line 1, near "--target"
(Missing operator before target?)
Unknown regexp modifier "/v" at -e line 1, at end of line
syntax error at -e line 1, near "--target"
Execution of -e aborted due to compilation errors.
# 2  
Old 10-31-2015
$repl_str contains a / that is also used as a separator.
Take another separator
's#$find_str#$repl_str#'
--
Why do you run another perl instance?
This User Gave Thanks to MadeInGermany For This Post:
# 3  
Old 10-31-2015
Another option is to escape the `/'
Code:
$repl_str = '--bundle_type=021 --target=\/dev\/disk1s2';

Code:
#!/usr/bin/perl
.
.
.
if( system("/usr/bin/perl -p -i -e 's/$find_str/$repl_str/' $csd_table_path") != 0 ) {
		print "\n\n\nERROR: adding new target to <$csd_table_path>!\n\n\n";
		exit 1;
	}
.
.
.

Running Perl inside Perl by invoking a system call is like running a virtual machine inside a virtual machine, to list the content of a file system. By the way, I do not believe it does what you think it does, neither.
This User Gave Thanks to Aia For This Post:

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