Perl /Sed replacement

perlsed

Below command is used to replace the password in a script but the perl command is adding a white space when it does the subsstitution

password=arche20
perl -i -p -e "s/^(password[]*=[ ]*).*$/\1 $passwd/" config.properties

cat config.properties
userid=ARCHE
password= arche20

It does the work but it seems to be adding a space after password when it replaces.
Could this be done without the space?
Got a sed way:

sed -i "s/password.*/password=$passwd/g"

Best Answer

You have a space after \1 in your replacement, just remove that and you should be good

perl -i -p -e "s/^(password[]*=[ ]*).*$/\1$passwd/" config.properties
                                          ^
                                          Removed space here

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Replace non-printable characters in perl and sed

That's a typical job for tr:

LC_ALL=C tr '\0-\10\13\14\16-\37' '[ *]' < in > out

In your case, it doesn't work with sed because you're in a locale where those ranges don't make sense. If you want to work with byte values as opposed to characters and where the order is based on the numerical value of those bytes, your best bet is to use the C locale. Your code would have worked with LC_ALL=C with GNU sed, but using sed (let alone perl) is a bit overkill here (and those \xXX are not portable across sed implementations while this tr approach is POSIX).

You can also trust your locale's idea of what printable characters are with:

tr -c '[:print:]\t\r\n' '[ *]'

But with GNU tr (as typically found on Linux-based systems), that only works in locales where characters are single-byte (so typically, not UTF-8).

In the C locale, that would also exclude DEL (0x7f) and all byte values above (not in ASCII).

In UTF-8 locales, you could use GNU sed which doesn't have the problem GNU tr has:

sed 's/[^[:print:]\r\t]/ /g' < in > out

(note that those \r, \t are not standard, and GNU sed won't recognize them if POSIXLY_CORRECT is in the environment (will treat them as backslash, r and t being part of the set as POSIX requires)).

It would not convert bytes that don't form valid characters if any though.

To replace % marks but not \% in Perl Regex

Like many, if not most, text parsing tools, perl can take input from the command line, there's no need for cat. You just need -e which lets you pass a script as a command line parameter and -n which means "run the script on each line of input". ALternatively, you can use the -p switch which means "run the script on each line of input, then print that line". These two commands are equivalent (but the second is a classic useless use of cat, use the first) :

perl -pe 's/foo/bar/' file
cat file | perl -pe 's/foo/bar/'

Now, if I understand correctly, you want to delete all LaTeX comments (though that's not what your question states). If so, a lookbehind is the easiest way:

perl -pe 's/(?<!\\)%.*//' file

Your regex should also work, you just need to keep the character you matched before the % and escape the backslash:

perl -pe 's/(^|[^\\]+)%.*/$1/' file

You can do the same thing with GNU sed:

sed -r 's/(^|[^\\])%.*/\1/' file

Best Answer

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Replace non-printable characters in perl and sed

To replace % marks but not \% in Perl Regex

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