Regex and lookahead to specific number of chars

Ruby
1

If I have a string such as
“I love ruby, i really do love ruby, oh yes i do blah blah blah”

and i want to insert an asterix at every 10th char whether its a
whitespace, digit whatever how do i go about it

i was trying

string.gsub!(/[.]{10}/, “*”)

but it doesnt work

any hints???

2

Le 25 février à 21:28, Adam A. a écrit :

and i want to insert an asterix at every 10th char whether its a
whitespace, digit whatever how do i go about it

i was trying

string.gsub!(/[.]{10}/, “*”)

What about string.gsub!(/(.{9})./, ‘\1*’) ?

Fred

3

On 25.02.2008 21:28, Adam A. wrote:

but it doesnt work

any hints???

Remove the square brackets around the dot. The dot is meta only outside
of them.

Cheers

robert

4

Thanks Fred and Robert for that.

I dont understand what the \1 does. I know if i remove it it removes all
the characters before the 10th and if i include it in keeps them. Why is
this?

5

oh and a totally newb question but when i try to substitute the asterix
with a newline \n it just prints it instead of creating a newline, why
is that???

i wish there some regex exercises out there instead of the usual
tutorial only stuff.

6

Adam A. wrote:

I dont understand what the \1 does. I know if i remove it it removes all
the characters before the 10th and if i include it in keeps them. Why is
this?

When you put a regexp in round brackets, what it matches is remembered.
In a replacement string, you can then refer to them as \1, \2, etc.

oh and a totally newb question but when i try to substitute the asterix
with a newline \n it just prints it instead of creating a newline, why
is that???

Since the replacement string is in single quotes, \n is not special so
represents the two characters \ and n. To get a newline, you need to
put \n in double quotes ("\n"). In double quotes, the backslash quotes
any character, so \1 would become the character with code 1. So in
double quotes, the \1 needs to be \1 so:

string.gsub!(/(.{9})./, “\1*”)

for the original example and:

string.gsub!(/(.{9})./, “\1\n”)

for the newline example.

7

Le 27 février à 23:40, Adam A. a écrit :

Thanks Fred and Robert for that.

I dont understand what the \1 does. I know if i remove it it removes all
the characters before the 10th and if i include it in keeps them. Why is
this?

It’s a reference to the first parenthesied group. To decompose the
expression :

( => begin group 1
.{9} => find exactly 9 characters
) => end group 1
. => followed by exactly one character

I replace that whole group by the first group (my 9 characters) and a
star.

There are best ways to do it (especially if you’re manipulating very big
strings, I think), but I think it’s the clearest.

Fred