Match and Replace quotes in a string

why can't u use a slight variation of your code as:

match w/

\x27([^x27]+?)\x27

then replace w/

&quote;$1&quote;

it will render the target text

"Test!" For "quote" "good then"

to

&quote;Test!&quote; For &quote;quote&quote; &quote;good then&quote;

your using of \b is inefective, because u need to handle multi-word strings like

"good then"

which have white spaces in them, so using \b loses its validity in this case.

SergeiZ at 2007-10-2 >

Gary7 wrote:

I'm using regex in vb.net to search through strings for quote characters, and replace them with a new string: "

If you really are just trying to replace every occurrence of the quote character with another string then using a regular expression seems to be overkill when you could just use String.Replace (String, String).

Dim test1 As String = """Test!"" For ""quote"" ""good then""" Dim test2 As String test2 = test1.Replace("""", """)

FrankBoyne at 2007-10-2 >

Hi Frank, thanks for this!

I am going to include some routines such as you have indicated, but in reality I need to test for several different characters and replace them. This includes varieties of characters as well, such as the differeing versions of quotations, etc.

Which will lead me to another question about search and replace:

Can one search for multiple characters of different types, and replace them each with diferent characters in one search pattern?

Thanks!

Gtary

Gary7 at 2007-10-2 >

Thanks for this, Sergei! I caught the hex - I think it captures single quotes ' . Anyway, that does what I need and I thank you.

How about if I want to also add another character to match, as well a seperate replacement string. Can this be done in one string, or should I trap them in another regex?

any ideas?

Thanks!

Gary7 at 2007-10-2 >

Hi Sergei,

I simply asked a question in principal. One could imagine any set of characters, I would think, but if not;

suppose I want to search for more than just quotes in a string. Maybe I need to also find all the single quotes and apostrophe's as well, and replace them with some other string of characters.

So with a single-item match, we used: \x22([^\x22]+?)\x22 which will find all the hex equivs of standard quotations.

But what if the quotes are in a "pretty" state - inverted at one end of the word -- and thus aren't recognized in the hex value x22?

Since "pretty quotes" would be replaced the same as regular quotes, it would stand to reason we only need to match the instances along with regular quotes, and change them both to the desired characters.

But what if I wanted to match ampersands and replace those with an html equivilent. Is this something that can be accomplished within the same expression, or would a new one need to be constructed?

That is the question, so do you need any more info?

Thanks!

Gary

Gary7 at 2007-10-2 >

Very nice!

Thank you Sergei for the discussion. it has done more than the countless articles, examples and sample chapters I've read over the past 4 days!

I'll be experimenting with this a bit. I'll certainly post back if needed!

Thanks again!

Gary7 at 2007-10-2 >

Hi Sergei; I promissed I'd have more questions, and of course I do!

So, running the expression you provided:

\x22(?<quotes_group>[^\x22]+?)\x22|(?<ampersand_group>&)

using this string:

"Test This Joe," "and Push & it & hard!"

returns this analysis:

Match 1: "Test This Joe," 0 16

Group "quotes_group": Test This Joe, 1 14

Group "ampersand_group" did not participate in the match

Match 2: "and Push & it & hard!" 17 23

Group "quotes_group": and Push & it & hard! 18 21

Group "ampersand_group" did not participate in the match

So group ampersand isn't finding the ampersand character.

Does this need to be built up with the brackets and caret?

I thought the actual ampersand symbol didn't need to be

escaped; no?

Thanks!

Gary7 at 2007-10-2 >