Regex examples

[eschwartz]

Originally Posted by signum

Yeah, it works, sorta. But it matches all paragraphs with quotation marks, whether or not they are "valid", i.e. matched or balanced. This was first reported by the OP in msg #428, above.

Well, my regex was not meant to find paragraphs where the only quotation marks were at the end. In fact, I specifically precluded the idea. You need a second regex for that.

But they should correctly have found #1 and #3, and I cannot see why they would find, for example,

<p>"Hello."</p>

So how exactly was my regex matching valid paragraphs, I wonder? This bit should make it impossible (unless there was random extra space at the end or some such?):

[^"”]</p>

[eschwartz]

Originally Posted by Buchstabensalat

Hi,

Thank you.

Using Sigil`s search (regex mode) your first string finds every single word in the text, the second string matches every uppercase word in my text and not only those that meet my search criteria ( uppercase letter(s) within a word). Is there a possiblity to match only those words with upercase letters in the word ?

Thanks again

Regards,

Buchstabensalat

I would use the following:

Find:

\b(\p{L})((?:\p{Ll}*\p{Lu})+\p{Ll}*)\b

Replace:

\1\L\2\E

NOTE: This uses unicode code points, on recommendation. Also, I have specifically saved the current case value of the first letter of the word, rather than assuming all words are lowercase.

Other than that, the only real difference from Doitsu's regex is that I specifically looked for uppercase letters inside the match, rather than applying an equalizer on all words -- this has the same result, but ignores matches where we wouldn't end up changing anything.

[eschwartz]

On an almost-related note, Kovid Goyal has just (as of 6 hours ago) begun work on function mode in calibre's editor!

[Buchstabensalat]

Originally Posted by eschwartz

I would use the following:

Find:

\b(\p{L})((?:\p{Ll}*\p{Lu})+\p{Ll}*)\b

Replace:

\1\L\2\E

...

works fine

[eschwartz]

Originally Posted by Buchstabensalat

works fine

You're welcome.

[rubeus]

Hi,

i'm currently working on combining paragraphs. I've a bunch of regex' which are working fine.

But for this:

...he went on,</i></p>
<p><i> shaking his head...

i'm unable to find a solution. Italic is just an example, it can be a <b>, <span> <img> something completely different.

What i can do is of course something like this:

,</i></p>\s+<p><i>

to replace it with

,

and have one for italic, one for bold (or even combine them with | but is it possible to have that a little bit more in general?

Thx

rubeus

[eschwartz]

Simple, a backreference.

Capture the first used tag, and match it later -- yes, backreferences work inside the regex, not just for replacing.

Instead of:

,</i></p>\s+<p><i>

Use:

,</(i|b|others)></p>\s+<p><\1>

Actually, you might want to think of matching tags with attributes as well.

,</(i|b|others)></p>\s+<p( .+)?><\1>

[rubeus]

Originally Posted by eschwartz

,</(i|b|others)></p>\s+<p><\1>

Thanks, and i will read this chapter

Originally Posted by eschwartz

Actually, you might want to think of matching tags with attributes as well.

,</(i|b|others)></p>\s+<p( .+)?><\1>

To be honest, i'm doing that, but was too lazy to write this down in my example

[eschwartz]

Originally Posted by rubeus

Thanks, and i will read this chapter

Any time.

To be honest, i'm doing that, but was too lazy to write this down in my example

[Steadyhands]

Have I over-complicated this?

My aim was to come up with one regex to find ‘s and <space>‘s that followed any Unicode text and could be followed by either a space or punctuation, then replace the curly quote and remove the extra space.
Find

((?<=([\p{Ll}]))‘s(?=([\p{P}|\s]))|(?<=([\p{Ll}]))\s‘s(?=([\p{P}|\s])))

Replace

’s

fooè ‘s foo foo‘s foo fo's foo ‘s foo ’s dog ‘scooter’

[eschwartz]

(?<=\p{Ll})\s?‘s(?=\p{P}|\s)

Mainly, just make the space optional which cuts the regex in half. You can also remove some of the groupings you used.

[Steadyhands]

Originally Posted by eschwartz

(?<=\p{Ll})\s?‘s(?=\p{P}|\s)

Mainly, just make the space optional which cuts the regex in half. You can also remove some of the groupings you used.

Thanks,

Changing it a little more I've started looking for things other than the s, allowing for uppercase and full stops.

(?<=\p{Ll}|\p{Lu}|\.)\s?‘(d|m|s|t|ll|re|ve)(?=\p{P}|\s)

[eschwartz]

(?<=\p{L}|\.)\s?‘(d|m|s|t|ll|re|ve)(?=\p{P}|\s)

Just look for letters, no need to look for "uppercase_letters OR lowercase_letters"

Do you anticipate any of these happening after a full stop? Perhaps that should be fixed... I can hear there being an uppercase word, e.g. someone screaming.

[Steadyhands]

Thanks again. The full stops count for the abbreviations ie A.B.C. ‘s

[eschwartz]

Oh, OK -- makes sense now.

And you're welcome.