Unwanted Para Break

[Paxman53]

Can anyone advise/assist?

I have a number of e-pub books that have unwanted para breaks in them

e.g.

<p class="calibre4">‘I do not mean to steal from them,’ he said at last,</p>

<p class="calibre4">‘although they have stolen life from some of us. But this time, if I am attacked... I shall fight back.’</p>

I can obviously close the gap manually when I spot them.

<p class="calibre4">‘I do not mean to steal from them,’ he said at last, ‘although they have stolen life from some of us. But this time, if I am attacked... I shall fight back.’</p>

Is there any method using Regex that can identify lower case characters at the beginning of a line and then move the line up?

[cybmole]

easy with regex

find </p>\s*<p class ="calibre4">([a-z])
replace \1

NB the replace line has a space in front of \1

works with v4.2 regex, I have not upgraded to 0.5.1 yet.

pretty sure there are other threads on how to do this, a varienat for example is to test end of previous line for not fullstop not quote not questionmark

[Paxman53]

Originally Posted by cybmole

easy with regex

find </p>\s*<p class ="calibre4">([a-z])
replace \1

NB the replace line has a space in front of \1

works with v4.2 regex, I have not upgraded to 0.5.1 yet.

pretty sure there are other threads on how to do this, a varienat for example is to test end of previous line for not fullstop not quote not questionmark

Tried this, unfortunately it didn't work : got - the search term was replaced 0 time(s) - although I could see an example right in front of me.

Specifically I put in:

Find:</p>\s*<p class ="calibre4">([a-z])
Replace: \1

I am using Sigil V5.0 - is the Regex version the same?

As to your variant I'm not even sure how to do that as I am virtually still a novice as far as regex is concerned - I can copy someone's else's suggestion (although not a clue as to what it means) and if it works that is great - if it doesn't I have no idea what to do

[theducks]

Originally Posted by cybmole

easy with regex

find </p>\s*<p class ="calibre4">([a-z])
replace \1

NB the replace line has a space in front of \1

works with v4.2 regex, I have not upgraded to 0.5.1 yet.

pretty sure there are other threads on how to do this, a varienat for example is to test end of previous line for not fullstop not quote not questionmark

This will catch about 98% of them. that do not end in punctuation.

My Wrap line Regex lightly tests the last character in the first line:

([\w",])</p>\s+<p class="calibre4">([\w"“…])

\1 \2

(there is a space between the \1 and \2)

This still misses:
...Dr.
Smith said...

...Mike
German ...

[cybmole]

in 0.4.2. you also have to tick th box that says regular expression in the find /replace options

as to what it means - find is looking for </p> followed by any amount of whitespace ( which includes new lines) , followed by the new style, followed by a lower case letter.

the replace \1 says "remember" that lower case letter you found earlier in the 1st set of ( ) - now reuse it after a space.

[cybmole]

the belt & braces approach ( & there's definitely a big thread on this somewhere ) is to run 2 tests.
1. for lines that begin with lower case
2. fot lines which end with " not the end of a sentence"

tricky cases are e.g.
Dr.
Roberts left the room and then I
followed him

also, beware of mangling any poetry or book extracts which are not using normal punctuation rules

see also
https://www.mobileread.com/forums/sho...d.php?t=114931

[Paxman53]

Thanks guys for all your advice - I am slowly getting to grips with it

I have now found some other threads relating to this (thanks Cybmole for the link)

I really should have looed harder but when you put in the search box and it comes up with 500 hits it kinda puts you off.

I am trying each Regex Expression on a test file to see the results and will then use the ones that work best for me.

But again many thanks it has been a great help.