Select bookmarked lines

guy038 · Dec 17, 2016, 2:30 PM

Hello, abuali huma,

You didn’t say if the regex must keep the Line 2 unchanged or if you want to wipe out this line !

Anyway, here are the appropriate regexes for each case :

1) If line 2 is unchanged :

From the original text :

FGHI;Ax#&;Dx#&
BCDE

the S/R, below :

SEARCH (?-si)^(FGHI)(.*)(?=\R(BCDE))

REPLACE \3\2\1

would give the result :

BCDE;Ax#&;Dx#&FGHI
BCDE

NOTES :

The first part (?-si) forces the dot ( . ) to match standard characters, only, and the regex engine to work, in a NON-insensitive way
The \R form represents any kind of EOL characters (\r\n ), (\n ) or (\r )
The (?=\R(BCDE)) syntax is called a positive look-ahead, that is to say a condition which must be verified to valid the overall regex, but which is never part of the final match. So the condition is “Does it exist, at the end of line 1, some EOL character(s), followed by the string BCDE ?”. The string BCDE is stored in group 3

2) If line 2 must be deleted, too :

From the original text :

FGHI;Ax#&;Dx#&
BCDE

The second S/R :

SEARCH (?-si)^(FGHI)(.*)\R(BCDE)

REPLACE \3\2\1

would give the final text :

BCDE;Ax#&;Dx#&FGHI

Best Regards,

guy038

abuali huma · Dec 17, 2016, 2:59 PM

@guy038 said:

Hello, abuali huma,

You didn’t say if the regex must keep the Line 2 unchanged or if you want to wipe out this line !

Anyway, here are the appropriate regexes for each case :

1) If line 2 is unchanged :

From the original text :
FGHI;Ax#&;Dx#&
BCDE
the S/R, below :

SEARCH (?-si)^(FGHI)(.*)(?=\R(BCDE))

REPLACE \3\2\1

would give the result :
BCDE;Ax#&;Dx#&FGHI
BCDE
NOTES :

The first part (?-si) forces the dot ( . ) to match standard characters, only, and the regex engine to work, in a NON-insensitive way

The \R form represents any kind of EOL characters (\r\n ), (\n ) or (\r )

The (?=\R(BCDE)) syntax is called a positive look-ahead, that is to say a condition which must be verified to valid the overall regex, but which is never part of the final match. So the condition is “Does it exist, at the end of line 1, some EOL character(s), followed by the string BCDE ?”. The string BCDE is stored in group 3

2) If line 2 must be deleted, too :

From the original text :
FGHI;Ax#&;Dx#&
BCDE
The second S/R :

SEARCH (?-si)^(FGHI)(.*)\R(BCDE)

REPLACE \3\2\1

would give the final text :
BCDE;Ax#&;Dx#&FGHI
Best Regards,

guy038

What if it is different text?
Example one
Line#1 こんにちは;Ax#&;Dx#&
Line#2 行く
Result
Line#1 行く ;Ax#&;Dx#&こんにちは

Example two
Line#18 細かい ;Ax#&;Dx#&
Line#19 自分を愛する
Result
Line#18 自分を愛する ;Ax#&;Dx#& 細かい

and so on, and yes 2nd line is supposed to be deleted

guy038 · Dec 17, 2016, 7:56 PM

Hi, abuali huma,

Sorry, I was absent for a while, because of an virus analysis on my laptop ! But, please be quiet : absolutely NO connection with our NodeBB site nor our discussion, too :-))

For a general regex, no problem at all ! I just interpreted that :

Group 1, beginning the first line, is supposed to contain ideographic characters, ONLY
Group 2 represents any range, after the last ideographic character, in the first line, till the end of the line
Group 3, standing for the second line, is supposed, also, to contain ideographic characters, ONLY

Remark :

Some additional space characters, NOT present in the original text, seem included in the replacement text, of your two examples :

Before the first semicolon, in your first example
After the ampersand character, in your second example

From my hypotheses, any space character would be stored in group 2, of course !. In addition, you may separate, in replacement, the three groups with a space character, as well !

So, a general S/R would be :

SEARCH (?-si)^([\x{3000}-\x{9faf}]+)(.*)\R([\x{3000}-\x{9faf}]+)

REPLACE \3\2\1

Notes :

The part [\x{3000}-\x{9faf}]+ tries to match the largest, non empty, range of ideographic characters, from beginning of a line ( ^ ), stored in group 1 OR after an EOL character (\R ), stored in group 3
During replacement, these three groups, on two consecutive lines, are, just, re-written, in a single line, into a different order !
As said above, the replacement regex may be changed into \3 \2 \1

Cheers,

guy038

hu ma · Dec 17, 2016, 8:16 PM

@guy038 said:

Hi, abuali huma,

Sorry, I was absent for a while, because of an virus analysis on my laptop ! But, please be quiet : absolutely NO connection with our NodeBB site nor our discussion, too :-))

For a general regex, no problem at all ! I just interpreted that :

Group 1, beginning the first line, is supposed to contain ideographic characters, ONLY

Group 2 represents any range, after the last ideographic character, in the first line, till the end of the line

Group 3, standing for the second line, is supposed, also, to contain ideographic characters, ONLY

Remark :

Some additional space characters, NOT present in the original text, seem included in the replacement text, of your two examples :

Before the first semicolon, in your first example

After the ampersand character, in your second example

From my hypotheses, any space character would be stored in group 2, of course !. In addition, you may separate, in replacement, the three groups with a space character, as well !

So, a general S/R would be :

SEARCH (?-si)^([\x{3000}-\x{9faf}]+)(.*)\R([\x{3000}-\x{9faf}]+)

REPLACE \3\2\1

Notes :

The part [\x{3000}-\x{9faf}]+ tries to match the largest, non empty, range of ideographic characters, from beginning of a line ( ^ ), stored in group 1 OR after an EOL character (\R ), stored in group 3

During replacement, these three groups, on two consecutive lines, are, just, re-written, in a single line, into a different order !

As said above, the replacement regex may be changed into \3 \2 \1

Cheers,

guy038

Thanks mate, I will give it a try once I get back on my pc after a while…

But as I understand from you, that is in group 1&3, if the line had mixed character set, the regex will not apply to that line ?

guy038 · Dec 17, 2016, 8:55 PM

abuali huma,

Seemingly, you consider in the original first line, two parts :

The first part , which, after replacement, will be moved to the end of this first line
The second part, which, after replacement, will follow the contents of the deleted second line

I, just, supposed that the limit, between these two parts, was the one between ideographic and NON-ideographic characters

However, any kind of limit may be considered. For instance, if you prefer, we may suppose that group 1 is all the characters of the first line, till a first semicolon or… anything else ! Just tell me which criterion to choose :-))

Cheers,

guy038

abuali huma · Dec 17, 2016, 9:08 PM

@guy038 said:

abuali huma,

Seemingly, you consider in the original first line, two parts :

The first part , which, after replacement, will be moved to the end of this first line

The second part, which, after replacement, will follow the contents of the deleted second line

I, just, supposed that the limit, between these two parts, was the one between ideographic and NON-ideographic characters

However, any kind of limit may be considered. For instance, if you prefer, we may suppose that group 1 is all the characters of the first line, till a first semicolon or… anything else ! Just tell me which criterion to choose :-))

Cheers,

guy038

Well let’s see,
group one should be all characters until the first semicolon
group two from the semicolon till line ends
group three should be all characters in the line

and there’s some lines “minoraty” that contains 5 or 7 groups, but I think I could just duplicate the process.

guy038 · Dec 18, 2016, 3:31 AM

Hi, abuali huma,

OK.! So, the general S/R becomes :

SEARCH (?-si)^(.+?)(;.*)\R(.+)

REPLACE \3\2\1

NOTES :

The group 2, ( (;.*) ), begins, as expected, with a semicolon, followed by any range, even empty, of standard character(s) ( Remember that the quantifier * is the short form for {0,} ! )
The group 1, ( (.+?) ), after the location beginning of line, looks for the shortest, NON-empty, range of standard character, till a semicolon.
Note that the form .+?, equivalent to .{1,}? is a lazy quantifier, which forces the regex engine to find the MINIMUM of standard characters till a semicolon => So, in your previous first example, group 1 is equal to the string こんにちは
On the contrary, if we have used the greedy quantifier, .+, equivalent to .{1,}, it would have forced the regex engine to find the MAXIMUM of standard characters, till a semicolon. So, in your previous first example, group 1 would have been equal to the string こんにちは;Ax#&, before the second semicolon !
The group 3, ( (.+) ), after the EOL character(s) of the first line, represents any NON-empty range of standard characters, of the second line
Remember that, as each group contains standard characters, they may, also, begins and/or ends with some space or tabulation character(s), except for group 2, which must begin with a semicolon

BTW, to know the meaning of the different quantifiers, just see my post to Casey, below :

https://notepad-plus-plus.org/community/topic/12905/how-to-remove-duplicate-row-in-find-result/4 NON-empty

Best Regards,

guy038

hu ma · Dec 20, 2016, 12:37 PM

@guy038 said:

Hi, abuali huma,

OK.! So, the general S/R becomes :

SEARCH (?-si)^(.+?)(;.*)\R(.+)

REPLACE \3\2\1

NOTES :

The group 2, ( (;.*) ), begins, as expected, with a semicolon, followed by any range, even empty, of standard character(s) ( Remember that the quantifier * is the short form for {0,} ! )

The group 1, ( (.+?) ), after the location beginning of line, looks for the shortest, NON-empty, range of standard character, till a semicolon.

Note that the form .+?, equivalent to .{1,}? is a lazy quantifier, which forces the regex engine to find the MINIMUM of standard characters till a semicolon => So, in your previous first example, group 1 is equal to the string こんにちは

On the contrary, if we have used the greedy quantifier, .+, equivalent to .{1,}, it would have forced the regex engine to find the MAXIMUM of standard characters, till a semicolon. So, in your previous first example, group 1 would have been equal to the string こんにちは;Ax#&, before the second semicolon !

The group 3, ( (.+) ), after the EOL character(s) of the first line, represents any NON-empty range of standard characters, of the second line

Remember that, as each group contains standard characters, they may, also, begins and/or ends with some space or tabulation character(s), except for group 2, which must begin with a semicolon

BTW, to know the meaning of the different quantifiers, just see my post to Casey, below :

https://notepad-plus-plus.org/community/topic/12905/how-to-remove-duplicate-row-in-find-result/4 NON-empty

Best Regards,

guy038

Thanks, that is perfect one, it saved me alot of time

abuali huma · Dec 21, 2016, 4:00 AM

Back again,
this time I want to swap two sentences or three separated by a symbol.

example1
Original

 this is text number two|This is text number one

Result

 This is text number one|this is text number two

Example2

 this is text number three|this is text number two|This is text number one

Result

 This is text number one|this is text number two|this is text number three

guy038 · Dec 21, 2016, 9:05 PM

Hi, abuali huma,

Not very difficult to achieve !

EXAMPLE 1 :

SEARCH ^(.+)\|(.+)

REPLACE \2|\1

Notes :

This first regex separates the two parts of text, before and after the symbol | , which must be escaped ( \| ) as this symbol, in search regexes, stands for the normal alternation symbol !
The part, before the symbol, is stored as group 1 and the part, located after, is stored as group 2
In replacement, we just reverse the order of these two groups

EXAMPLE 2

SEARCH ^(.+?)\|(.+)\|(.+)

REPLACE \3|\2|\1

Notes :

This second regex separates the three parts of text, created by the symbol |, which are stored as group 1, group 2 and group 3
The first part ^(.+?)\|, tries to match, from beginning of line ( ^) , the shortest non-null range of characters, till the FIRST separation symbol ( \| ), because of the lazy quantifier ( +?)
Note that if we would have written this first part ^(.+)\|, it would have matched the string this is text number three|this is text number two. Indeed, due to the greedy quantifier ( + ), it would have matched the tallest non-null range of characters, till the SECOND separation symbol !
Then for the two remaining parts, of the regex, (.+)\|(.+) just refer to the previous regex for explanations
In replacement, we just perform a permutation of these three groups

Cheers,

guy038