Select bookmarked lines
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Hi, hu ma,
Ah! I see ! You’re using CJK ideographic characters !
The different Unicode CJK scripts are :
- CJK Radicals Supplement - Phonetics and Symbols ( 2E80 - 2EFF ) - CJK Kangxi Radicals ( 2F00 - 2FDF ) - CJK Ideographic Description Characters ( 2FF0 - 2FFF ) - CJK Symbols and Punctuation ( 3000 - 303F ) - CJK Strokes ( 31C0 - 31EF ) - CJK Enclosed Letters and Months ( 3200 - 32FF ) - CJK Compatibility ( 3300 - 33FF ) - CJK Unified Ideographs Extension A ( 3400 - 4DB5 ) - CJK Unified Ideographs ( 4E00 - 9FD5 ) - CJK Compatibility Ideographs ( F900 - FAFF ) - CJK Compatibility Forms ( FE30 - FE4F ) - CJK Half-width Punctuation ( FF61 - FF64 ) - CJK Unified Ideographs Extension B ( 20000 - 2A6D6 ) - CJK Unified Ideographs Extension C ( 2A700 - 2B734 ) - CJK Unified Ideographs Extension D ( 2B740 - 2B91D ) - CJK Compatibility Ideographs Supplement ( 2F800 - 2FA1F )
As I have Asiatic languages, installed in my Win XP configuration, I chose the SimSun font to give you an example of my regex search/replacement, which does work, perfectly well !
In this example, the regex looks, successively, for the three following characters, enclosed by two angle brackets
<>
:-
The second character, of your range
[\x{3000}-\x{9faf}]
, known in my SimSun font, (、
), of Unicode code-point\x{3001}
-
A character, in the middle of your range
[\x{3000}-\x{9faf}]
, (栀
), of Unicode code-point\x{6800}
-
The last character, of your range
[\x{3000}-\x{9faf}]
, known in my SimSun font, (龥
), of Unicode code-point\x{9fa5}
Moreover, I added some random values :
-
The three characters 一倀怀, of Unicode code-points
\x{4e00}\x{5000}\x{6000}
, before the searched string -
The three characters 瀀耀退, of Unicode code-points
\x{7000}\x{8000}\x{9000}
, after the searched string
And, in replacement, I inserted the string Inserted灭Text, containing the ideographic character
灭
, of Unicode code-point\x{706d}
So, starting with the original text, with an UTF-8 encoding :
一倀怀<、>瀀耀退 一倀怀<栀>瀀耀退 一倀怀<龥>瀀耀退
The first S/R :
SEARCH
^.*<[\x{3000}-\x{9faf}]>
REPLACE
Inserted灭Text$0
gives the resulting text :
Inserted灭Text一倀怀<、>瀀耀退 Inserted灭Text一倀怀<栀>瀀耀退 Inserted灭Text一倀怀<龥>瀀耀退
And the second S/R :
SEARCH
<\x{3000}-\x{9faf}]>.*
REPLACE
$0Inserted灭Text
gives the final text :
一倀怀<、>瀀耀退Inserted灭Text 一倀怀<栀>瀀耀退Inserted灭Text 一倀怀<龥>瀀耀退Inserted灭Text
Cheers,
guy038
P.S. :
I just saw your recent post. Let’s me a couple of minutes to think about it. I’m back , soon !
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@guy038 said:
Hi, hu ma,
Ah! I see ! You’re using CJK ideographic characters !
The different Unicode CJK scripts are :
- CJK Radicals Supplement - Phonetics and Symbols ( 2E80 - 2EFF ) - CJK Kangxi Radicals ( 2F00 - 2FDF ) - CJK Ideographic Description Characters ( 2FF0 - 2FFF ) - CJK Symbols and Punctuation ( 3000 - 303F ) - CJK Strokes ( 31C0 - 31EF ) - CJK Enclosed Letters and Months ( 3200 - 32FF ) - CJK Compatibility ( 3300 - 33FF ) - CJK Unified Ideographs Extension A ( 3400 - 4DB5 ) - CJK Unified Ideographs ( 4E00 - 9FD5 ) - CJK Compatibility Ideographs ( F900 - FAFF ) - CJK Compatibility Forms ( FE30 - FE4F ) - CJK Half-width Punctuation ( FF61 - FF64 ) - CJK Unified Ideographs Extension B ( 20000 - 2A6D6 ) - CJK Unified Ideographs Extension C ( 2A700 - 2B734 ) - CJK Unified Ideographs Extension D ( 2B740 - 2B91D ) - CJK Compatibility Ideographs Supplement ( 2F800 - 2FA1F )
As I have Asiatic languages, installed in my Win XP configuration, I chose the SimSun font to give you an example of my regex search/replacement, which does work, perfectly well !
In this example, the regex looks, successively, for the three following characters, enclosed by two angle brackets
<>
:-
The second character, of your range
[\x{3000}-\x{9faf}]
, known in my SimSun font, (、
), of Unicode code-point\x{3001}
-
A character, in the middle of your range
[\x{3000}-\x{9faf}]
, (栀
), of Unicode code-point\x{6800}
-
The last character, of your range
[\x{3000}-\x{9faf}]
, known in my SimSun font, (龥
), of Unicode code-point\x{9fa5}
Moreover, I added some random values :
-
The three characters 一倀怀, of Unicode code-points
\x{4e00}\x{5000}\x{6000}
, before the searched string -
The three characters 瀀耀退, of Unicode code-points
\x{7000}\x{8000}\x{9000}
, after the searched string
And, in replacement, I inserted the string Inserted灭Text, containing the ideographic character
灭
, of Unicode code-point\x{706d}
So, starting with the original text :
一倀怀<、>瀀耀退 一倀怀<栀>瀀耀退 一倀怀<龥>瀀耀退
The first S/R :
SEARCH
^.*<[\x{3000}-\x{9faf}]>
REPLACE
Inserted灭Text$0
gives the resulting text :
Inserted灭Text一倀怀<、>瀀耀退 Inserted灭Text一倀怀<栀>瀀耀退 Inserted灭Text一倀怀<龥>瀀耀退
And the second S/R :
SEARCH
[\x{3000}-\x{9faf}]>.*
REPLACE
$0Inserted灭Text
gives the final text :
一倀怀<、>瀀耀退Inserted灭Text 一倀怀<栀>瀀耀退Inserted灭Text 一倀怀<龥>瀀耀退Inserted灭Text
Cheers,
guy038
P.S. :
I just saw your recent post. Let’s me a couple of minutes to think about it. I’m back , soon !
I see bunch of options here I could use, thanks for the information
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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
-
@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
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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
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@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 ?
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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
-
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@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 lineand there’s some lines “minoraty” that contains 5 or 7 groups, but I think I could just duplicate the process.
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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/4NON-empty
Best Regards,
guy038
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@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/4NON-empty
Best Regards,
guy038
Thanks, that is perfect one, it saved me alot of time
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Back again,
this time I want to swap two sentences or three separated by a symbol.example1
Originalthis 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
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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
-