@Dennis-M-Marks

as you writing java script and creating web pages I would say you need
to read this.

Cheers
Claudia

Hello, Jivko,

Once, you get all your results in the Find result windows

Select all your results, in the Find Result panel

Right click, inside the Find Result panel

Choose the Copy option

Open a New tab

Paste all the lines of your original text, which match your search

=> The string “Line ##:”, at beginning of each line, should not be present !

Best Regards,

guy038

@ניר-מולא

Wrong place to ask that. Try…Google…?!!

@Tal-Segal

I’m under the impression that you are looking for something like e.g. snippets plugin, could this be?

Cheers
Claudia

@Adriana-Maria-Castro-Sanchez

First, is launchff++ a commandline executable? If so,
call it with the commandline interpreter like

cmd /k launchff++ “$(FULL_CURRENT_PATH)”
(also consider using nppexec)

If it is a win gui application, checkout if launchff++ can be executed to run the gui
or if it is needed to put in the whole path.

Cheers
Claudia

Thanks perfect :)

@Sandra-Blebbings

first of all, make clear what you want.
You’re saying that the website is showing up and in the sub-clause you are saying it isn’t.
Then describe the steps you’ve done clearly - how should we know what causes the problem
if you don’t provide any useful information. Sorry.

Cheers
Claudia

@Brian-Edwards

what about clicking into the top window area (don’t know the name currently, the area where save as appears)?
It should open a menu where you can resize or restore (or something like that) the window.

Cheers
Claudia

Hello , Gonen Shoham,

Of course, it can -:))

I noticed that, finally, you would like to change, ONLY, the lines which begin with a digit, wouldn’t you ?

If so, you can achieve all your changes, in one go, with a regex Search-Replacement !

Move back to the very beginning of your file ( Ctrl + Origin )

Open the Replace dialog ( Ctrl + H )

In the Find what: zone, type (?-s)^(?=\d)(\w+).+\W+(\w+)\W+(\w+)$

In the Replace with: zone, type \1 \2 \3 , with a space character after digits 1 and 2

Click on the Replace All button

Et voilà !!

NOTES :

As usual, the in-line modifier (?-s) ensures you the any meta-character dot ( . ) will match standard characters, ONLY, and not EOL characters

Then, the part ^(?=\d) is called a positive look-ahead, that is to say a condition which verifies that, at beginning ( ^ ) of each line, a digit ( \d ) does occur ( Note that this condition must be true, in order to valid the overall search regex. However, these look-arounds do not consume any character and the regex engine position is, still, at the very beginning of each line )

So, the part (\w+) represents the first non-null range of word characters, of each line, which is stored as group 1, due to the parentheses

Then, the part .+ stands for any non-null range of standard characters, till the regex engine reaches :

A non-null list of NON-word characters ( \W+ ), followed by :

A non-null range of word characters ( (\w+) ), stored as group 2, followed by :

An other non-null list of NON-word characters \W+, followed by :

An other non-null range of word characters ( (\w+) ), stored as group 3, finally followed by the end of the line ( $ )

In replacement, we just re-write the contents of the three groups 1 ( first word ), 2 ( penultimate word ) and 3 ( last word ), separated by one space character.

REMARKS* :

The class \w represents, in the world-common ASCII range of characters, the class [A-Za-z0-9_], ( upper-case letters, lower-case letters, digits and the low line character ) But, if your present file has an Unicode encoding ( UTF-8 or UCS-2 ), the range of word characters ( \w ), in N++, is extended to any character, considered as letter or digit, in either, Latin, Greek, Cyrillic, Hebrew and Arab languages, correcty displayed with the Courrier New default N++ font !

And the inverse class, \W, stands for any NON-word character, that is to say, any chracter, which does NOT belong to the word characters range

So, from your example :

*JSP - 060 N N 1309,com.sapiens.mig.ri.v050m00.RI17483,70 1309,com.sapiens.mig.ri.v050m00.RI17483,70 *SQL - 812 N N #SQL01 *DDC - 045 N N *JRN - 055 N N

we would obtain the modified text, below :

*JSP - 060 N N 1309 RI17483 70 1309 RI17483 70 *SQL - 812 N N #SQL01 *DDC - 045 N N *JRN - 055 N N

Even if the non-selected lines begin with an othersymbol than the star, except for digits, AND/OR the different words are separated by several non-word characters, our S/R, always, behave as expected :-) For instance, if you consider the text, below, where I inserted numerous non-word characters, whose some space and tabulation characters, in the two lines, that have to be changed and where I changed the first character of the lines, which have to stay unchanged :

+JSP - 060 N N 1309;com---sapiens~mig&ri>>>>>v050m00 RI17483,@@@@@@@70 1309:com"""sapiens===mig/ri!v050m00 RI17483%%%70 *SQL - 812 N N #SQL01 DDC - 045 N N JRN - 055 N N

We, still, get the result, below !

+JSP - 060 N N 1309 RI17483 70 1309 RI17483 70 *SQL - 812 N N #SQL01 DDC - 045 N N JRN - 055 N N

Best Regards,

guy038

debug info please

Hello, Manorma Gautam,

Not very difficult !

From what you said, I deduced some points :

You want to keep the lines, which contain, either, the string vprn OR vpls

These two words begin by the two lowercase letters vp

These two words are always located at beginning of lines

All the lines, which do not contain these two words, must be deleted

So, follow the few steps, below :

Move back to the very beginning of your file ( Ctrl + Origin )

Open the Replace dialog (Ctrl + H )

In the Find what zone, type in (?-is)^(?!vp).+\R

Leave the Repalce with zone EMPTY

Uncheck, preferably, the Wrap around option

Select the Regular expression search mode

Click on the Replace All button

Et voilà !

NOTES :

The first part of the regex are in-line modifiers (?-is) to force the regex engine to consider the search :

In a non-insensitive way ( in the case you, previously, unchecked the March case option

With the dot symbol matching standard characters, exclusively, in the case you, previously, checked the . matches newline** option

The middle part ^(?!vp) looks, from beginning of each line, for a negative look-ahead ( a condition that must be true in order to satisfy the overall match but which is not part of the final regex. So, it verifies, for each line, if the exact string vp does NOT occur, at beginning of lines. If it’s the case :

The last part, of this regex, .+\R matches any non-null range of standard characters, from beginning of line, followed by any kind of EOL characters. That is to say any complete line, which does NOT begin with vp

All these complete lines are, of course, deleted, as the replacement part has been left EMPTY

REMARK : The important thing, to note, is that look-arounds ( look-behinds and look-aheads ) do NOT move the regex engine position of search. So, after evaluating the negative look-ahead (?!vp) , the regex engine position is, still, just before the first character of each line ! Therefore, the part .+ does match all the standard characters of each line !

Best Regards,

guy038

Quote from the Notepad++ website:

Note that the most of plugins (including Plugin Manager) are not yet available in x64

The Line Warp option fixed the problem for me, thanks for the help!

http://docs.notepad-plus-plus.org/index.php/Main_Page