moving year in a date from the end to the front.

Vegard Johansen

Hello

I’m new to this program but i have a large document with many dates like this “23032022”, and i want it to be like this: “20222303” - I hope there’s is anyone that could help me here?

Alan Kilborn

@Vegard-Johansen

A possible problem is if your data has other 8-digit numbers that aren’t dates, but if it does just let us know and we’ll adjust to that. Another assumption to make is that your dates are not adjacent to other alphanumeric data, e.g., you don’t have mydate23032022.

Keeping it simple for now… :

Find: (\d{2}\d{2})(\d{4})
Replace: ${2}${1}
Search mode: Regular expression

wonkawilly

Hi I had a similar problem and the solution I adopted was crating a some macros to perform the conversion in many direction. Here they are:

<Macro name="Date convert all dd-/.mm-/.yyyy → yyyy-mm-dd" Ctrl="yes" Alt="no" Shift="yes" Key="89">
            <Action type="3" message="1700" wParam="0" lParam="0" sParam="" />
            <Action type="3" message="1601" wParam="0" lParam="0" sParam="(\d{2})(\.|-|/)(\d{2})(\.|-|/)(\d{4})" />
            <Action type="3" message="1625" wParam="0" lParam="2" sParam="" />
            <Action type="3" message="1602" wParam="0" lParam="0" sParam="($5)-($3)-($1)" />
            <Action type="3" message="1702" wParam="0" lParam="768" sParam="" />
            <Action type="3" message="1701" wParam="0" lParam="1609" sParam="" />
        </Macro>
        <Macro name="Date convert dd.mm.yyyy → yyyy-mm-dd" Ctrl="no" Alt="yes" Shift="yes" Key="68">
            <Action type="3" message="1700" wParam="0" lParam="0" sParam="" />
            <Action type="3" message="1601" wParam="0" lParam="0" sParam="(\d{2})(\.)(\d{2})(\.)(\d{4})" />
            <Action type="3" message="1625" wParam="0" lParam="2" sParam="" />
            <Action type="3" message="1602" wParam="0" lParam="0" sParam="($5)-($3)-($1)" />
            <Action type="3" message="1702" wParam="0" lParam="768" sParam="" />
            <Action type="3" message="1701" wParam="0" lParam="1609" sParam="" />
        </Macro>
        <Macro name="Date convert yyyy-mm-dd → dd.mm.yyyy" Ctrl="yes" Alt="yes" Shift="yes" Key="89">
            <Action type="3" message="1700" wParam="0" lParam="0" sParam="" />
            <Action type="3" message="1601" wParam="0" lParam="0" sParam="(\d{4})(-)(\d{2})(-)(\d{2})" />
            <Action type="3" message="1625" wParam="0" lParam="2" sParam="" />
            <Action type="3" message="1602" wParam="0" lParam="0" sParam="($5).($3).($1)" />
            <Action type="3" message="1702" wParam="0" lParam="768" sParam="" />
            <Action type="3" message="1701" wParam="0" lParam="1609" sParam="" />
        </Macro>

Notice that they are made to match my standard date format that is yyyy-mm-dd. So if needed can be freely adapted to specific use cases.

gerdb42

Or, if you want to rearrange day and month as well you could use
Search: (\d{2})(\d{2})(\d{4})
Replace: ${3}${2}${1}

And if you like to learn about RegExes look here

Vegard Johansen

@gerdb42 Thanks for this:) worked perfectly :)

Vegard Johansen

@gerdb42 Hello again, I have another problem. in a large document i have some lines like this: “-4 497” and like this: “17 787” i want to remove the space? do you know how?

guy038

Hello, @Vegard-Johansen, @gerdb42, @alan-kilborn, @wonkawilly and All,

I think that a correct answer could be :

SEARCH (?<=\d)\x20+(?=\d)
REPLACE Leave EMPTY

Best Regards,

guy038

Vegard Johansen

@guy038 Im sorry but i didnt get that to work :(

Alan Kilborn

@Vegard-Johansen

Did you make sure Search Mode was set to Regular expression?

Maybe show what you did here.

Vegard Johansen

@Alan-Kilborn Hello

I have a large document that look like this:

“120000” “various” “646958” “20220323” “12” “firstname lastname” “-4 497” “20220407” “646958”

And i want to remove the space in: “-4 497”

Paul Wormer

@Vegard-Johansen Are the double quotes part of your input? Or did you put them in as a clarification of your question?

In general, it is clearest and easiest when you cut and paste the relevant part of your actual input into a codeblock (triggered by </>).

Vegard Johansen

@Paul-Wormer Hello

Its part of the input :)

Paul Wormer

@Vegard-Johansen

Find: "(-?\d+)\x20(\d+)"   Replace:  "\1\2"

Vegard Johansen

@Paul-Wormer did not work :( maybe i do something wrong :(

guy038

Hello, @Vegard-Johansen, @gerdb42, @alan-kilborn, @wonkawilly, @paul-wormer and All,

May be I should have given more details on my regex S/R !

So :

Start N++ and open your document
Open the Replace dialog ( Ctrl + H )
Uncheck all box options
Enter (?<=\d)\x20+(?=\d) in the Find what : zone
Verify that the Replace with : zone is EMPTY
Check the Wrap around box
Select the Regular expression search mode
Click on the Replace All button

=> All the Space characters, which exist within numbers, should have disappeared !

Save the changes

Best Regards,

guy038

Vegard Johansen

@guy038 said in moving year in a date from the end to the front.:

(?<=\d)\x20+(?=\d)

sorry but i cant understand what im doing wrong… But i cant get it to work :I

Paul Wormer

@Vegard-Johansen Did you include the double quotes in both the Find expression and the Replace expression? Did you have exactly one space between the digits? Did you check the regex mode box?

My regex is a sort of brute force solution, I capture quotes and digits before and after a single space and copy them without a space in between them. @guy038 has a more sophisticated solution. He matches one (or more) space(s) that are surrounded by one digit on the left and one digit on the right and replaces the matching space(s) by an empty string.

However, both methods work. I have tried them both.

PeterJones

@Vegard-Johansen said in moving year in a date from the end to the front.:

i cant understand what im doing wrong

Could you give us a screenshot of your “Replace” dialog when you’re trying to use one of those regular expressions? We can probably figure out what setting you forgot to include.

Vegard Johansen

@PeterJones @guy038

PeterJones

@Vegard-Johansen ,

Your Replace dialog looked right, so I went to your original post and (using my moderator powers) looked at the text you typed rather than what the forum rendered. When I copied from there and pasted into Notepad++, I saw

Please notice that the character between the -4 and the 497 is not a space; it is the special character NBSP (Non-breaking space). Because you said “i want to remove the space”, and did not accurately tell us what character was between there, we could not derive a regex that works. The \x20 in the regular expression only applies to the normal ASCII space character, not to a tab or to a NBSP or to any other space-like character.

If your Notepad++ is not showing the NBSP symbol, then make sure you are on a relatively-recent Notepad++ version (v8.5 or newer), and make sure that View > Show Symbol > Show Non-printing Characters is turned on (has a checkmark next to it):

But back to your question:

Since we now know it’s a non-breaking space, we can change @guy038’s regular expression to (?<=\d)\xA0+(?=\d), and that will match just if there’s a non-breaking space between those characters. If there might be any horizontal space (real space, tab, nbsp, or others), then use (?<=\d)\h+(?=\d) instead, to make it match on any horizontal space.

This shows the importance of knowing (and sharing) what the actual characters are.