Trim values of CSV using Notepad++ only if the value is not blank

Paul Marcelin Bejan 0
Viewed 63

I have the following CSV:

COLUMN_A;COLUMN_B;COLUMN_C;COLUMN_D;COLUMN_E;COLUMN_F;COLUMN_G;COLUMN_H;COLUMN_I;COLUMN_L;
01234;AB ;00001; ;100000001;   ;ABC;+0000000000099998,080;XYZ        ;

I would like to remove the white spaces only if the value is not blank. So the result will be like this:

COLUMN_A;COLUMN_B;COLUMN_C;COLUMN_D;COLUMN_E;COLUMN_F;COLUMN_G;COLUMN_H;COLUMN_I;COLUMN_L;
01234;AB;00001; ;100000001;   ;ABC;+0000000000099998,080;XYZ;

I know that I can use find ' ' replace '' so the space will be replaced by nothing. But in this way, I will remove all the spaces, and I want to maintain the string with spaces where there aren't any other character.

If I have to use regular expression (my first time), I think I need to concatenate the following expression:

[a-zA-Z]
[\s]
[;]

so I can use [a-zA-Z][\s][;] in the find box,
but I don't know how to replace what it finds with [a-zA-Z][;]

removing-whitespace notepad++ csv regex
3 Answers

You can distinguish the cases where the string of spaces is not prefixed with a semi-colon or white-space, and the cases where the string of spaces is not suffixed with a semi-colon or white-space.

For this you can use look-around:

Find what: \h+(?![;\s])|(?<![;\s])\h+
Replace with: (empty)
⦿ Regular expression

Replace all

Explanation

  • \h matches a horizontal white-space character (so not newline)
  • (?![;\s]) is a negative look ahead: what follows at the current position should not be one of those characters, but they are not included in the overall match.
  • (?<![;\s]) is a negative look behind: what precedes at the current position should not be one of those characters, but they are not included in the overall match.
trincot0
  • Ctrl+H
  • Find what: (?:;|\G(?!^))\K\h*([^\s;]+)\h*(?=[^;]*;)
  • Replace with: $1
  • TICK Wrap around
  • SELECT Regular expression
  • Replace all

Explanation:

(?:             # non capture group
    ;               # semi-colon
  |               # OR
    \G              # restart from last match position
    (?!^)           # not the beginning of line
)               # end group
\K              # forget all we have seen until this position
\h*             # 0 or more horizontal spaces
(               # group 1
    [^\s;]+         # 1 or more any character that is not a space or semi-colon
)               # end group
\h*             # 0 or more horizontal spaces
(?=             # positive lookahead, make sure we have after:
    [^;]*           # 0 or more any character that is not a semi-colon
    ;               # a semi-colon
)               # end lookahead

Screenshot (before):

enter image description here

Screenshot (after):

enter image description here

usertRy10

To answer my last question,

It is possible to capture a group by surrounding it with parentheses (round brackets) and each one will be numerated starting from 1, so it will be possible to use them by prefixing the number with a dollar.

In the find box:

([a-zA-Z])([ ]{1,})([;])

in the replace box

$1$3
Paul Marcelin Bejan0
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