Search and Replace with Regex

[jens ubert]

Hi, I’m trying to clean up some very old code and try to use some RegEx in Notepad++.

I have a mix of variables:

A: $somearr_name[some_key]
B: $somearr_name[‘some_key’]
C: $somearr_name[$some_key]
D: ‘$somearr_name[some_key]’

the goal is to have only:
$somearr_name[‘some_key’] after substitution.

somearr_name and some_key should stay untouched

$somearr_name[$some_key] © and ‘$somearr_name[some_key]’ (D) should not be touched at all.

[jens ubert]

My Expression to find the matches is:

[^‘]$(.*?)[^$][^’]

… but I have no idea, how to handle the replacement …

[Ekopalypse]

@jens-ubert

not really sure I get what you try to achieve but what about
find what:(?<!')(\$[\w]+\[)([\w]+)(?=\])
replace with:\1'\2'

[Alan Kilborn]

@jens-ubert

Maybe try:

Bring up the Replace window (ctrl+h).
Find what box: (\$\w+)\[(\w+)\](?!')
Replace with box: \1['\2']
Search mode: Regular expression
Wrap around: ticked
Press the Replace All button.

[jens ubert]

ohhhh … it looks pretty! I’ll try little more und will give some response.
Did not know about \1 … because of my poor english and knowledge about the terminology in this case: How is this kind of replacement called.

[Alan Kilborn]

@jens-ubert said:

How is this kind of replacement called

Hmmm, not sure exactly…maybe replacement by captured group number ? Go look it up and see if it has a better name.

[PeterJones]

@jens-ubert said:

How is this kind of replacement called.

The Boost documentation calls it a “backreference ”. In the replacement string, the $-based equivalent is called a “placeholder sequence ”.

[guy038]

Hello, @jens-ubert, @ekopalypse, @alan-kilborn, @peterjones, and All,

Jens, welcome to the N++ community !

Seemingly, I understand that your goal is to obtain the B) syntax, leaving the C) and D syntaxes untouched. In other words, you just would like that any A) syntax be replaced with the B syntax, wouldn’t you ?

Thus, we have to grab any A syntax, specifically, and change it into the B one

Comparing all syntaxes, it comes that only the A syntax does not contain any single quote character, ' nor the $ symbol, between square brackets. We can add the rule that the variable name, before the square brackets does not contain any single quote, too !

On the other hand, the part before square brackets does not change, when moving from the A to the B syntax. So, we just need to store the text, between square brackets, and surround it with single quotes, during replacement !

---

So, a possible regex would be :

SEARCH \s\$[^'\r\n]+\[\K([^'$\r\n]+)(?=\])

REPLACE '\1'

Notes :

• The beginning of the regex \s searches for the general kind of space character ( the Space or Tabulation chars, the \n or \r line break chars and some others… )

• Then the part \$ looks for the literal regex symbol $

• Now, the syntax [^'\r\n]+\[ searches the greatest non-null range of consecutive characters, different from a single quote and line-breaks chars, followed by a literal opening square bracket

• The special syntax \K cancels any match and resets the regex engine working position

• Thus, the part ([^'$\r\n]+) tries, now, to match the greatest non-null range of consecutive characters, either different from a single quote, a dollar and line-breaks chars, stored as group 1, due to the outer parentheses

• But ONLY IF  the look-ahead (?=\]) is true, i.e. if this range is followed with an ending literal square bracket !

• Finally, in replacement, the \1 syntax represents all group 1 contents, surrounded with single quotes

Best Regards,

guy038