Find/Replace - keeping a part of the find, remove everything else

Gabriel Mourão

To sum it up take this example im trying to reproduce:

This is the contents of document.txt:

![\vec{Q}=m \cdot \vec{\Delta v}](vec{Q}=m_!cdot_!vec{!Delta_v})

For the documents i have created, there are hundreds of patterns like these, and i want to dwindle them down to:

$\vec{Q}=m \cdot \vec{\Delta v}$

To do that, i would have to locate all structures just like this one inside the document, which i did with:

\!\[+[^]]+[A-Za-z{}=\/!_-\s]]\(+[^)]+[A-Za-z{}=\/()!_-\s])

The problem right now is that i have no idea how to a keep a part of the find, or if its even possible. Until now all the replaces i did were only literal, replacing a find with a select string. Is there a way to separate a part of the found characters as a variable and keep them on replacement?

Alan Kilborn

@gabriel-mourão

You want “capture groups” and substitutions using them.
See the user manual HERE.

Gabriel Mourão

@alan-kilborn I took a look at it, added a few tweaks (by that i mean just two parantheses) and now it works perfectly. Thanks alot.
For those who might be curious to how the solution looks like:

\!\[+([^]]+[A-Za-z{}=\/!_-\s])]\(+[^)]+[A-Za-z{}=\/()!_-\s]

edit: and for the replace:

\$$1\$

guy038

Hello, @gabriel-mourão, @alan-kilborn and All,

I’m sorry, Gabriel, but your final search regex seems invalid -:((

May be it’s some typos, introduced by the forum syntax, but one correct syntax is :

!\\[([^]]+[A-Za-z{}=\\/!_-\s])\\]\([^)]+[A-Za-z{}=\\/()!_-\s]

Using the free-spacing mode, (?x), this syntax can be split as below :

(?x)

!\\[
(
[^]]+
[A-Za-z{}=\\/!_-\s]
)
\\]

\(
[^)]+
[A-Za-z{}=\\/)!_-\s]

Note that :

On one hand :
- The part [^]]+ matches the chars range ![\vec{Q}=m \cdot \vec{\Delta v}
- The part [A-Za-z{}=\\/!_-\s] matches the single char ]
On the other hand :
- The part [^)]+ matches the chars range (vec{Q}=m_!cdot_!vec{!Delta_v}
- The part [A-Za-z{}=\\/)!_-\s] matches the single char ), at the end

I suppose that it’s not exactly what you want !

Your need could rather be expressed as :

First search for an exclamation mark and an opening square bracket ![
Then, search for the smallest range of chars…
Till an ending square bracket ]
Now, search for an opening parenthese (
Then, search for the smallest range of chars…
Till an ending parenthese )

This leads to this simple regex S/R :

SEARCH (?-s)!\\[(.+?)\\]$.+?$

REPLACE \$$1\$

Best Regards,

guy038