Search and replace in Visual Studio

Question

In Visual Studio, when I search within a selection, I want to replace the first instance (or second, third, etc.) of a match per line using regular expressions. How would I do this?

Search and replace

foo1 = foo1;
foo2 = foo2;
...
foo20 = foo20;

into the following.

foo1 = bar1;
foo2 = bar2;
...
foo20 = bar20;

score 83 · Answer 1 · edited Dec 17 '13 at 11:52

83

In Visual Studio 2012, capture groups and backreferences are used just like in C#. You can capture them with common parenthesis, and backreference them with $0, $1, etc. Hope it helps!

Note that the syntax $1 is used for find-replace, but \1 is used for backreferences in the search string.

edited Dec 17 '13 at 11:52

Oliver

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answered Jul 24 '12 at 17:31

SoManyGoblins

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I was about to draw the conclusion that backreferences had been removed from Visual Studio 2012 until I read this answer. It is not mentioned at all in the official documentation http://msdn.microsoft.com/en-us/library/2k3te2cs(v=vs.110).aspx – Klas Mellbourn Oct 12 '12 at 12:40
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Just to elaborate with an example for VS 2012, a reg. exp. for removing the surrounding quotes from text elements in an XML document: _Find box_: **>"(.*)"<** _Replace box_: **>$1<** – Ed Graham Dec 20 '12 at 15:09
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Huzzah for proper regular expressions in VS 2012! This is huge! – Cristian Diaconescu Apr 05 '13 at 11:24
Used for find-replace in VS2019 and it worked just fine! – A. J. Mar 15 '21 at 14:07

score 33 · Accepted Answer · edited Sep 21 '15 at 10:59

33

In Visual Studio 2010 and earlier, use regular expressions with back references

Visual Studio's regular expressions are completely different from what I've learned. Took me some time to figure out the correct answer.

Search for

{foo}{:d+} = \1\2

Replace with

\1\2 = bar\2

Back references are done by tagging with curly braces {foo}. :d+ is the same for \d+

Read more about VS RegEx here

edited Sep 21 '15 at 10:59

danio

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answered Oct 21 '11 at 11:02

Ranhiru Jude Cooray

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Would this regex not return? bar1 = bar1; bar2 = bar2; ... bar20 = bar20; – San Oct 21 '11 at 11:08
Nope it wouldn't. Did I miss anything? – Ranhiru Jude Cooray Oct 21 '11 at 11:09
Have you edited your Post? I remember an other version. This one is okay =) – San Oct 21 '11 at 11:14
i would rather not use digits in my regex. for example i might want to replace `redMonkey = redMonkey` with `redMonkey = blueMonkey` – Joseph Le Brech Oct 21 '11 at 11:26
Oh well, then it's just a matter of dropping the digit `:d+` from the regex :). And please clearly state your requirements. Do you know what `foo` is beforehand, or you just to search for `foo = foo` where you don't know what foo is ? – Ranhiru Jude Cooray Oct 21 '11 at 11:29
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@RanhiruCooray: I would drop the first part, with the "standard" regex solution. Or move it to the bottom--but by all means keep the part about the VS regex flavor being, um, *unique*. ;) – Alan Moore Oct 21 '11 at 13:08
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IN VS2013 this doesnt work at all. I have to replace curly {} with round () and it works. – Piotr Kula Apr 15 '14 at 15:02
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@ppumkin Same with VS 2012 for me. – mbomb007 Apr 08 '15 at 18:52
What the heck is this syntax?! – Mateen Ulhaq Mar 10 '16 at 22:09

score 3 · Answer 3 · answered Oct 28 '15 at 21:42

3

Here it is, type exactly as it is displayed here

Search: (\w+\d+\s*=\s*)[^\d]+(\d+);

Replace: $1bar$2;

Read more: Using Regular Expressions in Visual Studio

answered Oct 28 '15 at 21:42

BJovke

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userom · Answer 4 · 2019-09-06T18:52:06.507

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As simple as this in Visual Studio 2019 Find/Replace. I needed to replace FORTRAN IO format string to C++ format and used sub-expression and numbers of regexp.

Example: find: "f9.8", "f18.3", in line and replace with %9.8f, %18.3f
reg exp: Find = ( f)(\d+.\d+) Replace = %$2f

edited Sep 06 '19 at 18:52

answered Sep 06 '19 at 18:46

userom

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score 3 · Answer 5 · edited Sep 02 '13 at 18:37

3

I can be done without a regular expression as well:

Replace = foo with = bar.

If a regular expression is needed, one could use:

foo(\d*) = foo(\d*);

Replace with:

foo\1 = bar\2;

edited Sep 02 '13 at 18:37

Peter Mortensen

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answered Oct 21 '11 at 10:58

Jayendra

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This wouldn't actually work because `\d` is not an accepted character in Visual Studio. – Ranhiru Jude Cooray Oct 22 '11 at 06:29
3

Backreferences for replacing need to use `$1`, not `\1`, in vs 2012 at least. – mbomb007 Apr 08 '15 at 18:51
This is for older, pre-2012 versions of Visual Studio. For newer versions see the answer for this question: https://stackoverflow.com/questions/3147096/visual-studio-find-and-replace-regex/3147177#3147177 – paulwesterberg Jul 27 '17 at 16:20

score -2 · Answer 6 · answered Oct 21 '11 at 11:38

-2

If you would be more variable:

Regex.Replace(input, @"(?<=\= )[^;0-9]*(?=[0-9]*;)", replacewith);

This search for = and (anynumber); and replace that between.

Edit: The number is optional.

answered Oct 21 '11 at 11:38

San

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This question is about using the Find/Replace module of Visual Studio, not about VB. – mbomb007 Apr 08 '15 at 18:50

Search and replace in Visual Studio

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