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VS Code: Search-and-Replace Regex with Dollar-Sign

rfornal profile image bob.js Updated on ・2 min read

In a previous article about VS Code, Search-and-Replace using RegEx, I had a question that prompted me to create this article so that I can remember what I did ...

The Question

What if I need to replace something with $? For example, replace $var with $this->var = $var, the regex to search is easy, the problem is the replacement. I (have tried this) with $this->$1 = $$1 (and) it does not work properly. So how do we go about it?

The Solution

The solution, as I see it is actually in two parts. First, we need to properly select the content for replacement. Then, second, we need to define the replacement, accounting for the dollar-sign ...

Selecting the Content

When I started breaking down the question in my head, I saw some content like the following ...

$var;
$content;
$question;
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... the assumption being that this should be changed into ...

$this->var = $var;
$this->content = $content;
$this->question = $question;
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So, this means I need to capture the text after the dollar sign. As stated in the question, this is pretty straight forward. There are many ways to do this. I chose ...

\$([a-zA-Z0-9]*)
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Replacing the Content

The dollar-sign after the equal sign seems to be where the main concern of the question comes from. I will admit that it took me a few minutes to get my head around the issue of escaping it.

Original Attempt

First, I tried this replacement RegEx ...

$this->$1 = $$1
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... this is shown in the question and makes sense; however, the result is ...

$this->var = $1
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Escaping the Dollar-Sign

... then, I tried escaping the dollar-sign after the equal-sign ...

$this->$1 = \$$1
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... however, the result is not quite what I expected ...

$this->var = \$var
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Proper Escaping of the Dollar-Sign

While this is closer, there is clearly something wrong with this approach. I then started searching for how to properly escape the dollar-sign in a RegEx Search-and-Replace.

I found the following ...

You need to escape the $ replacement with $$ and then use the substitution value of $1.

The resulting expression should be: $$$1

The Solution

And, here's the answer ...

$this->$1 = $$$1
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... try it out, it works!

Discussion

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