从字符串中删除字符的最后一个实例-Javascript-重新审视

Question

According to the accepted answer from this question , the following is the syntax for removing the last instance of a certain character from a string (In this case I want to remove the last & ):

 function remove (string) { string = string.replace(/&([^&]*)$/, '$1'); return string; } console.log(remove("height=74&width=12&")); 

But I'm trying to fully understand why it works.

According to regex101.com,

/&([^&]*)$/

& matches the character & literally (case sensitive)

1st Capturing Group ([^&]*) Match a single character not present in the list below [^&]*

* Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)

& matches the character & literally (case sensitive)

$ asserts position at the end of the string, or before the line terminator right at the end of the string (if any)

So if we're matching the character & literally with the first &:

Then why are we also "matching a single character not present in the following list"?

Seems counter productive.

And then, " $ asserts position at the end of the string" - what does this mean? That it starts searching for matches from the back of the string first?

And finally, what is the $1 doing in the replaceValue? Why is it $1 instead of an empty string? ""

Answer 1

1- The solution for that problem I think is different to the solution you want:

That regex will replace the last "&" no matter where it is, in the middle or in the end of the string.

If you apply this regex to this two examples you will see that the first get "incorrectly" replaced:

height=74&width=12&test=1
height=74&width=12&test=1&

They get replaced as :

height=74&width=12test=1
height=74&width=12&test=1

So to really replace the last "&" the only thing you need to do is :

string.replace(/&$/, '');

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Now, if you want to replace the last ocurrence of "&" no matter where it is, I will explain that regex :

$1 Represents a (capturing group), everything inside those ([^&]*) are captured inside that $1. This is a oversimplification.

&([^&]*)$

& Will match a literal "&" then in the following capturing group this regex will look for any amount (0 to infinite) of characters (NOT EQUAL TO "&", explained latter) until the end of the string or line (Depending on the flag you use in the regex, /m for matching lines ). Anything captured in this capturing group will go to $1 when you apply the replacement.

So, If you apply this logic in your mind you will see that it will always match the last & and replace it with anything on its right that does not contain a single "&""

&(<nothing-like-a-&>*)<until-we-reach-the-end> replaced by anything found inside (<nothing-like-a-&>*) == $1. In this case because of the use of * , it means 0 or more times, sometimes the capturing group $1 will be empty.

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NOT EQUAL TO part: The regex uses a [^], in simple terms [] represents a group of independent characters, example: [ab] or [ba] represents the same, it will always look for "a" or "b". Inside this you can also look for ranges like 0 to 9 like this [0-9ba], it will always match anything from 0 to 9, a or b.

The "^" here [^] represents a negation of the content, so, it will match anything not in this group, like [^0-9] will always match anything that is not a number. In your regex [^&] it was used for looking for anything that is not a "&"