How does this code `/(\s+(\W)/g, '$1')` remove spaces?

Question

How does this code `/(\s+(\W)/g, '$1')` remove spaces?

let a = ' lots of    spaces     in this  !  '
console.log(a.replace(/\s+(\W)/g, '$1'))

logs out lots of spaces in this!

The regex above efficiently accomplishes my goal, but I am curious about the reasoning behind it. I have a grasp of the following:

s+ searches for 1 or more spaces
(\W) captures the non-alphanumeric characters
/g - global, searches/replaces all instances
$1 returns the previously captured alphanumeric character

The use of capture/$1 is what eliminates the space between the words This and !

I comprehend this, but I am puzzled about HOW all the other spaces are disappearing? I didn't specifically request for them to be removed (though I appreciate that they are).

I understand why this code

console.log(a.replace(/\s+/g, ' '));

works because it replaces 1 or more spaces between alphanumeric characters with a single space ' '.

I am genuinely baffled about HOW the initial RegEx /\s+(\W)/g, '$1' substitutes 1 or more spaces with a single space.

javascript regex

Answer 1

Answer №1

When you use a regular expression that includes \s+</, it means to match one or more spaces followed by one or more non-alphanumeric characters, and then replace that entire result with those non-alphanumeric characters. The key here is that \s+</ is greedy, so it will try to match as many characters as possible. This means it will match all the spaces it can within a string of spaces. However, because your regex also includes \W+</, which requires one or more non-word characters, the last part of the regex must match the final space since the next character after it is a letter.

For example, in the string a b, when using parentheses to mark the matches for \s+</ and \W+</, it will look like a( )( )b. This is the only valid way for the regex to match (\s+</ matches the first two spaces and \W+</ matches the last space). Then it's just a matter of simple substitution. By enclosing \W+</ in parentheses, it becomes the first and only capturing group, so replacing the match with $1 will replace it with the final space.

Another example is replacing the match in a !b, which would result in a( )(!)b (with ! being the last non-word character), leading to the final result being a!b.

Answer 2

When you use a regular expression that includes \s+</, it means to match one or more spaces followed by one or more non-alphanumeric characters, and then replace that entire result with those non-alphanumeric characters. The key here is that \s+</ is greedy, so it will try to match as many characters as possible. This means it will match all the spaces it can within a string of spaces. However, because your regex also includes \W+</, which requires one or more non-word characters, the last part of the regex must match the final space since the next character after it is a letter.

For example, in the string a b, when using parentheses to mark the matches for \s+</ and \W+</, it will look like a( )( )b. This is the only valid way for the regex to match (\s+</ matches the first two spaces and \W+</ matches the last space). Then it's just a matter of simple substitution. By enclosing \W+</ in parentheses, it becomes the first and only capturing group, so replacing the match with $1 will replace it with the final space.

Another example is replacing the match in a !b, which would result in a( )(!)b (with ! being the last non-word character), leading to the final result being a!b.

Answer 3

Answer №2

Let's analyze the given string 'aaa &bbb' and process it.

The result is 'aaa&bbb'

\s+ captures the 3 spaces before the ampersand
(\W) captures the ampersand
$1 represents the ampersand and replaces ' &' with '&'

The same concept applies to the spaces. By including one of the spaces in the (\W) capture group, you ensure it meets the criteria for replacement. This also explains why your exclamation mark remains untouched.

Answer 4

Let's analyze the given string 'aaa &bbb' and process it.

The result is 'aaa&bbb'

\s+ captures the 3 spaces before the ampersand
(\W) captures the ampersand
$1 represents the ampersand and replaces ' &' with '&'

The same concept applies to the spaces. By including one of the spaces in the (\W) capture group, you ensure it meets the criteria for replacement. This also explains why your exclamation mark remains untouched.

Answer 5

Answer №3

These are the matches listed. I have replaced spaces with ☹ for better visibility

 "☹☹☹☹(☹)",
  "☹☹☹☹(☹)",
  "☹☹(!)",
  "☹(☹)"

The code specifies to replace the match with the content of the capture group.

' lots of☹☹☹☹(☹)spaces☹☹☹☹(☹)in this☹☹(!)☹(☹)'

After replacement, it will be

' lots of☹spaces☹in this!☹'

Answer 6

These are the matches listed. I have replaced spaces with ☹ for better visibility

 "☹☹☹☹(☹)",
  "☹☹☹☹(☹)",
  "☹☹(!)",
  "☹(☹)"

The code specifies to replace the match with the content of the capture group.

' lots of☹☹☹☹(☹)spaces☹☹☹☹(☹)in this☹☹(!)☹(☹)'

After replacement, it will be

' lots of☹spaces☹in this!☹'

How does this code `/(\s+(\W)/g, '$1')` remove spaces?

Answer №1

Answer №2

Answer №3

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