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I have the following replace function
myString.replace(/\s\w(?=\s)/,"$1\xA0");
The aim is to take single-letter words (e.g. prepositions) and add a non-breaking space after them, instead of standard space.
However the above $1 variable doesn't work for me. It inserts text "$1 " instead of a part of original matched string + nbsp.
What is the reason for the observed behaviour? Is there any other way to achieve it?
592
To match a character having special meaning in regex, you need to use a escape sequence prefix with a backslash ( \ ). E.g., \. matches "." ; regex \+ matches "+" ; and regex \( matches "(" . You also need to use regex \\ to match "\" (back-slash).
In your specific example, the $1 will be the group (^| ) which is "position of the start of string (zero-width), or a single space character". So by replacing the whole expression with that, you're basically removing the variable theClass and potentially a space after it.
$1 is the first group from your regular expression, $2 is the second. Groups are defined by brackets, so your first group ($1) is whatever is matched by (\d+). You'll need to do some reading up on regular expressions to understand what that matches.
To use RegEx, the first argument of replace will be replaced with regex syntax, for example /regex/ . This syntax serves as a pattern where any parts of the string that match it will be replaced with the new substring. The string 3foobar4 matches the regex /\d. *\d/ , so it is replaced.
$1 doesn't work because you don't have any capturing subgroups.
The regular expression should be something like /\b(\w+)\s+/.
171
Seems you want to do something like this:
myString.replace(/\s(\w)\s/,"$1\xA0");
but that way you will loose the whitespace before your single-letter word. So you probably want to also include the first \s in the capturing group.
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