Is there a simple way to ignore the white space in a target string when searching for matches using a regular expression pattern? For example, if my search is for "cats", I would want "c ats" or "ca ts" to match. I can't strip out the whitespace beforehand because I need to find the begin and end index of the match (including any whitespace) in order to highlight that match and any whitespace needs to be there for formatting purposes.
You can easily trim unnecessary whitespace from the start and the end of a string or the lines in a text file by doing a regex search-and-replace. Search for ^[ \t]+ and replace with nothing to delete leading whitespace (spaces and tabs). Search for [ \t]+$ to trim trailing whitespace.
Turn on free-spacing mode to ignore whitespace between regex tokens and allow # comments. Turn on free-spacing mode to ignore whitespace between regex tokens and allow # comments, both inside and outside character classes.
regex ignore spacesTrim whitespaces around string, but not inside of string.
Use the String. replace() method to remove all whitespace from a string, e.g. str. replace(/\s/g, '') . The replace() method will remove all whitespace characters by replacing them with an empty string.
You can stick optional whitespace characters \s*
in between every other character in your regex. Although granted, it will get a bit lengthy.
/cats/
-> /c\s*a\s*t\s*s/
While the accepted answer is technically correct, a more practical approach, if possible, is to just strip whitespace out of both the regular expression and the search string.
If you want to search for "my cats", instead of:
myString.match(/m\s*y\s*c\s*a\*st\s*s\s*/g)
Just do:
myString.replace(/\s*/g,"").match(/mycats/g)
Warning: You can't automate this on the regular expression by just replacing all spaces with empty strings because they may occur in a negation or otherwise make your regular expression invalid.
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