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I get this behavior in both Chrome (Developer Tools) and Firefox (Firebug). Note the regex test returns alternating true/false values:
> var re = /.*?\bbl.*\bgr.*/gi;
undefined
> re
/.*?\\bbl.*\\bgr.*/gi
> re.test("Blue-Green");
true
> re.test("Blue-Green");
false
> re.test("Blue-Green");
true
> re.test("Blue-Green");
false
However, testing the same regex as a literal:
> /.*?\bbl.*\bgr.*/gi.test("Blue-Green");
true
> /.*?\bbl.*\bgr.*/gi.test("Blue-Green");
true
> /.*?\bbl.*\bgr.*/gi.test("Blue-Green");
true
> /.*?\bbl.*\bgr.*/gi.test("Blue-Green");
true
I can't explain this and it's making debugging very difficult. Can anyone explain this behavior?
295
The match() method retrieves the matches when matching a string against a regular expression. Use . test if you want a faster boolean check.
test() The test() method executes a search for a match between a regular expression and a specified string. Returns true or false .
If you need to know if a string matches a regular expression RegExp , use RegExp.prototype.test() . If you only want the first match found, you might want to use RegExp.prototype.exec() instead.
Use the test() method to check if a regular expression matches an entire string, e.g. /^hello$/. test(str) . The caret ^ and dollar sign $ match the beginning and end of the string. The test method returns true if the regex matches the entire string, and false otherwise.
/g (global) regexps will do that, yes.
See this question.
When you write a literal, you're getting a new regexp object every time, so losing the lastIndex state associated with the old object.
143
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