regular expression lazy pattern means that if the expression matches successfully, it will match as few characters as possible.
according to the literal meaning, take a look at my code as follows
var pattern=/a(\w*?)/;
var str="a123a";
console.log(str.replace(pattern,"$1")); I expect the output is" 1century, because of the lazy mode, matching a character can make it a success
, but the actual output is" 123a", which is the same as the greedy mode output
Why is this?
* means matching 0 to multiple times , so matches at least 0 characters. Because it is lazy matching , matching a empty string , that is, / a (\ code?) / and / a / are equivalent; $1 means str.replace (pattern,'$1') is equivalent to str.replace (/ a code,') , that is, replace the first a in the string with the empty string ; Note that / a / is not a global match, so only the first a is replaced; if you want to replace all a , you need to set pattern to global match pattern = / a (\ pattern?) / g ; $1 matches the first (), that is, the matching content in the group
w means matching all strings followed by means zero or any number of times? For example, do can match do or doo. Since an is followed by a value, it all matches
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