Question

Why does

var a = "1ab";
a = a.replace(/[^\d][a-z]/g, "");

remove “a” and “b”? I thought that it should remove only “b” because “a” is preceeded by a number which I excluded by [^\d].

Who can help me?

Answer 1

It removes them because your character class [^\d] means “anything that is not a digit.” The ^ means “not” and \d means “digit”. So your expression means “match anything that is not a digit followed by any lowercase letter”, which matches "ab".

To remove only lowercase letters that don’t have a digit in front of them, it’s probably easiest to do it with a capture group:

a = a.replace(/([\d][a-z])[a-z]/g, "$1");

That says: Match a digit followed by a lowercase letter (as a group) followed by a lowercase letter (outside the group), and replace it with just the group’s contents. If you want to remove "c" from "1abc", add a + after the last [a-z], e.g.: /([\d][a-z])[a-z]+/g.