简单的正则表达式，如果没有尾随数字，则将许多零变成单个零

Question

I want to remove any digits before '$' symbol.
I want to remove any '0' ',' digits after $ symbol.
But i don't want to remove '0's if string is mentioned only like '$00' or '$0' .

This is my Regex: home_value.replace(/,|(\\d*\\$0*)|(^0)*/g, ''); I want to remove all 0's and '$', ',' symbols from the below mentioned strings. But lets say if user enter '$0' or '$000' ..it should be convert to '0'

0045$002,34234 = 234234
$000300 = 300
$000387687 = 387687
$000 = 0
$00 = 0
$0 = 0

Answer 1

There's an easier way to do this than with one regex, since you're just after the dollar amount:

var stripExcessNumbers = function (n) {
    return parseInt(String(n).split("$")[1].replace(/[^\d]/g, ""), 10);
}

This should work so long as your numbers don't get too large.

test cases

 var stripExcessNumbers = function (n) { return parseInt(String(n).split("$")[1].replace(/[^\\d]/g, ""), 10); } var tests = "0045$002,34234 $000300 $000387687 $000 $00 $0".split(" ") , answers = [234234, 300, 387687, 0, 0, 0] for (var i = 0; i < tests.length; i++) { document.write("<p>" + stripExcessNumbers(tests[i]) + ", correct answer: " + answers[i] + "</p>"); } 

Answer 2

You can use a regex like this:

^.*\$(0)0*$|.*\$0*|,

Working demo

Look at the substitution section, you have this output:

234234
300
387687
0
0
0

Answer 3

s = '2342$08934200'
+s.substr(s.indexOf('$')+1) // 8934200

s = '$0000'
+s.substr(s.indexOf('$')+1) // 0