replace all special characters expect first occurrence of +

Question

I want to replace everything other than numeric char from a string, but if + appears at the starting of the string it should be ignored. for example

1234++!@#$%^&*()_+=-;',.><:" becomes 1234
+1234 becomes +1234
++++1234 becomes +1234
+1234++!@#$%^&*()_+=-;',.><:" becomes +1234
+1234++!@#$%^&*()_+=-;',.><:" becomes +1234
+1234ABCabc++!@#$%^&*()_+=-;',.><:" becomes +1234
1234ABCabc++!@#$%^&*()_+=-;',.><:" becomes 1234
Aa1234ABCabc++!@#$%^&*()_+=-;',.><:" becomes 1234
a+1234ABCabc++!@#$%^&*()_+=-;',.><:" becomes 1234
1+1234ABCabc++!@#$%^&*()_+=-;',.><:" becomes 11234

can you please suggest?

So you began with `if (str.startsWith('+') { ... }` right? then what happened? — Wyck, Sep 12 '22 at 19:58
so the nearest solution I found is str.replace(/\D/g, '') but this replaces the first + as well. I think we can have some regex that can ignore only first occurrence of + — Neeraj Pathak, Sep 12 '22 at 20:05
Many thanks for the help but unfortunately no so a123 should be 123 :( — Neeraj Pathak, Sep 12 '22 at 20:14
Your last example contradicts your stated objective. Do you want the first number found in a string with optional leading `+` ? Or do you want to extract all digits and retain only the very first `+` preceding the very first digit? — Peter Thoeny, Sep 12 '22 at 20:37

score 2 · Accepted Answer · answered Sep 12 '22 at 20:32

2

You can try:

str.replace(/((?<!^)\D)|(^[^+0-9])/g, '');

This replaces (with nothing):

any non-digit that is not at the start of the string.
any non-digit except + that is at the start of the string.

answered Sep 12 '22 at 20:32

Wyck

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many many thanks it works :) but with a warning Unsafe Regular Expressioneslint(security/detect-unsafe-regex) but I think it's ok to disable it :( Many many thanks for the help – Neeraj Pathak Sep 12 '22 at 20:44
in fairness, you could have used your original strategy of `result = str.replace(/\D/g, '')` But then followed up with `if (str.startsWith('+')) result = '+' + result;` as a fixup. You were almost there, and there would be no warning. – Wyck Sep 12 '22 at 20:51
yeah, I already have that solution but wanted to get rid of this extra 1 line of if check that can be saved. but I think I have to go back to that solution as our old friend Safari is not able to understand this nice regex and breaks with the error => SyntaxError: Invalid regular expression: invalid group specifier name – Neeraj Pathak Sep 12 '22 at 21:12

score 0 · Answer 2 · answered Sep 12 '22 at 20:33

This extracts numbers from a string, including an optional leading +:

var tests = [
  '1234++!@#$%^&*()_+=-;\',.><:"', // becomes 1234
  '+1234', // becomes +1234
  '++++1234', // becomes +1234
  '+1234++!@#$%^&*()_+=-;\',.><:"', // becomes +1234
  '+1234++!@#$%^&*()_+=-;\',.><:"', // becomes +1234
  '+1234ABCabc++!@#$%^&*()_+=-;\',.><:"', // becomes +1234
  '1234ABCabc++!@#$%^&*()_+=-;\',.><:"', // becomes 1234
  'Aa1234ABCabc++!@#$%^&*()_+=-;\',.><:"', // becomes 1234
  'a+1234ABCabc++!@#$%^&*()_+=-;\',.><:"', // becomes 1234
  '1+1234ABCabc++!@#$%^&*()_+=-;\',.><:"' // becomes 11234
];

tests.forEach(str => {
  console.log(str + ' => ' + str.replace(/^.*?(\+?[0-9]+).*$/, '$1'));
});

Output:

1234++!@#$%^&*()_+=-;',.><:" => 1234
VM1036:2 +1234 => +1234
VM1036:2 ++++1234 => +1234
2VM1036:2 +1234++!@#$%^&*()_+=-;',.><:" => +1234
VM1036:2 +1234ABCabc++!@#$%^&*()_+=-;',.><:" => +1234
VM1036:2 1234ABCabc++!@#$%^&*()_+=-;',.><:" => 1234
VM1036:2 Aa1234ABCabc++!@#$%^&*()_+=-;',.><:" => 1234
VM1036:2 a+1234ABCabc++!@#$%^&*()_+=-;',.><:" => +1234
VM1036:2 1+1234ABCabc++!@#$%^&*()_+=-;',.><:" => 1

Your described objective and last example contradict each other. Please be more specific what you need.

I don't see a contradiction. I interpret it as: Any non-digit that is not "a plus at the start of the string" should be removed. — Wyck, Sep 12 '22 at 20:36

replace all special characters expect first occurrence of +

2 Answers2