Remove part of string using regex and replace

Question

give a string like this (it can be any pattern, but the following format):

lastname/firstname/_/country/postalCode/_/regionId/city/addressFirst/addressSecond/_/phone

I am making a function that, when I pass some address parts, the function will return those parts and remove extras part that are not requested and mantaining maximum one spacing _ if many where removed:

FE :

exs :

 input : ["country", "postalCode"]
 return "country/postalCode

 input : ["lastname", "firstname", "regionId"]
 return "lastname/firstname/_/regionId"

 input : ["firstname", "country", "regionId", "city"]
 return "firstname/_/country/_/regionId/city"

 input : ["country", "regionId", "phone"]
 return "country/_/regionId/_/phone"

Now, My method is as follow :

  type AddressPart = "firstname" | "lastname" | ... | "phone";
  const allAddressParts = ["firstname", "lastname", ... ,"phone"];
  static getAddress(
    format = "lastname/firstname/_/country/postalCode/_/regionId/city/addressFirst/addressSecond/_/phone"
    parts: AddressPart[],
  ) {
    const toRemove = allAddressParts.filter((part) => !parts.includes(part));
    toRemove.forEach((part) => {
      format = format
        .replace(`_/${part}/_`, '_')
        .replace(new RegExp(part + '/?'), '');
    });
    return format;
  }

the above looks ok, but it fail at the end and start :

_/country/postalCode/_/regionId/city/addressFirst/addressSecond/_/

How can I remove /_/ or _/ if it is situated on start or at the end without relooping over the array ?

Try this string.replace(/^_\/$/, '')

Forbidden
– Forbidden

2021-12-03 06:59:49 +00:00
Commented Dec 3, 2021 at 6:59 — Forbidden
– Forbidden, Commented Dec 3, 2021 at 6:59

ProfDFrancis · Accepted Answer · 2021-12-03 07:11:57Z

2

You can use regexes on the final string to remove leading and trailing unwanted symbols

The first regex looks for:

Start of the string, with the ^ symbol
One or more "_/", using

parentheses to group the character pair,

the + sign to allow 1 or more, and

a backslash to escape the forward slash

And the second regex looks for:

a "/"
zero or more "_/", with the *
the end of the string with the $ symbol

s1 = "_/country/postalCode/_/regionId/city/addressFirst/addressSecond/_/"
s2 = s1.replace(/^(_\/)+/,"").replace(/\/(_\/)*$/,"")
console.log(s2)

You may be able to simplify

If you know for certain that there will no more than one "_" at the beginning and one at the end, you don't need the parentheses and +/* symbols.

edited Dec 3, 2021 at 7:11

answered Dec 3, 2021 at 7:03

ProfDFrancis

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2 Comments

Crocsx Over a year ago

I made up this .match("^/?_?/?(.*?)/?_?/?$") and catch group 1, would this also work ? :x

ProfDFrancis Over a year ago

It would work but it is not testing exactly the criteria you described. Your proposed version would also be looking for, and removing, the leading underscore in the string _underground/overground/wombling/free. For your future coding sanity it is advisable to look for and remove exactly what you know to be the extraneous text, because you will be able to recognise it when you come to edit the code.

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Remove part of string using regex and replace

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You can use regexes on the final string to remove leading and trailing unwanted symbols

You may be able to simplify

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1 Answer 1

You can use regexes on the final string to remove leading and trailing unwanted symbols

You may be able to simplify

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