Check whether '.' and '/' are escaped properly

Question

I'm currently using a function to check if a URL has correctly escaped two different characters: . and /.

To escape a character within a string, it requires 2 preceding \'s in the string (i.e.: \\. or \\/).

Here is the code I'm currently using. As you can see I'm matching the string with some regex, and then for every element returned I check if the first character of the string is \.

I feel this could be simplified, how can I make this better and avoid looping through all elements returned by match?

function isValidEscapedString(value) {
 return (value.match(/.?[./]/g)||[]).every((escape) => escape[0] === '\\');
}
console.log(isValidEscapedString('')); // true
console.log(isValidEscapedString('https://stackoverflow.com')); // false
console.log(isValidEscapedString('https:\/\/stackoverflow\.com')); // false
console.log(isValidEscapedString('https://stackoverflow\\.com')); // false
console.log(isValidEscapedString('https:\\/\\/stackoverflow\\.com')); //true

The examples in the code are working as intended.

• adiga: Why are you specifically escaping . and /? For RegExp constructor or something else?
• kemicofa: @adiga yes that's right. Somewhere in one of our other microservices it's used like that.
• adiga: There are other meta characters which can be present in a URL like -?[]()*+. You don't want to check for those?
• kemicofa: @adiga you're probably right. We're currently porting a legacy RoR project into Typscript and that's how it's been done for years. And would appear those other characters haven't been used yet. lol
• adiga: I have couple more questions. By, "without query params", your url will always be in the form of https://domain-name.com or can it have a path after that: https://domain-name.com/path? Do you get the escaped string as input or do you have access to unescaped string? If you have the unescaped string, there is already a function for escaping all the metacharacters.

Answer 1

What about, instead of search for every . and / and check the predecessor character, you write a regexp that find a . and / that don't have \\ before?

This way, you can avoid the loop and the short circuit to an array

function isValidEscaped(str) {
 const pattern = /[^\\]{2}[./]/g;
 return str.search(pattern) === -1; 
}