I need to sanitize some chars. I have a solution that works but I am wondering if there is a better solution that may be faster or better or if I flat out should be approaching this differently?

function escapeStr(_str){
    if (/\"|\'|\%/g.test(_str)) {
        _str = _str.replace(/"/g, "%22");
        _str = _str.replace(/'/g, "%27");
        _str = _str.replace(/%/g, "%25");
    return _str;

And vice versa:

function unescapeStr(_str){
    if (/\%22|\%27|\%25/g.test(_str)) {
        _str = _str.replace(/\%22/g, '"');
        _str = _str.replace(/\%27/g, "'");
        _str = _str.replace(/\%25/g, "%");
    return _str;
  • \$\begingroup\$ Why the if clause? Why not just replace, even if there are no matches? \$\endgroup\$ – abl Jan 27 '17 at 22:07
  • \$\begingroup\$ Could you provide some context about what you want to accomplish? Why do you want to handle those three characters specially? \$\endgroup\$ – 200_success Jan 27 '17 at 22:11
  • 4
    \$\begingroup\$ What collation? What issues? Sounds like it could be a known problem with an existing solution. \$\endgroup\$ – abl Jan 27 '17 at 22:27
  • 3
    \$\begingroup\$ If it's telling you that replace is undefined, it's because you trying to use it when on a _str value that isn't a string. You can replace the test call with typeof _str === 'string', and it should work the same. The problem is: Why is _str sometimes not a string... Anyway, that's all irrelevant, since, as abl said, this sounds like a known problem \$\endgroup\$ – Flambino Jan 28 '17 at 0:44
  • 1
    \$\begingroup\$ Thanks all. I just learned that this is great place to get really good opinions and hints on my code. Much appreciated! \$\endgroup\$ – Paul Jan 30 '17 at 15:34

Instead of hardcoding the characters, and their escape values specifically in your .replace call, you can pass a callback function for the replacement. I suggested that you dive into encodeURI:

So, something like:

function escapeStr(_str) {
  return _str.replace(/([%'"])/g, encodeURI)

function unescapeStr(_str) {
  return _str.replace(/(%(?:2[257]))/g, decodeURI)
const e = escapeStr, u = unescapeStr;
console.log( e("string") );
console.log( e("some\" char%s") );
console.log( u("string") );
console.log( u("some%22 char%25s") );

|improve this answer|||||

In first and second function

function escapeStr(_str){
    if (/\"|\'|\%/g.test(_str)) { // unnecessary if 
        _str = _str.replace(/"/g, "%22"); // unnecessary asigment
        _str = _str.replace(/'/g, "%27"); // unnecessary asigment
        _str = _str.replace(/%/g, "%25"); // unnecessary asigment
    return _str;

You can add if (value) this will prevent errors while passing null of something like this

This probably more complex and optimal:

const escapeStr = str => str ? str
  .replace(/%/g, "%25")
  .replace(/"/g, "%22")
  .replace(/'/g, "%27") : str

const unescapeStr = str => str ? str
  .replace(/\%25/g, "%")  
  .replace(/\%22/g, '"')
  .replace(/\%27/g, "'") : str

And if we want to go crazy use this:

const escapeStr = str => str ? str.replace(/[%'"]/g, encodeURI) : str

const unescapeStr = str => str ? str.replace(/\(%2[257])/g, decodeURI) : str

const testData = [`"hello"`,`'world'`,`2%3`,``, `xyz`, `%22hello%22`]

for (const data of testData) {
  console.log(`escapeStr -> ${escapeStr(data)}`)
  console.log(`escapeStr -> ${unescapeStr(data)}`)
|improve this answer|||||

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