正则表达式用于转义特殊字符

Question

Trying RegExp to generate regular expression for string matching. Got confused with escaping specified in the docs . Whenever using RegExp constructor type input string need to escaped using function provided in that docs.

Properly working with literal approach

 "test".match(/te*/gi)

Output:

[ 'te', 't' ]

When I tried same thing with RegExp constructor output varies.

rg="te*"
function escapeRegExp(str) {
   return str.replace(/[\-\[\]\/\{\}\(\)\*\+\?\.\\\^\$\|]/g, "\\$&");
}

exp = new RegExp(escapeRegExp(rg),'gi')
console.log("exp =>", exp);
console.log("test".match(exp));

Output:

exp => /te*/gi

null

Will it be there any backslashes before '*'? Is it properly escaped? Why the regex generated is different from literal expression? I couldn't figure out what I am missing from that doc.

I appreciate the help to point out mistake available here!

Answer 1

Because you don't want to escape the * - it's a regex character:

 const temp = "te*"; const reg = new RegExp(temp, "gi"); console.log("test".match(reg)); 

Answer 2

My doubt is all these need to be escaped before passing new RegExp() or only (backslashes \\) alone need to be escaped. Which one need to be escaped or not be escaped is not clear to me?

Your question is answered right at the start of the document section you refer to. Read that again:

If you need to use any of the special characters literally (actually searching for a '*', for instance), you must escape it …

Conversely, if you need any of the special characters to have its special meaning, you must not escape it.

Besides the above, any backslash which is to be placed in the string has to be doubled if assigned from a string literal .