I am trying to understand why writing regex as a string is not working but using it without string does work.
this is my example:
var patt = new RegExp("/test_.*/gi");
var res = patt.test("test_4");
console.log(res);
will return false
but this:
var patt = /test_.*/gi;
var res = patt.test("test_4");
console.log(res);
will return true
what is the difference
Your syntax of RegExp is wrong.
RegExp constructor to create new Regular Expression.RegExp constructor.. should to be escaped, if you want to match . literal.Usage
var patt = new RegExp("test_.*", "gi");
Demo
var patt = new RegExp("test_.*", "gi");
var res = patt.test("test_4");
document.write(res);
Demo
. ? not the same Regex as OP.
"\." in string. While it technically works, since the \. resolves into . in the string literal before going into RegExp constructor, it is confusing to the reader. literal, Thanks for clarification, I've updated answerThe regexp constructor does not need delimiters, also, flags are isolated in another argument which simplifies it to
var patt = new RegExp('test_.*', 'gi');
var res = patt.test("test_4");
console.log(res);
You do not need to include the / at the start and end of the regular expression when using the constructor and the flags need to be in the second argument. See the MDN Documentation for RegExp.
var patt = new RegExp( "test_.*", "gi" );
var res = patt.test("test_4");
console.log(res);You are not calling the constructor correctly, the falgs should be passed as a second parameter:
new RegExp('test_.*', 'gi');
As you can see in Documentation, it should respect this format:
new RegExp(pattern[, flags])
This is your demo:
var patt = new RegExp("test_.*", 'gi');
var res = patt.test("test_4");
alert(res);
gflag withtest