Require template literals instead of string concatenation JS-0246
Anti-pattern
Minor
5 months ago
—
2 years old
Unexpected string concatenation
5477 r["scroll" + a],
5478 e.body["offset" + a],
5479 r["offset" + a],
5480 r["client" + a], 5481 ))
5482 : void 0 === n
5483 ? S.css(e, t, i)Unexpected string concatenation
5476 e.body["scroll" + a],
5477 r["scroll" + a],
5478 e.body["offset" + a],
5479 r["offset" + a], 5480 r["client" + a],
5481 ))
5482 : void 0 === nUnexpected string concatenation
5475 Math.max(
5476 e.body["scroll" + a],
5477 r["scroll" + a],
5478 e.body["offset" + a], 5479 r["offset" + a],
5480 r["client" + a],
5481 ))Unexpected string concatenation
5474 ? ((r = e.documentElement),
5475 Math.max(
5476 e.body["scroll" + a],
5477 r["scroll" + a], 5478 e.body["offset" + a],
5479 r["offset" + a],
5480 r["client" + a],Unexpected string concatenation
5473 : 9 === e.nodeType
5474 ? ((r = e.documentElement),
5475 Math.max(
5476 e.body["scroll" + a], 5477 r["scroll" + a],
5478 e.body["offset" + a],
5479 r["offset" + a],
Description
In ES2015 (ES6), we can use template literals instead of string concatenation.
var str = `Hello, ${name}!`;
Bad Practice
var str = "Hello, " + name + "!";
var str = "Time: " + (12 * 60 * 60 * 1000);
Recommended
var str = "Hello World!";
var str = `Hello, ${name}!`;
var str = `Time: ${12 * 60 * 60 * 1000}`;
var str = "Hello, " + "World!";