Require template literals instead of string concatenation JS-0246
Anti-pattern
Minor
4 months ago
—
2 years old
Unexpected string concatenation
111 r && l.isElement(r)
112 ? "element"
113 : (s = r) === null || typeof s == "undefined"
114 ? "" + s 115 : {}.toString
116 .call(s)
117 .match(/\s([a-z]+)/i)[1]Unexpected string concatenation
4230 typeof e == "string")
4231 ) {
4232 if (typeof o[e] == "undefined")
4233 throw new TypeError('No method named "' + e + '"');4234 o[e](this);
4235 }
4236 });Unexpected string concatenation
4093 typeof e == "string")
4094 ) {
4095 if (typeof o[e] == "undefined")
4096 throw new TypeError('No method named "' + e + '"');4097 o[e]();
4098 }
4099 });Unexpected string concatenation
3949 typeof e == "string")
3950 ) {
3951 if (typeof n[e] == "undefined")
3952 throw new TypeError('No method named "' + e + '"');3953 n[e]();
3954 }
3955 });Unexpected string concatenation
3903 (this._activeTarget = t), this._clear();
3904 var e = this._selector.split(",").map(function (e) {
3905 return (
3906 e + '[data-target="' + t + '"],' + e + '[href="' + t + '"]'3907 );
3908 }),
3909 n = i.default(
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!";