Require template literals instead of string concatenation JS-0246
Anti-pattern
Minor
5 months ago
—
3 years old
Unexpected string concatenation
10480 (Toggle.prototype.setActive = function (value) {
10481 if (this.active != value) {
10482 (this.active = value),
10483 typedoc.$html.toggleClass('has-' + this.className, value),10484 this.$el.toggleClass('active', value)
10485 const transition =
10486 (this.active ? 'to-has-' : 'from-has-') + this.classNameUnexpected string concatenation
10222 })) &&
10223 (name = '<span class="parent">' + parent + '.</span>' + name),
10224 $results.append(
10225 '<li class="' +10226 row.classes +10227 '"><a href="' +10228 base +10229 row.url +10230 '" class="tsd-kind-icon">' +10231 name +10232 '</li>'10233 )
10234 }
10235 }Unexpected string concatenation
10218 )
10219 let parent = row.parent || '';
10220 (parent = parent.replace(new RegExp(query, 'i'), function (match) {
10221 return '<b>' + match + '</b>'10222 })) &&
10223 (name = '<span class="parent">' + parent + '.</span>' + name),
10224 $results.append(Unexpected string concatenation
10213 let name = row.name.replace(
10214 new RegExp(query, 'i'),
10215 function (match) {
10216 return '<b>' + match + '</b>'10217 }
10218 )
10219 let parent = row.parent || '';Unexpected string concatenation
10207 ($results.empty(), loadingState == SearchLoadingState.Ready && query)
10208 ) {
10209 let res = index.search('*' + query + '*')
10210 res.length === 0 && (res = index.search('*' + query + '~1*'))10211 for (let i = 0, c = Math.min(10, res.length); i < c; i++) {
10212 const row = search.data.rows[Number(res[i].ref)]
10213 let name = row.name.replace(
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!";