Require template literals instead of string concatenation JS-0246
Anti-pattern
Minor
a month ago
—
3 months old
Unexpected string concatenation
114 const licenceNumberIndex = Number.parseInt(licenceNumber.split('-')[1]) + 1
115
116 return (
117 currentYear.toString() +118 '-' +119 licenceNumberIndex.toString().padStart(digits, '0')120 )
121}
122Unexpected string concatenation
108 .get(licenceNumberLength) as string
109
110 if (!licenceNumber) {
111 return currentYear.toString() + '-' + '1'.padStart(digits, '0')112 }
113
114 const licenceNumberIndex = Number.parseInt(licenceNumber.split('-')[1]) + 1Unexpected string concatenation
77 ) + 1
78
79 return (
80 categorySlug + '-' + licenceNumberIndex.toString().padStart(digits, '0') 81 )
82}
83Unexpected string concatenation
67 .get(licenceNumberLength) as string)
68
69 if (!licenceNumber) {
70 return categorySlug + '-' + '1'.padStart(digits, '0') 71 }
72
73 const licenceNumberIndex =Unexpected string concatenation
42 const licenceNumberLength = categorySlug.length + 1 + digits
43
44 database.function('userFn_matchesFormat', (licenceNumber: string): 0 | 1 => {
45 return licenceNumber.startsWith(categorySlug + '-') ? 1 : 0 46 })
47
48 database.function('userFn_digits', (licenceNumber: string): string => {
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!";