Using a juggling-check, you can test both null and undefined in one hit:
if (x == null) {
If you use a strict-check, it will only be true for values set to null and won't evaluate as true for undefined variables:
if (x === null) {
You can try this with various values using this example:
var a: number;
var b: number = null;
function check(x, name) {
if (x == null) {
console.log(name + ' == null');
}
if (x === null) {
console.log(name + ' === null');
}
if (typeof x === 'undefined') {
console.log(name + ' is undefined');
}
}
check(a, 'a');
check(b, 'b');
Output
Answer from Fenton on Stack Overflow"a == null"
"a is undefined"
"b == null"
"b === null"
Using a juggling-check, you can test both null and undefined in one hit:
if (x == null) {
If you use a strict-check, it will only be true for values set to null and won't evaluate as true for undefined variables:
if (x === null) {
You can try this with various values using this example:
var a: number;
var b: number = null;
function check(x, name) {
if (x == null) {
console.log(name + ' == null');
}
if (x === null) {
console.log(name + ' === null');
}
if (typeof x === 'undefined') {
console.log(name + ' is undefined');
}
}
check(a, 'a');
check(b, 'b');
Output
"a == null"
"a is undefined"
"b == null"
"b === null"
if( value ) {
}
will evaluate to true if value is not:
nullundefinedNaN- empty string
'' 0false
typescript includes javascript rules.
Videos
TypeScript types are only enforced at compile time and do not exist at runtime. Given this, if I define a TypeScript function that accepts a string parameter, is it necessary to perform null checks? This question arises because if the API returns null, it could lead to runtime errors.
Consider the following code snippet:
const capitalise = (val: string) => {
// Question: Is this check redundant due to TypeScript's type checking?
if (!val) {
return '';
}
return val.toUpperCase();
};
// Assume 'name' is fetched from an API or another source that could return null
const name: string = /* fetch from API or another source */;
capitalise(name);What are the best practices in such scenarios to ensure runtime safety in TypeScript applications?
given
let formEditor = document.getElementById('formeditor');Problem is getElementById return type is HTMLElement|null
I know that formeditor always exists. And I do not want to do "formEditor!." every time.
is this my only option
let formEditor = document.getElementById('formeditor') as HTMLElement;