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
Since switching to TypeScript I have been using a lot of optional properties, for example:
type store = {
currentUserId?: string
}
function logout () {
store.currentUserId = undefined
}However my coworkers and I have been discussing whether null is a more appropriate type instead of undefined, like this:
type store = {
currentUserId: string | null
}
function logout () {
store.currentUserId = null
}It seems like the use of undefined in TypeScript differs slightly from in Javascript.
Do you guys/girls use undefined or null more often? And, which of the examples above do you think is better?
In TypeScript 2, you can use the undefined type to check for undefined values.
If you declare a variable as:
let uemail : string | undefined;
Then you can check if the variable uemail is undefined like this:
if(uemail === undefined)
{
}
From Typescript 3.7 on, you can also use nullish coalescing:
let x = foo ?? bar();
Which is the equivalent for checking for null or undefined:
let x = (foo !== null && foo !== undefined) ?
foo :
bar();
https://www.typescriptlang.org/docs/handbook/release-notes/typescript-3-7.html#nullish-coalescing
While not exactly the same, you could write your code as:
var uemail = localStorage.getItem("useremail") ?? alert('Undefined');
Hello guys, i need your help, i would like to know which you think is a better approach:
1 --------------------------------------------
const {content} = props;
if(content !== "undefined"){
//code
}
2 --------------------------------------------
const {content = null} = props;
if(!content){
//code
}
This code do not work at all, but not sure why.
type IsUndefined<T> = T extends undefined ? 1 : 0;
For example:
type IsTypeUndefined = IsUndefined<number | undefined>; // This returns: 1 | 0
Is there a way for check if a type is undefined in a conditional?