How to use optional chaining with array in Typescript?

You need to put a . after the ? to use optional chaining:

myArray.filter(x => x.testKey === myTestKey)?.[0]

Playground link

Using just the ? alone makes the compiler think you're trying to use the conditional operator (and then it throws an error since it doesn't see a : later)

Optional chaining isn't just a TypeScript thing - it is a finished proposal in plain JavaScript too.

It can be used with bracket notation like above, but it can also be used with dot notation property access:

const obj = {
  prop2: {
    nested2: 'val2'
  }
};

console.log(
  obj.prop1?.nested1,
  obj.prop2?.nested2
);

And with function calls:

const obj = {
  fn2: () => console.log('fn2 running')
};

obj.fn1?.();
obj.fn2?.();

Just found it after a little searching on the what's new page on official documentation

The right way to do it with array is to add . after ?

so it'll be like

myArray.filter(x => x.testKey === myTestKey)?.[0]

I'll like to throw some more light on what exactly happens with my above question case.

myArray.filter(x => x.testKey === myTestKey)?[0]

Transpiles to

const result = myArray.filter(x => x.testKey === myTestKey) ? [0] : ;

Due to which it throws the error since there's something missing after : and you probably don't want your code to be transpilled to this.

Thanks to Certain Performance's answer I learned new things about typescript especially the tool https://www.typescriptlang.org/play/index.html .

ECMA 262 (2020) which I am testing on Edge Chromium 84 can execute the Optional Chaining operator without TypeScript transpiler:

// All result are undefined
const a = {};

console.log(a?.b);
console.log(a?.["b-foo-1"]);
console.log(a?.b?.());

// Note that the following statements throw exceptions:
a?.(); // TypeError: a is not a function
a?.b(); // TypeError: a?.b is not a function