Safe way to check if a property of an object contains a substring

Question

There is a variable v in my program, and I want to check if its property p contains a sub string sub. I could write a code like follows:

if (v.p.indexOf('sub') !== -1) {
    // do something here
}

However, I have some doubts when seeing this code:

What if v is never declared
What if the property p does not exist in v?
What if v.p is null or undefined?
What if v.p is not a string

I want all the above cases not to raise errors in my code, and only do something here when v.p exists and contains a string that contains sub.

Does anyone know how to write this code correctly?

Thank you... But I prefer raw typescript or javascript code... — SoftTimur
– SoftTimur, Commented Jun 11, 2020 at 9:40
I'm writing TypeScript, if you are sure about v?.p?.indexOf, please post an answer... — SoftTimur
– SoftTimur, Commented Jun 11, 2020 at 9:41
typescriptlang.org/docs/handbook/release-notes/… Optional chaining — Isa
– Isa, Commented Jun 11, 2020 at 9:42

Isa · Accepted Answer · 2020-06-11 09:43:38Z

2

With TS you can use "Optional chaining"

For example

let x = foo?.bar.baz();

this is a way of saying that when foo is defined, foo.bar.baz() will be computed; but when foo is null or undefined, stop what we’re doing and just return undefined.”

More plainly, that code snippet is the same as writing the following.

let x = (foo === null || foo === undefined) ?
    undefined :
    foo.bar.baz();

See more https://www.typescriptlang.org/docs/handbook/release-notes/typescript-3-7.html Optional chaining

answered Jun 11, 2020 at 9:43

Isa

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3 Comments

codeandcloud Over a year ago

foo?.bar.baz() will return error TS2304: Cannot find name 'foo'. For optional chaining foo must not be undefined

user120242 Over a year ago

add a note about typeof foo for undeclared variables. especially if you're already doing TS, it should never be allowed, and your linter should force rejection

SoftTimur Over a year ago

note that use this ? when you are 100% sure it is typescript; t silently has trouble in JavaScript code.

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