[TS] Type narrowing string union?

I have the following setup;

const fn = (source: keyof typeof Obj | "$none") => {
  if (source in Obj) Obj[source];
  //                 ^^^^^^^^^^^
  // Element implicitly has an 'any' type because expression of type [...] can't be used to index type [...]
  // more code...
}

const fn = (source: keyof typeof Obj | "$none") => {
  if (source in Obj) Obj[source];
  //                 ^^^^^^^^^^^
  // Element implicitly has an 'any' type because expression of type [...] can't be used to index type [...]
  // more code...
}

I also have the following relevant compiler options;

"strict": true,
"alwaysStrict": true,
"keyofStringsOnly": true,
"moduleResolution": "node",
"module": "esnext",
"target": "esnext"

"strict": true,
"alwaysStrict": true,
"keyofStringsOnly": true,
"moduleResolution": "node",
"module": "esnext",
"target": "esnext"

Is there a way that I can prevent the error from occuring and have typescript correctly narrow

source

source

's type from

keyof typeof Obj | "$none"

keyof typeof Obj | "$none"

keyof typeof Obj

keyof typeof Obj

?

I would have thought that

if (source in Obj)

if (source in Obj)

then obviously

typeof source === keyof typeof Obj

typeof source === keyof typeof Obj

, if not directly then due to

Obj

Obj

's keys definitely including

source

source

.
I can't figure out how to narrow it, I've tried

.startsWith("$")

.startsWith("$")

but for some reason that's still not a type predicate to a

${arg0}${string}

${arg0}${string}

literal,

Object.keys(Obj).includes(source)

Object.keys(Obj).includes(source)

is just an ugly equivalent to the codeblock above,

source[0] === "$"

source[0] === "$"

doesn't work.

Nothing except

Obj[source as keyof typeof Obj]

Obj[source as keyof typeof Obj]

seems to work, which is not only ugly but entirely defeats the purpose of using typescript there.

[TS] Type narrowing string union?

I have the following setup;

const fn = (source: keyof typeof Obj | "$none") => {
  if (source in Obj) Obj[source];
  //                 ^^^^^^^^^^^
  // Element implicitly has an 'any' type because expression of type [...] can't be used to index type [...]
  // more code...
}

const fn = (source: keyof typeof Obj | "$none") => {
  if (source in Obj) Obj[source];
  //                 ^^^^^^^^^^^
  // Element implicitly has an 'any' type because expression of type [...] can't be used to index type [...]
  // more code...
}

I also have the following relevant compiler options;

"strict": true,
"alwaysStrict": true,
"keyofStringsOnly": true,
"moduleResolution": "node",
"module": "esnext",
"target": "esnext"

"strict": true,
"alwaysStrict": true,
"keyofStringsOnly": true,
"moduleResolution": "node",
"module": "esnext",
"target": "esnext"

Is there a way that I can prevent the error from occuring and have typescript correctly narrow

source

source

's type from

keyof typeof Obj | "$none"

keyof typeof Obj | "$none"

keyof typeof Obj

keyof typeof Obj

?

I would have thought that

if (source in Obj)

if (source in Obj)

then obviously

typeof source === keyof typeof Obj

typeof source === keyof typeof Obj

, if not directly then due to

Obj

Obj

's keys definitely including

source

source

.
I can't figure out how to narrow it, I've tried

.startsWith("$")

.startsWith("$")

but for some reason that's still not a type predicate to a

${arg0}${string}

${arg0}${string}

literal,

Object.keys(Obj).includes(source)

Object.keys(Obj).includes(source)

is just an ugly equivalent to the codeblock above,

source[0] === "$"

source[0] === "$"

doesn't work.

Nothing except

Obj[source as keyof typeof Obj]

Obj[source as keyof typeof Obj]

seems to work, which is not only ugly but entirely defeats the purpose of using typescript there.

[TS] Type narrowing string union?

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[TS] Type narrowing string union?

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