I'm destructing an object as following:
const { lang, ...params } = ctx.query;
But TS gives me an error in this line:
if (!value.includes(params[key])) return false;
How can I solve this issue?
CodePudding user response:
Use .includes(params[key] as string) (or .includes(<string>params[key])) to tell Typescript you know it's a string.
CodePudding user response:
first check if params[key] is defined if (params && params[key]) {...
CodePudding user response:
Typescript is understanding your property represented by params[key] has either a type of an array, or a string, the property is even undefined because maybe key does not exist in your object.
You could first check if the property exists
Then you could check if the type of your property is a string this way
if (params[key] && typeof params[key] === string && !value.includes(params[key])) return false;
Ideally you would have a parent if block like this
if(params[key] && typeof params[key] === string){
//The property exists and if of type string, do your thing with it.
if (!value.includes(params[key])) return false;
}else{
//Why it's not the type you expected?
//Do something, log or throw an error for example
}
This way Typescript knows it will only reach the child if when the params[key] property is really a string, and then the typescript checker/compiler will not complain about that uncertainty.

