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0 pointssimion3144y ago0 comments

>type TFooOptions

is that creating a JS class under the hood or is just a hint for the compiler?

What I do , but I don't work with recent PHP versions is to create this as a class, then to make this easy to construct objects from JSON i add a static function fromObject or fromJson that does it for me. Using real classes could prevent bugs when working with shit APIs like Microsoft TTS , where they have 2 endponts and one returns objects like voice: {name:"Bob"} and the other voice: {Name:"Bob"} so my code will wrap both type of responses into a well defined class.

I don't think there is a PHPDoc annotation just to hint what that object looks like but I will be happey to be shown a way.

0 comments

AltruisticGapHN4y ago

Simple object type in TypeScript, the "type checking" abstraction layer on top of JS, which actually doesn't even need to be compiled (eg. esbuild strips all typing from the code before compiling).

https://www.typescriptlang.org/docs/handbook/2/everyday-type...

Your approach is sensible. I remember using json_encode() indeed so I could declare some data in a php file with <<< HEREDOC in the Javascript short syntax (though short array syntax in php nowadays makes it less painful, JSON syntax still a bit less verbose).

simion314OP4y ago

I checked that example in TS playground and it does nothing when transpiling, so is only a compile type check. From my experience when I have an option object with many properties I would create a class both in PHP and JavaScript, usually functions that need an options object are very complex and many times you need to add even more options and having a class make it simple to do the changes and refactor.

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