Here's what the interface look like

 export interface Patient {
    doctors: [{
      id: null,
      gp: null,
     }]
  }

Here's my tuple

  linkedDoctorShort: Array<any> = []; // will contain only the ID and GP 

I tried some solutions that I had found on StackOverflow, but I still got the same error, especially when I want to save all the information:

  onSave() {
    const patientInfo: Patient = {
    doctors: this.linkedDoctorShort,
  }; 

Error message :

Property '0' is missing in type 'any[]' but required in type '[{ id: string; gp: boolean; }]'.

Thank you for your help

Here's what the interface look like

 export interface Patient {
    doctors: [{
      id: null,
      gp: null,
     }]
  }

Here's my tuple

  linkedDoctorShort: Array<any> = []; // will contain only the ID and GP 

I tried some solutions that I had found on StackOverflow, but I still got the same error, especially when I want to save all the information:

  onSave() {
    const patientInfo: Patient = {
    doctors: this.linkedDoctorShort,
  }; 

Error message :

Property '0' is missing in type 'any[]' but required in type '[{ id: string; gp: boolean; }]'.

Thank you for your help

• javascript
• typescript

3 Answers 3

linkedDoctorShort: Array<any> = []; is not a tuple. It is an array initialized with an empty array.

If you want this to be an array (doctors can have any number of elements) use an array type in the interface

 export interface Patient {
    doctors: Array<{
      id: null,
      gp: null,
     }>
  }

If you want only a single element (ie a tuple of length one). Then use that in the type of linkedDoctorShort and initialize it accordingly:

export interface Patient {
    doctors: [{
        id: null,
        gp: null, // are you sure these can only be null ? I think you mean soemthing like string | null
    }]
}

let linkedDoctorShort: [any] = [{ id: null, gp: null }]; //  Or better yes let linkedDoctorShort: [Patient['doctors'][0]] to keep type safety 

const patientInfo: Patient = {
    doctors: this.linkedDoctorShort,
}; 

Change your interface to:

export interface Patient {
  doctors: Array<{id: string;gp: boolean;}>;
}

But I'm not a huge fan of inline typing. I prefer a cleaner syntax, like:

export interface Doctor {
  id: string;
  gp: boolean;
}

export interface Patient {
  doctors: Doctor[];
}

despite the error you encounter, suggestion is clear type definition.

export interface Doctor {
    id: string | number;
    gp: boolean;    
}
export interface Patient {
    doctors: Doctor[];
}

Just as @jpavel said. then you can use the advantage of Typescript.

const linkedDoctorShort:Doctor[] = [];
onSave() {
    const patientInfo: Patient = {
    doctors: this.linkedDoctorShort,
}; 

you won't miss it