This specific feature is unique to TypeScript. Upon initializing the saveData
variable with a value, TypeScript automatically deduces that it should only have two properties: userPartyRoleId
, notes
, and nothing else.
If you encounter this issue, there are several potential solutions available.
Solution 1
The simplest resolution involves including speciality
during the initialization of saveData
. This action will ensure that the correct type is assigned to saveData
.
const saveData = {
userPartyRoleId: Number(this.userPartyRoleId),
notes: this.notes,
speciality: this.spec
};
Solution 2
Alternatively, consider utilizing the spread operator as suggested in the comments. By doing so, you can create a new object with the appropriate type and assign it to a separate variable.
const saveData = {
userPartyRoleId: Number(this.userPartyRoleId),
notes: this.notes
};
const processedData = {
...saveData,
speciality: this.spec
};
Solution 3
This approach ensures that your code compiles without errors and runs smoothly. By utilizing the any
type, you can instruct TypeScript to skip type-checking for saveData
.
const saveData: any = {
userPartyRoleId: Number(this.userPartyRoleId),
notes: this.notes
};
saveData.speciality = this.spec;
While this solution resolves the immediate issue, it sacrifices TypeScript's strict-typing functionality, which serves as its primary purpose.