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bugSomething isn't workingSomething isn't working
Description
Describe the bug
ifError() is not working for methods of entities.
When we call an entity (or root entity) method with ifError(), the .value member is not being added in the fail check.
See example below:
To Reproduce
For the following Entity in bl:
Entity ModuleEntity {
static create(props: ModuleProps): (OK(ModuleEntity), Errors())
{}
public rename(name: ModuleNameVO): (OK(void), Errors()) {
this.name = name;
}
}
when we call the rename method in a command handler, for example:
const name = ModuleNameVO.create({name: command.name}).ifError();
module.rename(name).ifError();
In typescript will be generated the following code in the command handler:
const result_713354 = module.value.rename(name.value);
if (result_713354.isFail()) {
return fail(result_713354);
}
Expected behavior
The generated code should have the .value in fail function:
const result_713354 = module.value.rename(name.value);
if (result_713354.isFail()) {
return fail(result_713354.value);
}
Temporary Solution
Instead of (OK(void), Errors()) we can declare void as return type of the method:
public rename(name: ModuleNameVO): void {
this.name = name;
}
In this case in the command handler in bl we will write:
const name = ModuleNameVO.create({name: command.name}).ifError();
module.rename(name);
and the typescript code generated will be:
module.value.rename(name.value);
without the fail check.
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bugSomething isn't workingSomething isn't working