answers.md

Chapter 6 Questions and Answers

1. This select is incomplete. How would you complete it so that it says "Pleased to meet you, I'm " and then the NPC's name?

You can do it with concatenation using ++:

select NPC {
  name,
  greeting := "Pleased to meet you. I'm " ++ .name ++ '.'
};

2. How would you update Mina's places_visited to include Romania if she went to Castle Dracula for a visit?

Here is one way:

update Person filter .name = 'Mina Murray'
set {
  places_visited += (select Place filter .name = 'Romania')
};

You can of course go with update NPC and select Country if you prefer.

Also, here is the same thing using with:

with
  mina := (select NPC filter .name = 'Mina Murray'),
  romania := (select Country filter .name = 'Romania'),
update mina
set {
  places_visited += romania
};

3. With the set {'W', 'J', 'C'}, how would you display all the Person types with a name that contains any of these capital letters?

It looks like this:

with letters := {'W', 'J', 'C'}
select Person {
  name
} filter .name like '%' ++ letters ++ '%';

And should display these characters we've inserted so far:

{
  Object {name: 'Vampire Woman 1'},
  Object {name: 'Vampire Woman 2'},
  Object {name: 'Vampire Woman 3'},
  Object {name: 'Jonathan Harker'},
  Object {name: 'Count Dracula'},
}

The key is that like takes a string, so you can concatenate % on the left and right with ++.

4. How would you display this same query as JSON?

Getting JSON output is super easy by casting with <json>, but where does it go? You can't put it in front of select, and <json>Person isn't an expression either, so this won't work:

with letters := {'W', 'J', 'C'}
select <json>Person {
  name
} filter .name like '%' ++ letters ++ '%';

You need to wrap the select in brackets, cast with <json> and then select that:

with letters := {'W', 'J', 'C'}
select <json>(
  select Person {
    name
  } filter .name like '%' ++ letters ++ '%'
);

Or you can use with to do this:

with
  letters := {'W', 'J', 'C'},
  P := (
    select Person filter .name like '%' ++ letters ++ '%'
  )
select <json>P { name };

So you're selecting the casted-to-JSON version of the result of select Person.

5. How would you add ' the Great' to every Person type?

Easy, just update without filter:

update Person
set {
  name := .name ++ ' the Great'
};

Now their names are 'Vampire Woman 1 the Great', 'Mina Murray the Great', and so on.

Bonus question: to undo this, just set name to the same string minus the last 10 characters using [0:-10]:

update Person
set {
  name := .name[0:-10]
};