Mitch Fish asks:
The most performant way to get a set of the ID is already well known:
// ids: [001000000000001AAA, 001000000000002BBB, ...] Set<Id> ids = new Map<Id,SObject>(records).keySet();
We can also build sets from ANY string field, in a one-liner without a loop:
// names: [Matt, Neil, Chatter, Autoproc, ...] Set<String> names = new Map<String,SObject>([ SELECT Name Id FROM User GROUP BY Name ]).keySet();
How it works
The first thing to note is that we are using an Aggregate Query and not just ordinary SOQL. We combine special behaviours available to aggregate queries to build the set:
Step 1 – GROUP BY to collect unique values
Exactly per the documentation, you can use the GROUP BY option in a SOQL query to avoid iterating through individual query results. That is, you specify a group of records instead of processing many individual records.
Step 2 – Alias the field as Id
The default identifier for an aggregated field is
expr0 but we will alias the result as
Id. It doesn’t matter what the field contains, so long as it’s a string. We will exploit this special alias in the next step.
Step 3 – AggregateResult inherits from sObject
The methods available on sObject are also available on AggregateResult. This also applies to Lists and Maps; those same methods can be called on collections of aggregate results. Here is another example using
// emails: [email@example.com, firstname.lastname@example.org, ...] Set<String> emails = new Map<String,SObject>([ SELECT Email Id FROM Contact WHERE Email != null GROUP BY Email ]).keySet();