在续集中，如何选择与多行的不同值匹配的记录？

Question

As an example, I have the following table:

A | B
------
1 | 2
1 | 3
1 | 4
2 | 3
2 | 4
3 | 3

I want to select all values in B that have the value 1 AND 2 in the A Column. So in the above example I should get as a result (3,4) because only 3 and 4 have for column A the values 1 and 2.

How would I do this in sequelize?

This is what I tried:

 db.myModel.findAll({
    where: {
      A: whatImSearching
    },
    attributes: ['A', 'B']
  })

with whatImSearching = [1,2]

But that solution returns results where only one of the values match, ie I would get as a result (2,3,4). But I should only get (3,4).

Answer 1

Note: This is how it can be done in SQL, like @daf mentioned in comments, something on similar lines can be implemented in Sequelize. I've kept this around as a reference

You can use group by with conditional having like this

SELECT B 
FROM Table1 GROUP BY B
HAVING SUM(CASE WHEN A IN(1,2) THEN 1 ELSE 0 END) = 2

EDIT

Assuming your have duplicate values. this should be better

SELECT B 
FROM Table1 GROUP BY B
HAVING SUM(CASE WHEN A = 1 THEN 1 ELSE 0 END) = 1
    AND SUM(CASE WHEN A = 2 THEN 1 ELSE 0 END) = 1

Answer 2

Edit: I'm not immediately seeing a pure Sequelize way to do this, however the below should get where you want with the small hit of bringing back some extra rows from the database.

Assuming that you're doing the query using a model like:

var model = sequelize.define('model', { 
    A: { type: Sequelize.TEXT }, 
    B: { type: Sequelize.TEXT }
}, {
    tableName: 'SomeTable',
    timestamps: false
})

you can do:

var search = [1,2];
model.findAll({
    where: { A: { '$in': search } },
    attributes: ['A', 'B']
}).then(function(instances) {
  var unwrapped = instances.map(function(i) { 
    return i.get();
  });
  var withCounts = unwrapped.reduce(function(acc, cur) {
    var thisB = cur.B;
    var accItem = acc.filter(function(i) { 
      return i.B === thisB; 
    })[0];
    if (!accItem) {
      accItem = { B: thisB, count: 0 };
      acc.push(accItem);
    }
    accItem.count++;
  }, [])
  var hasAll = withCounts.filter(function(i) {
    return i.count === search.length;
  }).map(function(i) {
    return i.B;
  });
})