Why MongoDB cannot use a compound index that is mu

问题:

Consider the below Mongo index strategy and the query,

Index:

db.collec.ensureIndex({a:1,b:1,c:1});

Query:

db.collec.find({"a":"valueA"},{"_id":0,"a":1,"c":1}).sort({"c":-1}).limit(150)

The explain on the above query returns:

/* 0 */
{
    "cursor" : "BtreeCursor a_1_b_1_c_1",
    "isMultiKey" : false,
    "n" : 150,
    "nscannedObjects" : 178,
    "nscanned" : 178,
    "nscannedObjectsAllPlans" : 279,
    "nscannedAllPlans" : 279,
    "scanAndOrder" : true,
    "indexOnly" : true,
    "nYields" : 0,
    "nChunkSkips" : 0,
    "millis" : 1,
    "indexBounds" : {
        "a" : [ 
            [ 
                "valueA", 
                "valueA"
            ]
        ],
        "b" : [ 
            [ 
                {
                    "$minElement" : 1
                }, 
                {
                    "$maxElement" : 1
                }
            ]
        ],
        "c" : [ 
            [ 
                {
                    "$minElement" : 1
                }, 
                {
                    "$maxElement" : 1
                }
            ]
        ]
    }
}

The question here is Its clearly indicated that the query runs completely on Index(as "indexOnly" : true). But why the "scanAndOrder" : true
According to Btree index model, c is at the tail of the index so it can be utilized to sort. No?

Why its not used?

回答1:

This is correct and also documented.

As to why: The index looks essentially like this tree:

• A: "value A"
  • B : "ABC"
    • C: 435
    • C: 678
  • B : "BCD"
    • C: 123
    • C: 993

As you can see, the ordering is correct and ascending, but if you'd take the values of c in-order without limiting to a subset of fixed b, you'd get [435, 678, 123, 993], which is not correct, so scanAndOrder is required.

Unfortunately, indexes without index intersectioning are very inflexible.

回答2:

If you use below index scan and order will be false.

db.collec.ensureIndex({a:1,c:-1,b:1});

Check this out