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Diffstat (limited to 'jstests')
-rw-r--r-- | jstests/core/plan_cache_replanning.js | 120 |
1 files changed, 0 insertions, 120 deletions
diff --git a/jstests/core/plan_cache_replanning.js b/jstests/core/plan_cache_replanning.js deleted file mode 100644 index 82fe85a8794..00000000000 --- a/jstests/core/plan_cache_replanning.js +++ /dev/null @@ -1,120 +0,0 @@ -/** - * This test will attempt to create a scenario where the plan cache entry for a given query shape - * oscillates. It achieves this by creating two indexes, A and B, on a collection, and interleaving - * queries which are "ideal" for index A with queries that are "ideal" for index B. -*/ -(function() { - "use strict" - - load('jstests/libs/analyze_plan.js'); // For getPlanStage(). - const coll = db.plan_cache_replanning; - - function getPlansForCacheEntry(query) { - let key = {query: query, sort: {}, projection: {}}; - let res = coll.runCommand("planCacheListPlans", key); - assert.commandWorked(res, `planCacheListPlans(${tojson(key)}) failed`); - assert(res.hasOwnProperty("plans"), - `plans missing from planCacheListPlans(${tojson(key)}) failed`); - - return res; - } - - function planHasIxScanStageForKey(planStats, keyPattern) { - const stage = getPlanStage(planStats, "IXSCAN"); - if (stage === null) { - return false; - } - - return bsonWoCompare(keyPattern, stage.keyPattern) == 0; - } - - const queryShape = {a: 1, b: 1}; - - // Carefully construct a collection so that some queries will do well with an {a: 1} index - // and others with a {b: 1} index. - for (let i = 1000; i < 1100; i++) { - assert.commandWorked(coll.insert({a: 1, b: i})); - } - - for (let i = 1000; i < 1100; i++) { - assert.commandWorked(coll.insert({a: i, b: 2})); - } - - // This query will be quick with {a: 1} index, and far slower {b: 1} index. With the {a: 1} - // index, the server should only need to examine one document. Using {b: 1}, it will have to - // scan through each document which has 2 as the value of the 'b' field. - const aIndexQuery = {a: 1099, b: 2}; - // Opposite of 'aIndexQuery'. Should be quick if the {b: 1} index is used, and slower if the - // {a: 1} index is used. - const bIndexQuery = {a: 1, b: 1099}; - - assert.commandWorked(coll.createIndex({a: 1})); - assert.commandWorked(coll.createIndex({b: 1})); - - // Run a query where the {b: 1} index will easily win. - assert.eq(1, coll.find(bIndexQuery).itcount()); - - // The plan cache should now hold an inactive entry. - let entry = getPlansForCacheEntry(queryShape); - let entryWorks = entry.works; - assert.eq(entry.isActive, false); - assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true); - - // Re-run the query. The inactive cache entry should be promoted to an active entry. - assert.eq(1, coll.find(bIndexQuery).itcount()); - entry = getPlansForCacheEntry(queryShape); - assert.eq(entry.isActive, true); - assert.eq(entry.works, entryWorks); - assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true); - - // Now we will attempt to oscillate the cache entry by interleaving queries which should use - // the {a:1} and {b:1} index. When the plan using the {b: 1} index is in the cache, running a - // query which should use the {a: 1} index will perform very poorly, and trigger - // replanning (and vice versa). - - // The {b: 1} plan is currently in the cache. Run the query which should use the {a: 1} - // index. The current cache entry will be deactivated, and then the cache entry for the {a: 1} - // will overwrite it (as active). - assert.eq(1, coll.find(aIndexQuery).itcount()); - entry = getPlansForCacheEntry(queryShape); - assert.eq(entry.isActive, true); - assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {a: 1}), true); - - // Run the query which should use the {b: 1} index. - assert.eq(1, coll.find(bIndexQuery).itcount()); - entry = getPlansForCacheEntry(queryShape); - assert.eq(entry.isActive, true); - assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true); - - // The {b: 1} plan is again in the cache. Run the query which should use the {a: 1} - // index. - assert.eq(1, coll.find(aIndexQuery).itcount()); - entry = getPlansForCacheEntry(queryShape); - assert.eq(entry.isActive, true); - assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {a: 1}), true); - - // The {a: 1} plan is back in the cache. Run the query which would perform better on the plan - // using the {b: 1} index, and ensure that plan gets written to the cache. - assert.eq(1, coll.find(bIndexQuery).itcount()); - entry = getPlansForCacheEntry(queryShape); - entryWorks = entry.works; - assert.eq(entry.isActive, true); - assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true); - - // Now run a plan that will perform poorly with both indices (it will be required to scan 500 - // documents). This will result in replanning (and the cache entry being deactivated). However, - // the new plan will have a very high works value, and will not replace the existing cache - // entry. It will only bump the existing cache entry's works value. - for (let i = 0; i < 500; i++) { - assert.commandWorked(coll.insert({a: 3, b: 3})); - } - assert.eq(500, coll.find({a: 3, b: 3}).itcount()); - - // The cache entry should have been deactivated. - entry = getPlansForCacheEntry(queryShape); - assert.eq(entry.isActive, false); - assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true); - - // The works value should have doubled. - assert.eq(entry.works, entryWorks * 2); -})(); |