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mongodb - How do I check if an index is being used

I have a mongodb replica set with a lot of databases, collections & indexes.

We did a lot of refactor and optimization and, of course, I have a lot of "creative queries" from the consumers.

I would like to clean up the unused indexes. just wanna save some space.

How can I check if an index is being used? I can afford to check index by index and drop the unused ones.

Running an "explain" in all the possible queries is not an option :)

EDIT: SOLUTION BASED ON THE ACCEPTED ANSWER

The script was bugged. I am not a javascript expert, but I put the corrected script. I hope will be useful for someone:

DB.prototype.indexStats = function() {
  var queries = [];
  var collections = db.getCollectionNames();

  var findQuery = function(q) {
    for(entryIdx in queries) {
      if(q == queries[entryIdx].query) {
        return entryIdx;
      }
    }
    return -1;
  }

  for(cIdx in collections) {
    var cName = collections[cIdx];
    var nsName = db.getName()+"."+cName;
    if(cName.indexOf("system") == -1) {
      var i = 1;
      var count = db.system.profile.count({ns:nsName});
      print('scanning profile {ns:"'+nsName+'"} with '+count+' records... this could take a while...');
      db.system.profile.find({ns:nsName}).addOption(16).batchSize(10000).forEach(function(profileDoc) {           
        if(profileDoc.query && !profileDoc.query["$explain"]) { 
          var qIdx = findQuery(profileDoc.query);
          if(qIdx == -1 && profileDoc.query["query"] ) {
            var size = queries.push({query:profileDoc.query, count:1, index:""});                   
            var explain = db[cName].find(queries[size-1].query).explain();
            if(profileDoc.query && profileDoc.query["query"]) {
              queries[size-1].sort = profileDoc.query["orderby"];
              if(queries[size-1].sort) {
                explain = db[cName].find(queries[size-1].query.query).sort(queries[size-1].sort).explain();
              }
            }
            queries[size-1].cursor = explain.cursor;
            queries[size-1].millis = explain.millis;
            queries[size-1].nscanned = explain.nscanned;
            queries[size-1].n = explain.n;
            queries[size-1].scanAndOrder = explain.scanAndOrder ? true : false;
            if(explain.cursor && explain.cursor != "BasicCursor") {
              queries[size-1].index = explain.cursor.split(" ")[1];             
            } else {
              print('warning, no index for query {ns:"'+nsName+'"}: ');
              printjson(profileDoc.query);
              print('... millis: ' + queries[size-1].millis);
              print('... nscanned/n: ' + queries[size-1].nscanned + '/' + queries[size-1].n);
              print('... scanAndOrder: ' + queries[size-1].scanAndOrder);
            }
          } else if ( qIdx != -1 ) {
            queries[qIdx].count++;
          }
        }
      });
    }
  }

  for(cIdx in collections) {
    var cName = collections[cIdx];
    if(cName.indexOf("system") == -1) {
      print('checking for unused indexes in: ' + cName);
      for(iIdx in db[cName].getIndexes()) {
        var iName = db[cName].getIndexes()[iIdx].name;
        if(iName.indexOf("system") == -1) {
          var stats = db[cName].stats();
          var found = false;
          for(qIdx in queries) {
            if(queries[qIdx].index == iName) {
              found = true;
              break;
            }
          }
          if(!found) {
            print('this index is not being used: ');
            printjson(iName);
          }
        }
      }
    }
  }
}
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1 Reply

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by (71.8m points)

The simplest solution to this is to use the mongodb inbuilt $indexStats

Using the Mongo console run -

db.collection.aggregate([ { $indexStats: { } } ])

Using PyMongo -

from pymongo import MongoClient
collection = MongoClient()[db_name][collection_name]
index_stats = collection.aggregate([{'$indexStats':{}}])

for index_info in index_stats:
    print index_info

Apologies for re-opening an old question. This shows up on the first page of google searches and the only answer is to use a snippet of unmaintained code.


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