查询布尔字段为"not true"（例如false或不存在）

Question

我确信我在MongoDB查询中遗漏了一些非常基本的东西，似乎无法得到这个简单的条件。

考虑这个集合

> db.tests.find()
{ "_id" : ObjectId("..."), "name" : "Test1" , "deleted" : true}
{ "_id" : ObjectId("..."), "name" : "Test2" , "deleted" : false}
{ "_id" : ObjectId("..."), "name" : "Test3" }

我只想查询所有“未删除”的项目

我知道如何查找“已删除”标志设置为true的项目：

> db.tests.find({deleted:true})
{ "_id" : ObjectId("..."), "name" : "Test1" , "deleted" : true}

但是如何查找所有未被“删除”（例如，否定上述查询，或者换句话说，任何没有“已删除”字段的项目，或者具有值false

我通过猜测尝试了什么（请不要笑...）

> db.tests.find({$not : {deleted: true}})

（不返回任何结果）

> db.tests.find({$not : {$eq:{deleted:true}}})

错误：{“$err”：“无效运算符：$eq”，“代码”：10068}

> db.tests.find({deleted:{$not: true}})

错误：{“$err”：“无效使用$not”，“代码”：13041}

> db.tests.find({deleted:{$not: {$eq:true}}})

错误：{“$err”：“无效使用$not”，“代码”：13034}

我错过了什么？

Answer 1

强尼克有最好的答案。选择器中的$是最短、最干净的IMO。

这将测试“假”或“不存在”。并且可以被索引。

db.tests.find({$or:[{deleted:false},{deleted:{$exists:false}}]})

使用索引的示例。

((function(){
    print("creating collection 'testx' and inserting 50 trues, 50 falses, 50 non-existents");
    db.testx.drop();
    db.testx.ensureIndex({deleted:1});
    for (var i=0;i<50;i++){
        db.testx.insert({i:i,deleted:false});
    };
    for (var i=0;i<50;i++){
        db.testx.insert({i:i,deleted:true});
    };
    for (var i=0;i<50;i++){
        db.testx.insert({i:i});
    };
    var res0 = db.testx.find().explain();
    var res1 = db.testx.find({deleted:false}).explain();
    var res2 = db.testx.find({deleted:true}).explain();
    var res3 = db.testx.find({deleted:{$exists:false}}).explain();
    var res4 = db.testx.find({$or:[{deleted:false},{deleted:{$exists:false}}]}).explain();
    var res5 = db.testx.find({$or:[{deleted:true},{deleted:{$exists:false}}]}).explain();
    var res6 = db.testx.find({deleted:{$in:[false,null]}}).explain();
    print("res0: all objects                      ("+res0["n"]+" found, "+res0["nscannedObjects"]+" scanned)");
    print("res1: deleted is false                 ("+res1["n"]+" found, "+res1["nscannedObjects"]+" scanned)");
    print("res2: deleted is true                  ("+res2["n"]+" found, "+res2["nscannedObjects"]+" scanned)");
    print("res3: deleted is non-existent          ("+res3["n"]+" found, "+res3["nscannedObjects"]+" scanned)");
    print("res4: deleted is false or non-existent ("+res4["n"]+" found, "+res4["nscannedObjects"]+" scanned)");
    print("res5: deleted is true or non-existent  ("+res5["n"]+" found, "+res5["nscannedObjects"]+" scanned)");
    print("res6: deleted is in [false,null]       ("+res5["n"]+" found, "+res5["nscannedObjects"]+" scanned)");
})())

这应该打印出来

creating collection 'testx' and inserting 50 trues, 50 falses, 50 non-existents
res0: all objects                      (150 found, 150 scanned)
res1: deleted is false                 (50 found, 50 scanned)
res2: deleted is true                  (50 found, 50 scanned)
res3: deleted is non-existent          (50 found, 50 scanned)
res4: deleted is false or non-existent (100 found, 100 scanned)
res5: deleted is true or non-existent  (100 found, 100 scanned)
res6: deleted is in [false,null]       (100 found, 100 scanned)

Answer 2

为完整起见，另一种方法是使用$in：

js prettyprint-override">db.test.find({deleted: {$in: [null, false]}})

将null包含在数组中会拉入缺少deleted字段的文档。此查询可以使用当前2.6版本中{deleted:1}上的索引。6 MongoDB发布。

Answer 3

db.tests.find({deleted: {$ne: true}})

其中，$ne表示“不相等”。（关于mongodb运营商的文档）