如何在Mongo中查询这个? (不为null) [英] How do you query this in Mongo? (is not null)
问题描述
db.mycollection.find(HAS IMAGE URL)
这会返回带有IMAGE URL键的所有文档,但它们可能仍然为空值。
db.mycollection.find({IMAGE URL:{$ exists:true}});
这将返回所有包含名为IMAGE URL >非空值。
db.mycollection.find({IMAGE URL:{$ ne:null}} );
此外,根据docs,$ exists目前不能使用索引, 。
编辑:由于对此答案感兴趣而添加一些示例这些插入:
db.test.insert({num:1,check:check value}) ;
db.test.insert({num:2,check:null});
db.test.insert({num:3});
这将返回所有三个文档:
db.test.find();
这将只返回第一个和第二个文档:
db.test.find({check:{$ exists:true}});
这将只返回第一个文档:
db.test.find({check:{$ ne:null}});
这将仅返回第二个和第三个文档:
db.test.find({check:null})
db.mycollection.find(HAS IMAGE URL)
This will return all documents with a key called "IMAGE URL", but they may still have a null value.
db.mycollection.find({"IMAGE URL":{$exists:true}});
This will return all documents with both a key called "IMAGE URL" and a non-null value.
db.mycollection.find({"IMAGE URL":{$ne:null}});
Also, according to the docs, $exists currently can't use an index, but $ne can.
Edit: Adding some examples due to interest in this answer
Given these inserts:
db.test.insert({"num":1, "check":"check value"});
db.test.insert({"num":2, "check":null});
db.test.insert({"num":3});
This will return all three documents:
db.test.find();
This will return the first and second documents only:
db.test.find({"check":{$exists:true}});
This will return the first document only:
db.test.find({"check":{$ne:null}});
This will return the second and third documents only:
db.test.find({"check":null})
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