问题描述
在我的mongodb集合中，我有一个time_stamp = "2013-06-30 23:58:37 928" .我只需要在日期上使用" $ match "，例如time_stamp =" 2013-06-30 ".那么如何获得这样的子字符串呢?
In my mongodb collection, I have a time_stamp="2013-06-30 23:58:37 928".I need to use "$match" with only the date, like time_stamp="2013-06-30". So how can I get the substring like that ?
以前我尝试过使用$ substr，但是它显示错误"errmsg":异常:无效的运算符:$ substr"
Previously I've tried with $substr, but it shows an error "errmsg" : "exception: invalid operator: $substr"
推荐答案
我认为您尝试使用聚合框架进行查询，因为您尝试了& $ substr 运算符.我创建了一个简单的示例，展示了如何使用 $ substr 来获得所需的结果在聚合框架上.
I think you are trying to make query using aggregation framework since you tried $match & $substr operators. I have created a simple example to show how you can use $substr to achive result you wanted on aggregation framework.
我已将以下数据插入MongoDB.
I have inserted following data into the MongoDB.
{ "_id" : ObjectId("528b343881d4fe2cfe0b1b25"), "time_stamp" : "2013-06-30 23:58:37 928" }
{ "_id" : ObjectId("528b343b81d4fe2cfe0b1b26"), "time_stamp" : "2013-06-30 23:58:37 928" }
{ "_id" : ObjectId("528b344c81d4fe2cfe0b1b27"), "time_stamp" : "2013-06-30 12:58:37 928" }
{ "_id" : ObjectId("528b344f81d4fe2cfe0b1b28"), "time_stamp" : "2013-06-30 12:58:23 928" }
{ "_id" : ObjectId("528b345381d4fe2cfe0b1b29"), "time_stamp" : "2013-06-31 12:58:23 928" }
{ "_id" : ObjectId("528b345981d4fe2cfe0b1b2a"), "time_stamp" : "2013-07-31 12:58:23 933" }
我编写了以下代码，使用 $ substr 运算符按日期进行分组.
I wrote following code to group by date by using $substr operator.
db.myObject.aggregate(
{$project : {new_time_stamp : {$substr : ["$time_stamp",0, 10]}}},
{$group:{_id:"$new_time_stamp", "count": {$sum:1}}}
);
这篇关于与mongodb聚合中的子字符串匹配的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！