Mongodb mongo DB非法语法
以下是我的数据示例:Mongodb mongo DB非法语法,mongodb,mongodb-query,Mongodb,Mongodb Query,以下是我的数据示例: {"CAND_NAME": "WARREN, ELIZABETH", "CAND_PTY_AFFILIATION": "DEM", "CAND_OFFICE_ST":"MA","CONNECTED_ORG_NM": null, "cmte_received": {"CAND_NAME": "WARREN, ELIZABETH", "CAND_PTY_AFFILIATION": "DEM", "CAND_OFFICE_ST":"MA","CONNECTED_ORG_NM":
{"CAND_NAME": "WARREN, ELIZABETH", "CAND_PTY_AFFILIATION": "DEM", "CAND_OFFICE_ST":"MA","CONNECTED_ORG_NM": null, "cmte_received":
{"CAND_NAME": "WARREN, ELIZABETH", "CAND_PTY_AFFILIATION": "DEM", "CAND_OFFICE_ST":"MA","CONNECTED_ORG_NM": null, "cmte_received":
我正在按交易金额查找收到的前五名cmte_
,以下是我的代码:
db.Contrib_Data.aggregate({ $group: {$“cmte_received”},
totalcontri : { $sum : $"TRANSACTION_AMT" } } ,{ $sort: {totalcontri: -1},{$limit: 5})
由于一个非法字符,它一直给我一个语法错误。您使用的是
$group
管道,没有\u id
,因此首先,您应该查看的官方文档,并且您没有传递值{$“cmte\u received”}
和$“TRANSACTION\u AMT”
格式正确,应为{“$cmte\u received”}
和“$TRANSACTION\u AMT”
您对适当管道的查询应如下所示:
db.Contrib_Data.aggregate(
{
$group:
{
_id: "$cmte_received",
"totalcontri" : { $sum : "$TRANSACTION_AMT" }
}
},
{
$sort:
{ "totalcontri": -1 }
},
{
$limit:5
}
)
我还建议您经历MongoDB的管道阶段Yep,
{$“cmte_received”}
和$“TRANSACTION_AMT”
不可用JS对象和字符串。如果您想按cmte\u received
分组,则可能可以将1st更正为{u id:$cmte\u received”}
。其次是“$TRANSACTION\u AMT”
,作为有效字符串。此外,在进行聚合时,文档通常对我很有帮助: