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法律研究使用聚合的 MongoDB 查询来替换值？
使用聚合框架和 $literal 操作符。我们首先使用文档创建一个集合 -
> db.replaceValueDemo.insertOne(
{
_id : 100,
"EmployeeName" :"Chris",
"EmployeeOtherDetails": {
"EmployeeDesignation" : "HR",
"EmployeeAge":27
}
}
);
{ "acknowledged" : true, "insertedId" : 100 }
> db.replaceValueDemo.insertOne(
{
_id : 101,
"EmployeeName" :"David",
"EmployeeOtherDetails": {
"EmployeeDesignation" : "Tester",
"EmployeeAge":26
}
}
);
{ "acknowledged" : true, "insertedId" : 101 }以下是使用 find() 方法显示集合中所有文档的查询 -
> db.replaceValueDemo.find().pretty();
这将产生以下输出 -
{
"_id" : 100,
"EmployeeName" : "Chris",
"EmployeeOtherDetails" : {
"EmployeeDesignation" : "HR",
"EmployeeAge" : 27
}
}
{
"_id" : 101,
"EmployeeName" : "David",
"EmployeeOtherDetails" : {
"EmployeeDesignation" : "Tester",
"EmployeeAge" : 26
}
}以下是替换值的方法 -
> db.replaceValueDemo.aggregate([{
"$project": {
"_id": 1,
"EmployeeOtherDetails": {
EmployeeAge: 1,
EmployeeDesignation : { $literal: "Developer" }
}
}
}]);这将产生以下输出 -
{ "_id" : 100, "EmployeeOtherDetails" : { "EmployeeAge" : 27, "EmployeeDesignation" : "Developer" } }
{ "_id" : 101, "EmployeeOtherDetails" : { "EmployeeAge" : 26, "EmployeeDesignation" : "Developer" } }
更新于： 30-7-2019
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