Question

我的收藏集具有如下文档结构：

{ AppName: 'test',
AppEnv: 'Development'
AppTech:'Java'
AppVersion: '1_0',

AppName: 'test',
AppEnv: 'SQA'
AppTech:'Java'
AppVersion: '1_0',

AppName: 'test22',
AppEnv: 'Development'
AppTech:'Java'
AppVersion: '1_2',

AppName: 'test22',
AppEnv: 'SQA'
AppTech:'Java'
AppVersion: '1_1',

AppName: 'test22',
AppEnv: 'Production'
AppTech:'Java'
AppVersion: '1_0',

AppName: 'test2',
AppEnv: 'Development'
AppTech:'.NET'
AppVersion: '1_0'

}

我需要基于提供的AppName / AppEnv / AppTech获取AppVersion。例如：

基于所有AppTech为“ Java”的AppEnv的AppVersion（带有一些标志）。结果可能如下所示：

{
AppName: 'test', 
AppTech: 'Java',
VersionInfo: [AppEnv: {'Development','SQA'}, AppVersion: {'1_0','1_1'}],
hasVersionDiff: 'false',

AppName: 'test22', 
AppTech: 'Java',
VersionInfo: [AppEnv: {'Development','SQA','Production}, AppVersion: {'1_2','1_1','1_0'}],
hasVersionDiff: 'true'

}

有人可以建议我如何开始为此编写查询吗？

Answer 1

您可以尝试这个

let query = {AppTech: 'Java'};
let result = await Collection.aggregate([
{ $match: query },
{ $group: { 
    _id: '$AppName', 
    AppTech: {$first: '$AppTech' },
    AppEnv: {$push: '$AppEnv'},
     AppVersion: {$push: '$AppVersion'}} 
},
{
    $project: {
        AppName: '$_id',
        AppTech: '$AppTech',
        VersionInfo: { AppEnv: '$AppEnv', AppVersion: '$AppVersion' }
    }
}

]）

我更改了输出格式，因为无法实现您期望的格式。而且我没有包含hasVersionDiff字段，您可以稍后再添加。

 {
    AppName: 'test', 
    AppTech: 'Java',
    VersionInfo: {
        AppEnv: ['Development','SQA'], 
        AppVersion: ['1_0','1_1']
    },
},
{
    AppName: 'test22', 
    AppTech: 'Java',
    VersionInfo: {
        AppEnv: ['Development','SQA','Production'],
        AppVersion: ['1_2','1_1','1_0']
    }
}

Answer 2

您可以使用MongoTemplate。

假设您的课程是这样的：

public class OutPutClass {
    String appName;
    String appTech;
    VersionInfo versionInfo;

    public class VersionInfo{

        List<String> appEnvs;
        List<String> appVersions;
    }
}

和mongoTemplate查询：

final Criteria criteria =
            Criteria.where("appTech").is("Java");
    Aggregation aggregation = Aggregation.newAggregation(
            Aggregation.match(criteria),
            Aggregation.group("AppName")
                    .first("AppName")
                    .as("appName")
                    .first("AppTech")
                    .as("appTech")
                    .push("AppEnv").as("versionInfo.appEnvs")
                    .push("AppVersion").as("versionInfo.appVersions")
    );

mongoTemplate.aggregate(
                        aggregation,
                        DocumentClass.class,
                        OutputClass.class
                );

Mongo聚集条件$ match，$ group和$ project

2 个答案: