Spring/Java Mongo query: How to add values of a field from multiple documents as an array while grouping/projectingHow to remove a field completely from a MongoDB document?mongodb group values by multiple fieldsHow to merge array field in document in Mongo aggregationSpring data mongo db query on inner fields of embedded document (DBRef)Mongo Query On Multiple Fields Of Sub-DocumentGroup by array of document in Spring Mongo DbMongo aggregate group by multiple valuesmongo group and project for nested documentsProject as nested document in spring mongoQuery on Embedded Array Of Document in mongo based on multiple fields in spring
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Spring/Java Mongo query: How to add values of a field from multiple documents as an array while grouping/projecting
How to remove a field completely from a MongoDB document?mongodb group values by multiple fieldsHow to merge array field in document in Mongo aggregationSpring data mongo db query on inner fields of embedded document (DBRef)Mongo Query On Multiple Fields Of Sub-DocumentGroup by array of document in Spring Mongo DbMongo aggregate group by multiple valuesmongo group and project for nested documentsProject as nested document in spring mongoQuery on Embedded Array Of Document in mongo based on multiple fields in spring
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
My records look like below:
domain: 'ABC',
technology: 'Java',
appName: 'app1',
environment: 'Dev',
version : '1_1',
domain: 'ABC',
technology: 'Java',
appName: 'app1',
environment: 'SQA',
version : '1_1',
domain: 'ABC',
technology: 'Java',
appName: 'app2',
environment: 'Dev',
version : '1_1',
domain: 'DEF',
technology: '.NET',
appName: 'app3',
environment: 'SQA',
version: '1_3'
Final result is to get the list of apps's environment and version grouped by domain and tech(1 to 1 mapping between these two). Environment/Domain/Tech will be input. if that environment does not exist, then I still want to show on result, with version as N/A.
[
"domain": "ABC",
"technology": "Java",
"apps": [
"appName": "app1",
"environment": ["Dev","SQA"],
"version": ["1_1", "1_1"],
"hasVersionDiff": false
,
"appName": "app2",
"environment": ["Dev","SQA"],
"version": ["1_1", "N/A"],
"hasVersionDiff": true ]
,
"domain": "DEF",
"technology": ".NET",
"apps": [
"appName": "app3",
"environment": ["Dev","SQA"],
"version": ["N/A", "1_1"],
"hasVersionDiff": true
]
]
So far I have tried below:
Aggregation agg = Aggregation.newAggregation(
Aggregation.match(cr),
Aggregation.group("domain").first("technology").as("technology")
.addToSet(new BasicDBObject()
put("technology", "$technology");
put("domain", "$domain");
put("environment","$environment");
put("version", "$version");
put("appName", "$appName");
).as("apps"));
But this is giving me result as below, where for each app it is showing me two diff record for each env, also I am not able figure out how to apply hasVersionDiff value for them:
[
"domain": "ABC",
"technology": "Java",
"apps": [
"appName": "app1",
"environment": ["DEV"],
"version": ["1_1"],
"hasVersionDiff": false
,
"appName": "app1",
"environment": ["SQA"],
"version": ["1_1"],
"hasVersionDiff": false
]
etc..
My model class is as below:
CompareResult has domain, technology and ArrayList apps
App has appName, List of environments, list of versions and hasVersionDiff boolean.
mongodb mongodb-query aggregation-framework spring-data-mongodb
asked Mar 25 at 14:39
user2262292user2262292
297 bronze badges
add a comment |
My records look like below:
domain: 'ABC',
technology: 'Java',
appName: 'app1',
environment: 'Dev',
version : '1_1',
domain: 'ABC',
technology: 'Java',
appName: 'app1',
environment: 'SQA',
version : '1_1',
domain: 'ABC',
technology: 'Java',
appName: 'app2',
environment: 'Dev',
version : '1_1',
domain: 'DEF',
technology: '.NET',
appName: 'app3',
environment: 'SQA',
version: '1_3'
Final result is to get the list of apps's environment and version grouped by domain and tech(1 to 1 mapping between these two). Environment/Domain/Tech will be input. if that environment does not exist, then I still want to show on result, with version as N/A.
[
"domain": "ABC",
"technology": "Java",
"apps": [
"appName": "app1",
"environment": ["Dev","SQA"],
"version": ["1_1", "1_1"],
"hasVersionDiff": false
,
"appName": "app2",
"environment": ["Dev","SQA"],
"version": ["1_1", "N/A"],
"hasVersionDiff": true ]
,
"domain": "DEF",
"technology": ".NET",
"apps": [
"appName": "app3",
"environment": ["Dev","SQA"],
"version": ["N/A", "1_1"],
"hasVersionDiff": true
]
]
So far I have tried below:
Aggregation agg = Aggregation.newAggregation(
Aggregation.match(cr),
Aggregation.group("domain").first("technology").as("technology")
.addToSet(new BasicDBObject()
put("technology", "$technology");
put("domain", "$domain");
put("environment","$environment");
put("version", "$version");
put("appName", "$appName");
).as("apps"));
But this is giving me result as below, where for each app it is showing me two diff record for each env, also I am not able figure out how to apply hasVersionDiff value for them:
[
"domain": "ABC",
"technology": "Java",
"apps": [
"appName": "app1",
"environment": ["DEV"],
"version": ["1_1"],
"hasVersionDiff": false
,
"appName": "app1",
"environment": ["SQA"],
"version": ["1_1"],
"hasVersionDiff": false
]
etc..
My model class is as below:
CompareResult has domain, technology and ArrayList apps
App has appName, List of environments, list of versions and hasVersionDiff boolean.
mongodb mongodb-query aggregation-framework spring-data-mongodb
asked Mar 25 at 14:39
user2262292user2262292
297 bronze badges
add a comment |
My records look like below:
domain: 'ABC',
technology: 'Java',
appName: 'app1',
environment: 'Dev',
version : '1_1',
domain: 'ABC',
technology: 'Java',
appName: 'app1',
environment: 'SQA',
version : '1_1',
domain: 'ABC',
technology: 'Java',
appName: 'app2',
environment: 'Dev',
version : '1_1',
domain: 'DEF',
technology: '.NET',
appName: 'app3',
environment: 'SQA',
version: '1_3'
Final result is to get the list of apps's environment and version grouped by domain and tech(1 to 1 mapping between these two). Environment/Domain/Tech will be input. if that environment does not exist, then I still want to show on result, with version as N/A.
[
"domain": "ABC",
"technology": "Java",
"apps": [
"appName": "app1",
"environment": ["Dev","SQA"],
"version": ["1_1", "1_1"],
"hasVersionDiff": false
,
"appName": "app2",
"environment": ["Dev","SQA"],
"version": ["1_1", "N/A"],
"hasVersionDiff": true ]
,
"domain": "DEF",
"technology": ".NET",
"apps": [
"appName": "app3",
"environment": ["Dev","SQA"],
"version": ["N/A", "1_1"],
"hasVersionDiff": true
]
]
So far I have tried below:
Aggregation agg = Aggregation.newAggregation(
Aggregation.match(cr),
Aggregation.group("domain").first("technology").as("technology")
.addToSet(new BasicDBObject()
put("technology", "$technology");
put("domain", "$domain");
put("environment","$environment");
put("version", "$version");
put("appName", "$appName");
).as("apps"));
But this is giving me result as below, where for each app it is showing me two diff record for each env, also I am not able figure out how to apply hasVersionDiff value for them:
[
"domain": "ABC",
"technology": "Java",
"apps": [
"appName": "app1",
"environment": ["DEV"],
"version": ["1_1"],
"hasVersionDiff": false
,
"appName": "app1",
"environment": ["SQA"],
"version": ["1_1"],
"hasVersionDiff": false
]
etc..
My model class is as below:
CompareResult has domain, technology and ArrayList apps
App has appName, List of environments, list of versions and hasVersionDiff boolean.
mongodb mongodb-query aggregation-framework spring-data-mongodb
asked Mar 25 at 14:39
user2262292user2262292
297 bronze badges
My records look like below:
domain: 'ABC',
technology: 'Java',
appName: 'app1',
environment: 'Dev',
version : '1_1',
domain: 'ABC',
technology: 'Java',
appName: 'app1',
environment: 'SQA',
version : '1_1',
domain: 'ABC',
technology: 'Java',
appName: 'app2',
environment: 'Dev',
version : '1_1',
domain: 'DEF',
technology: '.NET',
appName: 'app3',
environment: 'SQA',
version: '1_3'
Final result is to get the list of apps's environment and version grouped by domain and tech(1 to 1 mapping between these two). Environment/Domain/Tech will be input. if that environment does not exist, then I still want to show on result, with version as N/A.
[
"domain": "ABC",
"technology": "Java",
"apps": [
"appName": "app1",
"environment": ["Dev","SQA"],
"version": ["1_1", "1_1"],
"hasVersionDiff": false
,
"appName": "app2",
"environment": ["Dev","SQA"],
"version": ["1_1", "N/A"],
"hasVersionDiff": true ]
,
"domain": "DEF",
"technology": ".NET",
"apps": [
"appName": "app3",
"environment": ["Dev","SQA"],
"version": ["N/A", "1_1"],
"hasVersionDiff": true
]
]
So far I have tried below:
Aggregation agg = Aggregation.newAggregation(
Aggregation.match(cr),
Aggregation.group("domain").first("technology").as("technology")
.addToSet(new BasicDBObject()
put("technology", "$technology");
put("domain", "$domain");
put("environment","$environment");
put("version", "$version");
put("appName", "$appName");
).as("apps"));
But this is giving me result as below, where for each app it is showing me two diff record for each env, also I am not able figure out how to apply hasVersionDiff value for them:
[
"domain": "ABC",
"technology": "Java",
"apps": [
"appName": "app1",
"environment": ["DEV"],
"version": ["1_1"],
"hasVersionDiff": false
,
"appName": "app1",
"environment": ["SQA"],
"version": ["1_1"],
"hasVersionDiff": false
]
etc..
My model class is as below:
CompareResult has domain, technology and ArrayList apps
App has appName, List of environments, list of versions and hasVersionDiff boolean.
mongodb mongodb-query aggregation-framework spring-data-mongodb
mongodb mongodb-query aggregation-framework spring-data-mongodb
asked Mar 25 at 14:39
user2262292user2262292
297 bronze badges
asked Mar 25 at 14:39
user2262292user2262292
297 bronze badges
asked Mar 25 at 14:39
user2262292user2262292
297 bronze badges
asked Mar 25 at 14:39
user2262292user2262292
297 bronze badges
asked Mar 25 at 14:39
user2262292user2262292
297 bronze badges
297 bronze badges
add a comment |
add a comment |
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