I have a linq to sql query and I am pulling data from it fine but unsure how to add a group or count.
Linq Code
public IEnumerable<COMPLAINT> GetAllCount()
{
try
{
return from m in _context.Comp
join p in _context.Checklists on m.ID equals p.C_ID
orderby m.Received_DT descending
select m;
}
catch (Exception ex)
{
_logger.LogError("Could not get complaint", ex);
return null;
}
}
results
[
{
"comP_ID": 909,
"received_DT": "2018-01-22T00:00:00",
"Outcome": "Unsatisfactory",
"RecFinding": "Decline"
},
{
"comP_ID": 909,
"received_DT": "2018-01-22T00:00:00",
"Outcome": "Unsatisfactory",
"RecFinding": "Sustained"
},
{
"comP_ID": 909,
"received_DT": "2018-01-22T00:00:00",
"Outcome": "satisfactory",
"RecFinding": "Decline"
},
{
"comP_ID": 909,
"received_DT": "2018-01-22T00:00:00",
"Outcome": "satisfactory",
"RecFinding": "Sustained"
},
]
Question
I am fairly new to the LINQ To SQL and also writing Fluent API's to return the data but I am unsure how do I get the return of only one variable grouped by results or and count to look like this:
[
{
"Decline": 2,
"Sustained": 2,
},
]
Or
[
{
"Unsatisfactory": 2,
"satisfactory": 2,
},
]
I have tried a few group into g then select new methods, but I am really confused on how to incorporate it into the linq code. Thank you.
Related
I have a column in my database that stores a json string listing the weapons used in each game activity, what I need to be able to do is return the 'values'->'uniqueWeaponKills'->'basic'->'value' when the 'referenceId' key = 1994645182, and 0 if the key,value pair is not in the column.
Example 'weapons' column data
{
"weapons": [
{
"values": {
"uniqueWeaponKills": {
"basic": {
"value": 14,
"displayValue": "14"
}
},
"uniqueWeaponPrecisionKills": {
"basic": {
"value": 0,
"displayValue": "0"
}
},
"uniqueWeaponKillsPrecisionKills": {
"basic": {
"value": 0,
"displayValue": "0%"
}
}
},
"referenceId": 1994645182
},
{
"values": {
"uniqueWeaponKills": {
"basic": {
"value": 2,
"displayValue": "2"
}
},
"uniqueWeaponPrecisionKills": {
"basic": {
"value": 1,
"displayValue": "1"
}
},
"uniqueWeaponKillsPrecisionKills": {
"basic": {
"value": 0.5,
"displayValue": "50%"
}
}
},
"referenceId": 1853180924
}
]
}
Edit 1:
Using the suggestion from Kendle I got to the following query, I haven't seen a way to dynamically look in each of the array elements instead of having to specify the one to look at.
Query
select weapons::json->'weapons'->1->'values'->'uniqueWeaponKills'->'basic'->>'value' as "uniqueWeaponKills",
weapons::json->'weapons'->1->'referenceId' as "weaponId"
from activities
where (weapons::json->'weapons'->1->>'referenceId')::BIGINT = 1687353095;
You could try
SELECT
weapons::json->'values'->> 'uniqueWeaponKills'->>'basic' ->>'value'
FROM table_name
WHERE
weapons::json->'referenceId' = 1994645182;
See also How to parse JSON in postgresql
I think I found the solution I am looking for using json_array_elements()
SELECT obj->'values'->'uniqueWeaponKills'->'basic'->>'value' as "uniqueWeaponKills"
FROM activities a, json_array_elements(a.weapons#>'{weapons}') obj
WHERE (obj->>'referenceId')::BIGINT = 1687353095;
SELECT prop_type , count(prop_type) As no_of_properties
from prop_type
JOIN prop_for_rent USING (prop_type)
GROUP BY prop_type;
prop_type and prop_for_rent are both tables. Then "prop_type" used for the JOIN and GROUP BY is an attribute name in both.
The expected result is for the query to count the number of properties in each property type.
And this is the result of the sql version of it
Screenshot of sql query result
So i need it to display the same prop_type and then the number of properties in each type in mongodb
db.prop_type.aggregate([
{
"$lookup": {
"from": "prop_for_rent",
"localField": "prop_type",
"foreignField": "prop_type",
"as": "prop_for_rent_docs"
}
},
{
"$unwind": {
path: "$prop_for_rent_docs",
preserveNullAndEmptyArrays: true
}
},
{
"$group": {
"_id": "$prop_type",
"no_of_properties": {
"$sum": 1
}
}
}
])
mongoplayground
I am new to MongoDB. I need to convert this SQL code to MongoDB
select TOP 5 r.regionName, COUNT(c.RegionID)
from region as r,
company as c
where c.RegionID = r._id
group by r.regionName
order by COUNT(c.RegionID) DESC;
Option 1. You can use the aggregation framework with $lookup, $group, $project , $sort and $limit stages, but this seems like a wrong approach since the true power to change relation database with mongoDB is the denormalization and avoidance of join ($lookup) like queries.
Option 2. You convert your multi-table relational database schema to document model and proceed with simple $group, $project, $sort and $limit stage aggregation query for the above task.
Since you have not provided any mongodb document examples it is hard to provide how your queries will look like ...
Despite of my comment I try to give a translation (not tested):
db.region.aggregate([
{
$lookup: // left outer join collections
{
from: "company",
localField: "_id",
foreignField: "RegionID",
as: "c"
}
},
{ $match: { c: { $ne: [] } } }, // remove non-matching documents (i.e. INNER JOIN)
{ $group: { _id: "$regionName", regions: { $addToSet: { "$c.RegionID" } } } }, // group and get distinct regions
{ $project: { regionName: "$_id", count: { $size: "$regions" } , _id: 0} } // some cosmetic and count
{ $sort: { regionName: 1 } }, // order result
{ $limit: 5 } // limit number or returned documents
])
I have the following query to be converted to DSL and executed on ES. I could not find a suitable aggregation and filter over results of aggregation available out-of-the-box in ES. As an alternative, I am fetching the 'group by count' for each id from ES and filtering the result as a part of my application logic, which is not efficient. Can you suggest any more suitable solution?
select distinct id from index where colA = "something" group by id having count(*) > 10;
index mapping
id : (string)
colA: (string)
Terms aggregation: to get distinct Ids.
Bucket selector: to return ids with doc count more than 10
{
"query": {
"bool": {
"filter": [
{
"term": {
"colA.keyword": "something" --> where clause
}
}
]
}
},
"aggs": {
"distinct_id": {
"terms": { --> group by
"field": "id.keyword",
"size": 10
},
"aggs": {
"ids_having_count_morethan_10": {
"bucket_selector": { --> having
"buckets_path": {
"count": "_count"
},
"script": "params.count>10"
}
}
}
}
}
}
Goal:
This sql and its result should be the same result from mongoDB's query code.
In order words, same result but for mongoDB.
Problem:
How to you make a lookup connection in relation to People and Role in Mongo DB's query code?
Info:
I'm new in mongo DB
SQL code
SELECT
a.*,
'.' AS '.',
b.*,
'.' AS '.',
c.*
FROM
[db1].[dbo].[People_Course_Grade] a
INNER JOIN [db1].[dbo].[People] b on a.PeopleId = b.PeopleId
INNER JOIN [db1].[dbo].[Role] c on b.RoleId = c.RoleId
Json data:
Role:
[{"RoleId":1,"Name":"Student"},{"RoleId":2,"Name":"Teacher"}]
People_Course_Grade:
[{"People_Course_GradeId":1,"PeopleId":1,"CourseId":1},
{"People_Course_GradeId":2,"PeopleId":2,"CourseId":1},
{"People_Course_GradeId":3,"PeopleId":3,"CourseId":2},
{"People_Course_GradeId":4,"PeopleId":1,"CourseId":2}]
Course:
[{"CourseId":1,"Name":"Java"},{"CourseId":2,"Name":"Java II"},
{"CourseId":3,"Name":"Statistik 1"}]
db.People_Course_Grade.aggregate([
{
$lookup:{
from: "People",
localField: "people_id",
foreignField: "_id",
as: "people"
}
},
{ $unwind:"$people" },
{
$project:{
course_id : 1,
people_id : 1,
// grade_id : 1,
Name : "$people.Name",
}
}
]);
You need to start with double $lookup since you have three collections. Then you can use $arrayElemAt to always get single element from lookup's result. To flatten your structure you can use $replaceRoot with $mergeObjects (promotes all the fields from people and course to root level.
db.People_Course_Grade.aggregate([
{
$lookup:{
from: "Role",
localField: "PeopleId",
foreignField: "RoleId",
as: "people"
}
},
{
$lookup:{
from: "Course",
localField: "CourseId",
foreignField: "CourseId",
as: "course"
}
},
{
$replaceRoot: {
newRoot: {
$mergeObjects: [
"$$ROOT",
{ $arrayElemAt: [ "$people", 0 ] },
{ $arrayElemAt: [ "$course", 0 ] },
]
}
}
},
{
$project: {
people: 0,
course: 0
}
}
])
Mongo Playground
$arrayElemAt can always be replaced with $unwind like you tried. You also have a naming conflict on name field so probably you need to run $project to rename one of those fields - otherwise you'll get only one of them in final result.