From MySQL to Mongoose: aggregate document with maximum multiplication of two values in all rows
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1
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For given table:
id | val_a | val_b | mult | category
------------------------------------
0 | 1 | 2 | 9 | A
1 | 0 | 3 | 5 | B
2 | 0.5 | 4 | 6 | C
following query in MySQL
SELECT * FROM table WHERE category IN ('A','B','C')
AND IF( val_a > 0, val_a, val_b ) * mult = (SELECT MAX( IF( val_a > 0, val_a, val_b ) * mult ) FROM table WHERE category IN ('A','B','C')) LIMIT 1;
will find multiplications of val_a * mult for every row (if val_a == 0 : val_b * mult)
0) 1*9 = 9
1) 3*5 = 15
2) 0.5*4 = 2
and then find the maximum of all answers, and finally return row with ID:1
1) 15
How do I aggregate the same in Mongoose? Even partial answer will help to move forward and to learn.
Tried to start with http://www.querymongo.com/ but without success.
mongodb mongoose mongodb-query
asked Nov 10 at 19:32
Pumych
5921819
add a comment |
up vote
1
down vote
favorite
For given table:
id | val_a | val_b | mult | category
------------------------------------
0 | 1 | 2 | 9 | A
1 | 0 | 3 | 5 | B
2 | 0.5 | 4 | 6 | C
following query in MySQL
SELECT * FROM table WHERE category IN ('A','B','C')
AND IF( val_a > 0, val_a, val_b ) * mult = (SELECT MAX( IF( val_a > 0, val_a, val_b ) * mult ) FROM table WHERE category IN ('A','B','C')) LIMIT 1;
will find multiplications of val_a * mult for every row (if val_a == 0 : val_b * mult)
0) 1*9 = 9
1) 3*5 = 15
2) 0.5*4 = 2
and then find the maximum of all answers, and finally return row with ID:1
1) 15
How do I aggregate the same in Mongoose? Even partial answer will help to move forward and to learn.
Tried to start with http://www.querymongo.com/ but without success.
mongodb mongoose mongodb-query
asked Nov 10 at 19:32
Pumych
5921819
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For given table:
id | val_a | val_b | mult | category
------------------------------------
0 | 1 | 2 | 9 | A
1 | 0 | 3 | 5 | B
2 | 0.5 | 4 | 6 | C
following query in MySQL
SELECT * FROM table WHERE category IN ('A','B','C')
AND IF( val_a > 0, val_a, val_b ) * mult = (SELECT MAX( IF( val_a > 0, val_a, val_b ) * mult ) FROM table WHERE category IN ('A','B','C')) LIMIT 1;
will find multiplications of val_a * mult for every row (if val_a == 0 : val_b * mult)
0) 1*9 = 9
1) 3*5 = 15
2) 0.5*4 = 2
and then find the maximum of all answers, and finally return row with ID:1
1) 15
How do I aggregate the same in Mongoose? Even partial answer will help to move forward and to learn.
Tried to start with http://www.querymongo.com/ but without success.
mongodb mongoose mongodb-query
asked Nov 10 at 19:32
Pumych
5921819
For given table:
id | val_a | val_b | mult | category
------------------------------------
0 | 1 | 2 | 9 | A
1 | 0 | 3 | 5 | B
2 | 0.5 | 4 | 6 | C
following query in MySQL
SELECT * FROM table WHERE category IN ('A','B','C')
AND IF( val_a > 0, val_a, val_b ) * mult = (SELECT MAX( IF( val_a > 0, val_a, val_b ) * mult ) FROM table WHERE category IN ('A','B','C')) LIMIT 1;
will find multiplications of val_a * mult for every row (if val_a == 0 : val_b * mult)
0) 1*9 = 9
1) 3*5 = 15
2) 0.5*4 = 2
and then find the maximum of all answers, and finally return row with ID:1
1) 15
How do I aggregate the same in Mongoose? Even partial answer will help to move forward and to learn.
Tried to start with http://www.querymongo.com/ but without success.
mongodb mongoose mongodb-query
mongodb mongoose mongodb-query
asked Nov 10 at 19:32
Pumych
5921819
asked Nov 10 at 19:32
Pumych
5921819
asked Nov 10 at 19:32
Pumych
5921819
asked Nov 10 at 19:32
Pumych
5921819
asked Nov 10 at 19:32
Pumych
5921819
5921819
add a comment |
add a comment |
1 Answer
1
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up vote
1
down vote
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You can use $match with $in to apply filtering condition and then $multiply with $cond to calculate the value based on your formula. Finally you can add $sort with $limit to get MAX value. Try:
db.col.aggregate([
$match:
category: $in: [ "A", "B", "C" ]
,
$project:
_id: 1,
value:
$multiply: [ $cond: [ $ne: [ "$val_a", 0 ] , "$val_a", "$val_b" ] , "$mult" ]
,
$sort: value: -1
,
$limit: 1
])
Outputs:
"_id" : 1, "value" : 15
answered Nov 10 at 21:40
mickl
10.8k51535
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can use $match with $in to apply filtering condition and then $multiply with $cond to calculate the value based on your formula. Finally you can add $sort with $limit to get MAX value. Try:
db.col.aggregate([
$match:
category: $in: [ "A", "B", "C" ]
,
$project:
_id: 1,
value:
$multiply: [ $cond: [ $ne: [ "$val_a", 0 ] , "$val_a", "$val_b" ] , "$mult" ]
,
$sort: value: -1
,
$limit: 1
])
Outputs:
"_id" : 1, "value" : 15
answered Nov 10 at 21:40
mickl
10.8k51535
add a comment |
up vote
1
down vote
accepted
You can use $match with $in to apply filtering condition and then $multiply with $cond to calculate the value based on your formula. Finally you can add $sort with $limit to get MAX value. Try:
db.col.aggregate([
$match:
category: $in: [ "A", "B", "C" ]
,
$project:
_id: 1,
value:
$multiply: [ $cond: [ $ne: [ "$val_a", 0 ] , "$val_a", "$val_b" ] , "$mult" ]
,
$sort: value: -1
,
$limit: 1
])
Outputs:
"_id" : 1, "value" : 15
answered Nov 10 at 21:40
mickl
10.8k51535
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can use $match with $in to apply filtering condition and then $multiply with $cond to calculate the value based on your formula. Finally you can add $sort with $limit to get MAX value. Try:
db.col.aggregate([
$match:
category: $in: [ "A", "B", "C" ]
,
$project:
_id: 1,
value:
$multiply: [ $cond: [ $ne: [ "$val_a", 0 ] , "$val_a", "$val_b" ] , "$mult" ]
,
$sort: value: -1
,
$limit: 1
])
Outputs:
"_id" : 1, "value" : 15
answered Nov 10 at 21:40
mickl
10.8k51535
You can use $match with $in to apply filtering condition and then $multiply with $cond to calculate the value based on your formula. Finally you can add $sort with $limit to get MAX value. Try:
db.col.aggregate([
$match:
category: $in: [ "A", "B", "C" ]
,
$project:
_id: 1,
value:
$multiply: [ $cond: [ $ne: [ "$val_a", 0 ] , "$val_a", "$val_b" ] , "$mult" ]
,
$sort: value: -1
,
$limit: 1
])
Outputs:
"_id" : 1, "value" : 15
answered Nov 10 at 21:40
mickl
10.8k51535
answered Nov 10 at 21:40
mickl
10.8k51535
answered Nov 10 at 21:40
mickl
10.8k51535
answered Nov 10 at 21:40
mickl
10.8k51535
10.8k51535
add a comment |
add a comment |
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