I'm unable to convert MS Access query to SQL SERVER Query, with changing the group by columns because it will effect in the final result. The purpose of this query is to calculate the Creditor and debtor of accounts of projects.
I tried rewriting with 'CTE' but couldn't get any good result.. I hope someone could help me.. Thanks in advance...
this is the query I want to convert:
SELECT Sum(ZABC.M) AS M, Sum(ZABC.D) AS D, ZABC.ACC_NUMBER, ZABC.PROJECT_NUMBER, [M]-[D] AS RM, [D]-
[M] AS RD
FROM ZABC
GROUP BY ZABC.ACC_NUMBER, ZABC.PROJECT_NUMBER
ORDER BY ZABC.PROJECT_NUMBER;
The problem with the query are [M] and [D] in the select clause: these columns should either be repeated in the group by clause, or surrounded by an aggregate function. Your current group by clause gives you one row per (acc_number, project_number) tuple: you need to choose which computation you want for D and M, that may have several different values per group.
You did not explain the purpose of the original query. Maybe you meant:
SELECT
Sum(ZABC.M) AS M,
Sum(ZABC.D) AS D,
ZABC.ACC_NUMBER,
ZABC.PROJECT_NUMBER,
Sum(ZABC.M) - SUM(ZABC.D) AS RM,
SUM(ZABC.D) - SUM(ZABC.M) AS RD
FROM ZABC
GROUP BY ZABC.ACC_NUMBER, ZABC.PROJECT_NUMBER
ORDER BY ZABC.PROJECT_NUMBER;
There is a vast variety of aggregate functions available for you to pick from, such as MIN(), MAX(), AVG(), and so on.
Related
I'm using PostgreSQL 10 and trying to run this query. I started with a CTE which I am referencing as 'query.'
SELECT
ROW_NUMBER()OVER() AS my_new_id,
query.geom AS geom,
query.pop AS pop,
query.name,
query.distance AS dist,
query.amenity_size,
((amenity_size)/(distance)^2) AS attract_score,
SUM((amenity_size)/(distance)^2) AS tot_attract_score,
((amenity_size)/(distance)^2) / SUM((amenity_size)/(distance)^2) as marketshare
INTO table_mktshare
FROM query
WHERE
distance > 0
GROUP BY
query.name,
query.amenity_size,
query.geom,
query.pop,
query.distance
The query runs but the problem lies in the 'markeshare' column. It returns the same answer with or without the SUM operator and returns one, which appears to make both the attract_score and the tot_attract_score the same. Why is the SUM operator read the same as the expression above it?
This is occurring specifically because each combination of columns in the group by clause uniquely identifies one row in the table. I don't know if this is intentional, but more normally, one would expect something like this:
SELECT ROW_NUMBER() OVER() AS my_new_id,
query.geom AS geom, query.pop AS pop, query.name,
SUM((amenity_size)/(distance)^2) AS tot_attract_score,
INTO table_mktshare
FROM query
WHERE distance > 0
GROUP BY query.name, query.geom, query.pop;
This is not your intention, but it does give a flavor of what's expected.
I'm trying to write a few Oracle SQL scripts for an assignment. I've managed to get all of it to work, except for one part. To summarize, I have to display data from 2 tables if the average of 1 column in table A is greater than the average of another column in table B. I realize you cannot include AVG functions in a WHERE clause or HAVING clause since it seems unable to properly access the data (from what I've read). When I exclude this clause, the script executes properly, so I'm confident there are no other errors.
I've tried writing it as follows but the error I get is ORA-00936: missing expression and it is just before the > sign. I thought this may be due to improper bracket placing but none of my attempts resolved this. Here is my attempt:
SELECT l.l_category, SUM(r.r_sold), AVG(l.l_cost)
FROM promos l
INNER JOIN sales r
ON r.promo_id = l.promo_id
GROUP BY l.l_category
HAVING (SELECT AVG(l.l_cost) OVER (PARTITION BY l.l_cost)) >
(SELECT AVG(r.r_sold) OVER (PARTITION BY r.r_sold));
I tried doing this without the OVER (PARTITION BY ...) as well as putting it into a WHERE clause but it didn't resolve the error. I'm pretty sure I need to put it into a SELECT statement somehow but I'm at a loss.
You do not need to use the OVER clause when applying the aggregate functions in the HAVING clause. Just use the aggregate functions on their own.
SELECT l.l_category, SUM(r.r_sold), AVG(l.l_cost)
FROM promos l
INNER JOIN sales r
ON r.promo_id = l.promo_id
GROUP BY l.l_category
HAVING HAVING AVG(l.l_cost) > AVG(r.r_sold)
Based on the bigquery query reference, currently Quantiles do not allow any kind of grouping by another column. I am mainly interested in getting medians grouped by a certain column. The only work around I see right now is to generate a quantile query per distinct group member where the group member is a condition in the where clause.
For example I use the below query for every distinct row in column-y if I want to get the desired result.
SELECT QUANTILE( <column-x>, 1001)
FROM <table>
WHERE
<column-y> == <each distinct row in column-y>
Does the big query team plan on having some functionality to allow grouping on quantiles in the future?
Is there a better way to get what I am trying to get here?
Thanks
With the recently announced percentile_cont() window function you can get medians.
Look at the example in the announcement blog post:
http://googlecloudplatform.blogspot.com/2013/06/google-bigquery-bigger-faster-smarter-analytics-functions.html
SELECT MAX(median) AS median, room FROM (
SELECT percentile_cont(0.5) OVER (PARTITION BY room ORDER BY data) AS median, room
FROM [io_sensor_data.moscone_io13]
WHERE sensortype='temperature'
)
GROUP BY room
While there are efficient algorithms to compute quantiles they are somewhat memory intensive - trying to do multiple quantile calculations in a single query gets expensive.
There are plans to improve QUANTILES, but I don't know what the timeline is.
Do you need median? Can you filter outliers and do an average of the remainder?
If your per-group size is fixed, you may be able to hack it using combination of order, nest and nth. For instance, if there are 9 distinct values of f2 per value of f1, for median:
select f1,nth(5,f2) within record from (
select f1,nest(f2) f2 from (
select f1, f2 from table
group by f1,f2
order by f2
) group by f1
);
Not sure if the sorted order in subquery is guaranteed to survive the second group, but it worked in a simple test I tried.
I was having a problem getting mulitple sums from multiple tables. Short story, my answer was solved in the "sql sum data from multiple tables" thread on this site. But where it came up short, is that now I'd like to only show sums that are greater than a certain amount. So while I have sub-selects in my select, I think I need to use a HAVING clause to filter the summed amounts that are too low.
Example, using the code specified in the link above (more specifically the answer that the owner has chosen as correct), I would only like to see a query result if SUM(AP2.Value) > 1500. Any thoughts?
If you need to filter on the results of ANY aggregate function, you MUST use a HAVING clause. WHERE is applied at the row level as the DB scans the tables for matching things. HAVING is applied basically immediately before the result set is sent out to the client. At the time WHERE operates, the aggregate function results are not (and cannot) be available, so you have to use a HAVING clause, which is applied after the main query is complete and all aggregate results are available.
So... long story short, yes, you'll need to do
SELECT ...
FROM ...
WHERE ...
HAVING (SUM_AP > 1500)
Note that you can use column aliases in the having clause. In technical terms, having on a query as above works basically exactly the same as wrapping the initial query in another query and applying another WHERE clause on the wrapper:
SELECT *
FROM (
SELECT ...
) AS child
WHERE (SUM_AP > 1500)
You could wrap that query as a subselect and then specify your criteria in the WHERE clause:
SELECT
PROJECT,
SUM_AP,
SUM_INV
FROM (
SELECT
AP1.[PROJECT],
(SELECT SUM(AP2.Value) FROM AP AS AP2 WHERE AP2.PROJECT = AP1.PROJECT) AS SUM_AP,
(SELECT SUM(INV2.Value) FROM INV AS INV2 WHERE INV2.PROJECT = AP1.PROJECT) AS SUM_INV
FROM AP AS AP1
INNER JOIN INV AS INV1 ON
AP1.[PROJECT] = INV1.[PROJECT]
WHERE
AP1.[PROJECT] = 'XXXXX'
GROUP BY
AP1.[PROJECT]
) SQ
WHERE
SQ.SUM_AP > 1500
So I got this statement, which works fine:
SELECT MAX(patient_history_date_bio) AS med_date, medication_name
FROM biological
WHERE patient_id = 12)
GROUP BY medication_name
But, I would like to have the corresponding medication_dose also. So I type this up
SELECT MAX(patient_history_date_bio) AS med_date, medication_name, medication_dose
FROM biological
WHERE (patient_id = 12)
GROUP BY medication_name
But, it gives me an error saying:
"coumn 'biological.medication_dose' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.".
So I try adding medication_dose to the GROUP BY clause, but then it gives me extra rows that I don't want.
I would like to get the latest row for each medication in my table. (The latest row is determined by the max function, getting the latest date).
How do I fix this problem?
Use:
SELECT b.medication_name,
b.patient_history_date_bio AS med_date,
b.medication_dose
FROM BIOLOGICAL b
JOIN (SELECT y.medication_name,
MAX(y.patient_history_date_bio) AS max_date
FROM BIOLOGICAL y
GROUP BY y.medication_name) x ON x.medication_name = b.medication_name
AND x.max_date = b.patient_history_date_bio
WHERE b.patient_id = ?
If you really have to, as one quick workaround, you can apply an aggregate function to your medication_dose such as MAX(medication_dose).
However note that this is normally an indication that you are either building the query incorrectly, or that you need to refactor/normalize your database schema. In your case, it looks like you are tackling the query incorrectly. The correct approach should the one suggested by OMG Poinies in another answer.
You may be interested in checking out the following interesting article which describes the reasons behind this error:
But WHY Must That Column Be Contained in an Aggregate Function or the GROUP BY clause?
You need to put max(medication_dose) in your select. Group by returns a result set that contains distinct values for fields in your group by clause, so apparently you have multiple records that have the same medication_name, but different doses, so you are getting two results.
By putting in max(medication_dose) it will return the maximum dose value for each medication_name. You can use any aggregate function on dose (max, min, avg, sum, etc.)