Query that checks for subset to mysql query
SELECT PA.PATIENT_ID, PA.FIRSTNAME, PA.LASTNAME, PH.*
FROM pharmacies as PH, patients as PA
WHERE
(SELECT D.DRUG_ID
FROM drugs as D, sells as S
WHERE D.DRUG_ID = S.DRUG_ID AND PH.PHARMACY_ID = S.PHARMACY_ID and PA.TOWN = PH.TOWN )
contains
(SELECT D.DRUG_ID
FROM prescriptions as PR, drugs as D
WHERE PR.PATIENT_ID = PA.PATIENT_ID AND PR.DRUG_ID = D.DRUG_ID)
I have the above query and I would like to write it for Mysql. I want to find for all the patients, the pharmacies that are in the same town, and contain all their drugs.
mysql
asked Feb 28 '17 at 1:08
Andrew HantzosAndrew Hantzos
11
bumped to the homepage by Community♦ 15 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
SELECT PA.PATIENT_ID, PA.FIRSTNAME, PA.LASTNAME, PH.*
FROM pharmacies as PH, patients as PA
WHERE
(SELECT D.DRUG_ID
FROM drugs as D, sells as S
WHERE D.DRUG_ID = S.DRUG_ID AND PH.PHARMACY_ID = S.PHARMACY_ID and PA.TOWN = PH.TOWN )
contains
(SELECT D.DRUG_ID
FROM prescriptions as PR, drugs as D
WHERE PR.PATIENT_ID = PA.PATIENT_ID AND PR.DRUG_ID = D.DRUG_ID)
I have the above query and I would like to write it for Mysql. I want to find for all the patients, the pharmacies that are in the same town, and contain all their drugs.
mysql
asked Feb 28 '17 at 1:08
Andrew HantzosAndrew Hantzos
11
bumped to the homepage by Community♦ 15 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Shouldn'tDRUG_IDbe a column inprescriptions?
– Rick James
Feb 28 '17 at 3:01
add a comment |
SELECT PA.PATIENT_ID, PA.FIRSTNAME, PA.LASTNAME, PH.*
FROM pharmacies as PH, patients as PA
WHERE
(SELECT D.DRUG_ID
FROM drugs as D, sells as S
WHERE D.DRUG_ID = S.DRUG_ID AND PH.PHARMACY_ID = S.PHARMACY_ID and PA.TOWN = PH.TOWN )
contains
(SELECT D.DRUG_ID
FROM prescriptions as PR, drugs as D
WHERE PR.PATIENT_ID = PA.PATIENT_ID AND PR.DRUG_ID = D.DRUG_ID)
I have the above query and I would like to write it for Mysql. I want to find for all the patients, the pharmacies that are in the same town, and contain all their drugs.
mysql
asked Feb 28 '17 at 1:08
Andrew HantzosAndrew Hantzos
11
SELECT PA.PATIENT_ID, PA.FIRSTNAME, PA.LASTNAME, PH.*
FROM pharmacies as PH, patients as PA
WHERE
(SELECT D.DRUG_ID
FROM drugs as D, sells as S
WHERE D.DRUG_ID = S.DRUG_ID AND PH.PHARMACY_ID = S.PHARMACY_ID and PA.TOWN = PH.TOWN )
contains
(SELECT D.DRUG_ID
FROM prescriptions as PR, drugs as D
WHERE PR.PATIENT_ID = PA.PATIENT_ID AND PR.DRUG_ID = D.DRUG_ID)
I have the above query and I would like to write it for Mysql. I want to find for all the patients, the pharmacies that are in the same town, and contain all their drugs.
mysql
mysql
asked Feb 28 '17 at 1:08
Andrew HantzosAndrew Hantzos
11
asked Feb 28 '17 at 1:08
Andrew HantzosAndrew Hantzos
11
edited Feb 28 '17 at 9:33
Andrew Hantzos
asked Feb 28 '17 at 1:08
Andrew HantzosAndrew Hantzos
11
asked Feb 28 '17 at 1:08
Andrew HantzosAndrew Hantzos
11
asked Feb 28 '17 at 1:08
Andrew HantzosAndrew Hantzos
11
11
bumped to the homepage by Community♦ 15 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 15 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Shouldn'tDRUG_IDbe a column inprescriptions?
– Rick James
Feb 28 '17 at 3:01
add a comment |
Shouldn'tDRUG_IDbe a column inprescriptions?
– Rick James
Feb 28 '17 at 3:01
Shouldn't
DRUG_ID be a column in prescriptions?– Rick James
Feb 28 '17 at 3:01
Shouldn't
DRUG_ID be a column in prescriptions?– Rick James
Feb 28 '17 at 3:01
add a comment |
1 Answer
1
active
oldest
votes
That is a challenge!
By comparing now many drugs a patient needs to how many a pharmacy can provide of what he needs, I think this gives a list of patient:pharmacy matches.
SELECT have.patient_id, have.pharmacy_id
FROM ( SELECT s.pharmacy_id, pr.patient_id, COUNT(*) AS ct
FROM sells AS s
JOIN prescriptions AS pr USING(drug_id)
GROUP BY s.pharmacy_id, pr.patient_id ) AS have
JOIN ( SELECT patient_id, COUNT(*) AS ct
FROM prescriptions
GROUP BY patient_id ) AS need
ON need.patient_id = have.patient_id
WHERE need.ct = have.ct
That can then be used as a derived table to get the other specifics:
SELECT PA.PATIENT_ID, PA.FIRSTNAME, PA.LASTNAME, PH.*
FROM ( the-above-query ) AS x
JOIN pharmacies as PH USING(pharmacy_id)
JOIN patients as PA USING(patient_id)
Another approach might involve using LEFT JOIN to discover which pharmacies cannot supply a needed drug for a given patient, then use another LEFT JOIN to discover which other pharmacies there are for that patient.
answered Feb 28 '17 at 5:56
Rick JamesRick James
43.6k22259
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
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active
oldest
votes
That is a challenge!
By comparing now many drugs a patient needs to how many a pharmacy can provide of what he needs, I think this gives a list of patient:pharmacy matches.
SELECT have.patient_id, have.pharmacy_id
FROM ( SELECT s.pharmacy_id, pr.patient_id, COUNT(*) AS ct
FROM sells AS s
JOIN prescriptions AS pr USING(drug_id)
GROUP BY s.pharmacy_id, pr.patient_id ) AS have
JOIN ( SELECT patient_id, COUNT(*) AS ct
FROM prescriptions
GROUP BY patient_id ) AS need
ON need.patient_id = have.patient_id
WHERE need.ct = have.ct
That can then be used as a derived table to get the other specifics:
SELECT PA.PATIENT_ID, PA.FIRSTNAME, PA.LASTNAME, PH.*
FROM ( the-above-query ) AS x
JOIN pharmacies as PH USING(pharmacy_id)
JOIN patients as PA USING(patient_id)
Another approach might involve using LEFT JOIN to discover which pharmacies cannot supply a needed drug for a given patient, then use another LEFT JOIN to discover which other pharmacies there are for that patient.
answered Feb 28 '17 at 5:56
Rick JamesRick James
43.6k22259
add a comment |
That is a challenge!
By comparing now many drugs a patient needs to how many a pharmacy can provide of what he needs, I think this gives a list of patient:pharmacy matches.
SELECT have.patient_id, have.pharmacy_id
FROM ( SELECT s.pharmacy_id, pr.patient_id, COUNT(*) AS ct
FROM sells AS s
JOIN prescriptions AS pr USING(drug_id)
GROUP BY s.pharmacy_id, pr.patient_id ) AS have
JOIN ( SELECT patient_id, COUNT(*) AS ct
FROM prescriptions
GROUP BY patient_id ) AS need
ON need.patient_id = have.patient_id
WHERE need.ct = have.ct
That can then be used as a derived table to get the other specifics:
SELECT PA.PATIENT_ID, PA.FIRSTNAME, PA.LASTNAME, PH.*
FROM ( the-above-query ) AS x
JOIN pharmacies as PH USING(pharmacy_id)
JOIN patients as PA USING(patient_id)
Another approach might involve using LEFT JOIN to discover which pharmacies cannot supply a needed drug for a given patient, then use another LEFT JOIN to discover which other pharmacies there are for that patient.
answered Feb 28 '17 at 5:56
Rick JamesRick James
43.6k22259
add a comment |
That is a challenge!
By comparing now many drugs a patient needs to how many a pharmacy can provide of what he needs, I think this gives a list of patient:pharmacy matches.
SELECT have.patient_id, have.pharmacy_id
FROM ( SELECT s.pharmacy_id, pr.patient_id, COUNT(*) AS ct
FROM sells AS s
JOIN prescriptions AS pr USING(drug_id)
GROUP BY s.pharmacy_id, pr.patient_id ) AS have
JOIN ( SELECT patient_id, COUNT(*) AS ct
FROM prescriptions
GROUP BY patient_id ) AS need
ON need.patient_id = have.patient_id
WHERE need.ct = have.ct
That can then be used as a derived table to get the other specifics:
SELECT PA.PATIENT_ID, PA.FIRSTNAME, PA.LASTNAME, PH.*
FROM ( the-above-query ) AS x
JOIN pharmacies as PH USING(pharmacy_id)
JOIN patients as PA USING(patient_id)
Another approach might involve using LEFT JOIN to discover which pharmacies cannot supply a needed drug for a given patient, then use another LEFT JOIN to discover which other pharmacies there are for that patient.
answered Feb 28 '17 at 5:56
Rick JamesRick James
43.6k22259
That is a challenge!
By comparing now many drugs a patient needs to how many a pharmacy can provide of what he needs, I think this gives a list of patient:pharmacy matches.
SELECT have.patient_id, have.pharmacy_id
FROM ( SELECT s.pharmacy_id, pr.patient_id, COUNT(*) AS ct
FROM sells AS s
JOIN prescriptions AS pr USING(drug_id)
GROUP BY s.pharmacy_id, pr.patient_id ) AS have
JOIN ( SELECT patient_id, COUNT(*) AS ct
FROM prescriptions
GROUP BY patient_id ) AS need
ON need.patient_id = have.patient_id
WHERE need.ct = have.ct
That can then be used as a derived table to get the other specifics:
SELECT PA.PATIENT_ID, PA.FIRSTNAME, PA.LASTNAME, PH.*
FROM ( the-above-query ) AS x
JOIN pharmacies as PH USING(pharmacy_id)
JOIN patients as PA USING(patient_id)
Another approach might involve using LEFT JOIN to discover which pharmacies cannot supply a needed drug for a given patient, then use another LEFT JOIN to discover which other pharmacies there are for that patient.
answered Feb 28 '17 at 5:56
Rick JamesRick James
43.6k22259
answered Feb 28 '17 at 5:56
Rick JamesRick James
43.6k22259
answered Feb 28 '17 at 5:56
Rick JamesRick James
43.6k22259
answered Feb 28 '17 at 5:56
Rick JamesRick James
43.6k22259
43.6k22259
add a comment |
add a comment |
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Shouldn't
DRUG_IDbe a column inprescriptions?– Rick James
Feb 28 '17 at 3:01