Query that checks for subset to mysql query












0















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.










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  • Shouldn't DRUG_ID be a column in prescriptions?

    – Rick James
    Feb 28 '17 at 3:01
















0















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.










share|improve this question
















bumped to the homepage by Community 3 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't DRUG_ID be a column in prescriptions?

    – Rick James
    Feb 28 '17 at 3:01














0












0








0








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.










share|improve this question
















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






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edited Feb 28 '17 at 9:33







Andrew Hantzos

















asked Feb 28 '17 at 1:08









Andrew HantzosAndrew Hantzos

11




11





bumped to the homepage by Community 3 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 3 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'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

















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










1 Answer
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oldest

votes


















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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.






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    1 Answer
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    1 Answer
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    0














    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.






    share|improve this answer




























      0














      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.






      share|improve this answer


























        0












        0








        0







        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.






        share|improve this answer













        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.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Feb 28 '17 at 5:56









        Rick JamesRick James

        43.6k22259




        43.6k22259






























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