Query on matching json array keys












0














I have a json field in mysql called column_info. Here's an example value:



[{"id": ["132605", "132750", "132772", "132773", "133065", "133150", "133185", "133188", "133271", "133298"]}, 
{"number": ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10"]},
{"id": ["1", "1", "1", "1", "1", "1", "1", "1", "1", "1"]}
]


Is it possible to do a query that does the following:



keys = ['id', 'number', 'id']
SELECT * FROM `column_info` WHERE JSON(??) = keys









share|improve this question



























    0














    I have a json field in mysql called column_info. Here's an example value:



    [{"id": ["132605", "132750", "132772", "132773", "133065", "133150", "133185", "133188", "133271", "133298"]}, 
    {"number": ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10"]},
    {"id": ["1", "1", "1", "1", "1", "1", "1", "1", "1", "1"]}
    ]


    Is it possible to do a query that does the following:



    keys = ['id', 'number', 'id']
    SELECT * FROM `column_info` WHERE JSON(??) = keys









    share|improve this question

























      0












      0








      0







      I have a json field in mysql called column_info. Here's an example value:



      [{"id": ["132605", "132750", "132772", "132773", "133065", "133150", "133185", "133188", "133271", "133298"]}, 
      {"number": ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10"]},
      {"id": ["1", "1", "1", "1", "1", "1", "1", "1", "1", "1"]}
      ]


      Is it possible to do a query that does the following:



      keys = ['id', 'number', 'id']
      SELECT * FROM `column_info` WHERE JSON(??) = keys









      share|improve this question













      I have a json field in mysql called column_info. Here's an example value:



      [{"id": ["132605", "132750", "132772", "132773", "133065", "133150", "133185", "133188", "133271", "133298"]}, 
      {"number": ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10"]},
      {"id": ["1", "1", "1", "1", "1", "1", "1", "1", "1", "1"]}
      ]


      Is it possible to do a query that does the following:



      keys = ['id', 'number', 'id']
      SELECT * FROM `column_info` WHERE JSON(??) = keys






      python mysql sql






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Dec 27 '18 at 20:03









      David L

      30213




      30213
























          2 Answers
          2






          active

          oldest

          votes


















          2














          Try:



          SELECT `column_info`
          FROM
          `table`,
          (SELECT @`keys` := '["id", "number", "id"]') `init`
          WHERE (
          SELECT
          JSON_ARRAYAGG(
          JSON_UNQUOTE(
          JSON_EXTRACT(
          JSON_KEYS(`der`.`keys`),
          '$[0]'
          )
          )
          )
          FROM
          JSON_TABLE(
          `column_info`,
          '$[*]'
          COLUMNS(
          `keys` JSON PATH '$'
          )
          ) `der`
          ) = CAST(@`keys` AS JSON);


          See db-fiddle.






          share|improve this answer























          • wow, that's pretty complex for doing a json query. Thanks for sharing. Unfortunately, I'm using mysql5.7, which doesn't support some of the above, but I tested it in the db-fiddle and it's pretty neat.
            – David L
            Dec 28 '18 at 3:01





















          0














          What version of MySQL are your running?



          If it's v8, you can try JSON_CONTAINS()



          https://dev.mysql.com/doc/refman/8.0/en/json-search-functions.html






          share|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            Try:



            SELECT `column_info`
            FROM
            `table`,
            (SELECT @`keys` := '["id", "number", "id"]') `init`
            WHERE (
            SELECT
            JSON_ARRAYAGG(
            JSON_UNQUOTE(
            JSON_EXTRACT(
            JSON_KEYS(`der`.`keys`),
            '$[0]'
            )
            )
            )
            FROM
            JSON_TABLE(
            `column_info`,
            '$[*]'
            COLUMNS(
            `keys` JSON PATH '$'
            )
            ) `der`
            ) = CAST(@`keys` AS JSON);


            See db-fiddle.






            share|improve this answer























            • wow, that's pretty complex for doing a json query. Thanks for sharing. Unfortunately, I'm using mysql5.7, which doesn't support some of the above, but I tested it in the db-fiddle and it's pretty neat.
              – David L
              Dec 28 '18 at 3:01


















            2














            Try:



            SELECT `column_info`
            FROM
            `table`,
            (SELECT @`keys` := '["id", "number", "id"]') `init`
            WHERE (
            SELECT
            JSON_ARRAYAGG(
            JSON_UNQUOTE(
            JSON_EXTRACT(
            JSON_KEYS(`der`.`keys`),
            '$[0]'
            )
            )
            )
            FROM
            JSON_TABLE(
            `column_info`,
            '$[*]'
            COLUMNS(
            `keys` JSON PATH '$'
            )
            ) `der`
            ) = CAST(@`keys` AS JSON);


            See db-fiddle.






            share|improve this answer























            • wow, that's pretty complex for doing a json query. Thanks for sharing. Unfortunately, I'm using mysql5.7, which doesn't support some of the above, but I tested it in the db-fiddle and it's pretty neat.
              – David L
              Dec 28 '18 at 3:01
















            2












            2








            2






            Try:



            SELECT `column_info`
            FROM
            `table`,
            (SELECT @`keys` := '["id", "number", "id"]') `init`
            WHERE (
            SELECT
            JSON_ARRAYAGG(
            JSON_UNQUOTE(
            JSON_EXTRACT(
            JSON_KEYS(`der`.`keys`),
            '$[0]'
            )
            )
            )
            FROM
            JSON_TABLE(
            `column_info`,
            '$[*]'
            COLUMNS(
            `keys` JSON PATH '$'
            )
            ) `der`
            ) = CAST(@`keys` AS JSON);


            See db-fiddle.






            share|improve this answer














            Try:



            SELECT `column_info`
            FROM
            `table`,
            (SELECT @`keys` := '["id", "number", "id"]') `init`
            WHERE (
            SELECT
            JSON_ARRAYAGG(
            JSON_UNQUOTE(
            JSON_EXTRACT(
            JSON_KEYS(`der`.`keys`),
            '$[0]'
            )
            )
            )
            FROM
            JSON_TABLE(
            `column_info`,
            '$[*]'
            COLUMNS(
            `keys` JSON PATH '$'
            )
            ) `der`
            ) = CAST(@`keys` AS JSON);


            See db-fiddle.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Dec 27 '18 at 22:58

























            answered Dec 27 '18 at 21:37









            wchiquito

            11.7k22032




            11.7k22032












            • wow, that's pretty complex for doing a json query. Thanks for sharing. Unfortunately, I'm using mysql5.7, which doesn't support some of the above, but I tested it in the db-fiddle and it's pretty neat.
              – David L
              Dec 28 '18 at 3:01




















            • wow, that's pretty complex for doing a json query. Thanks for sharing. Unfortunately, I'm using mysql5.7, which doesn't support some of the above, but I tested it in the db-fiddle and it's pretty neat.
              – David L
              Dec 28 '18 at 3:01


















            wow, that's pretty complex for doing a json query. Thanks for sharing. Unfortunately, I'm using mysql5.7, which doesn't support some of the above, but I tested it in the db-fiddle and it's pretty neat.
            – David L
            Dec 28 '18 at 3:01






            wow, that's pretty complex for doing a json query. Thanks for sharing. Unfortunately, I'm using mysql5.7, which doesn't support some of the above, but I tested it in the db-fiddle and it's pretty neat.
            – David L
            Dec 28 '18 at 3:01















            0














            What version of MySQL are your running?



            If it's v8, you can try JSON_CONTAINS()



            https://dev.mysql.com/doc/refman/8.0/en/json-search-functions.html






            share|improve this answer


























              0














              What version of MySQL are your running?



              If it's v8, you can try JSON_CONTAINS()



              https://dev.mysql.com/doc/refman/8.0/en/json-search-functions.html






              share|improve this answer
























                0












                0








                0






                What version of MySQL are your running?



                If it's v8, you can try JSON_CONTAINS()



                https://dev.mysql.com/doc/refman/8.0/en/json-search-functions.html






                share|improve this answer












                What version of MySQL are your running?



                If it's v8, you can try JSON_CONTAINS()



                https://dev.mysql.com/doc/refman/8.0/en/json-search-functions.html







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Dec 27 '18 at 20:33









                Thomas Skubicki

                365




                365






























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