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mysql group by show all rows


By : Gazet
Date : May 04 2020, 05:15 PM

I have a MySQL db with names and medical prescriptions. What I what to do is loop through all prescriptions belonging to a specific name and show them in a table, and then create a new table with the next name and a new set of prescriptions.

When I using GROUP BY, I able to display the name of a patient and the first table row of each group. But I do not know how to populate rows from the sub_fields within the groups. I have tried with GROUP_CONCAT, and it allows me to show all rows of a column from the group as comma-separated, but still I need to populate all the data from the sub rows like this:

NAME1: - START1_1 - END1_1 - DRUGID1_1

  • START1_2
  • END1_2
  • DRUGID1_2

  • START1_3

  • END1_3
  • DRUGID1_3

NAME2: - START2_1 - END2_1 - DRUGID2_1

  • START2_2
  • END2_2
  • DRUGID2_2

  • START2_3

  • END2_3
  • DRUGID2_3

I tried with the following code, using Group by, and I have also tried using GROUP_CONCAT, but I do not know how to handle the output.

$med = "SELECT  *  FROM {$wpdb->prefix}medicin_skema GROUP BY elevnavn";

 foreach ($wpdb->get_results($med, ARRAY_A ) as $row) {
     echo '<br>'.$row['elevnavn'].' ';
     echo '<br>'.$row['start_dato'].' ';
     echo '<br>'.$row['slut_dato'].' ';
     echo '<br><br>'.$row['drugid'].' '; 
 }
Answer :

 you are group by the patient's name - so you will get one record PER PATIENT NAME.

The brute force approach is to select a distinct array of patient names; then select each of their prescriptions ordered by start date or / end date or whatever sensible option you want - then do your creation?

A sub-select in essence (I assume elevnavn is NAME)

<?php

SELECT distinct(elevnavn) from {$wpdb->prefix}medicin_skema while($row = mysql_fetch_array($result)) { $name=$row[0]; $newquery="SELECT * from {$wpdb->prefix}medicin_skema where elevnavn='".$name"'"; execute the query while($row = mysql_fetch_array($result)) { retrieve dates, drug id etc new create query } }

You can do a insert (select from) but you will need to get the sub-select correct first and use a left join to get ALL the records for each name.



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