Joining a One-to-Many Relationship: A Guide

Assuming that Bill is a member of groups A, B, and C, and we obtain a set from our subquery that includes Group A, Bill should be included in the result set alongside groups A, B, and C. To simplify the process, we can use the following approach to retrieve all the groups that individuals in the provided group list belong to. For example, suppose the table has data for month 1 as shown below: Monthly Results and Issue Results. By examining the Monthly Results table, we can deduce that the only information connected with the 5000 copies is in month 1.


My objective is to combine 3 tables –
tables … months
, monthly_results, and issue_results. While each month has one record in monthly_results, it could have zero to multiple records in issue_results. To avoid over-counting monthly totals, I need to ensure that only one instance of monthly_results is included in the result set when there are multiple issue_results for a given month. However, setting up a Left Join works when there are no issue_results for a month.

For example, here are tables

Months 1,2,3

The monthly figures for the number of copies sold are as follows: 5000 in the first month, 2500 in the second month, and 1000 in the third month.

The outcome of
Issue, Month
orders is as follows:
– Order 1 resulted in 750 units.
– Order 2 resulted in 500 units.
– Order 3 resulted in 250 units.

Upon merging the tables, my desired outcome is as follows:

The table below shows the distribution details of a certain publication over the course of four months. The first column represents the month number, followed by the issue number, the number of copies printed, and the number of orders received for that particular issue. In month 1, issue 1, a total of 5000 copies were printed and 750 orders were received. In month 1, issue 2, the number of copies printed is unknown, but 500 orders were received. In month 2, the publication printed 2500 copies, but the issue number is not specified. In month 3, issue 3, 1000 copies were printed and 250 orders were received.

By utilizing Left Outer Joins, the second month can be retrieved without any difficulty in terms of correct population. However, there is an issue of duplicate monthly outcomes in the second month 1.

The data consists of four sets of values. The first set includes three numbers: 1, 1, and 5000, followed by 750. The second set includes only two numbers: 1 and 2, followed by 5000. The third set includes two numbers: 2 and 500, followed by 2500. The fourth set includes three numbers: 3, 3, and 1000, followed by 250.

We welcome and value any recommendations you may have.

Solution 1:

Insufficient data in the provided tables prevents the identification of the specific issue associated with the 5000 copies in month 1. The desired outcome is to link the 5000 copies with issue 1, not issue 2, but no information in the tables allows for such differentiation. To clarify:

The following are the rows of your table containing details about the first month.

Monthly Results:

    (Month, Copies)

Issue Results:

    (Issue, Month, Orders)

By analyzing the Monthly Results table, it becomes evident that the sole data linked with the 5000 copies pertains to month 1. Hence, upon merging with the Issue Results, both the issues that correspond to month 1 are connected to the 5000 copies.

In order to obtain the desired outcome, you can either modify your initial table to include the necessary columns
include issue
(Month, Issue, Copies), or specify the criteria for selecting a particular issue in case multiple issues are present in a month. A helpful suggestion by sgeddes proposes selecting the first issue of the month as the recipient of the copies.

Solution 2:

The methods to accomplish this may vary depending on the type of RDBMS you are using.

Here is a MySQL solution:

SELECT M.Month, 
  @copies:=IF(@prevMonth=M.Month,NULL,M.Copies) copies,
FROM MonthResults M
  LEFT JOIN IssueResults I ON M.Month = I.Month
  JOIN (SELECT @copies:=0) t;

Sample Fiddle Demo

This approach is designed to be compatible with most RDBMS solutions.

SELECT M.Month, 
FROM MonthResults M
  LEFT JOIN IssueResults I ON M.Month = I.Month
      SELECT Min(I.Issue) minIssue,
        M.Month, M.Copies
      FROM MonthResults M LEFT JOIN IssueResults I ON M.Month=I.Month
      GROUP BY M.Month, M.Copies
) t ON M.Month = t.Month AND (I.Issue = t.minIssue OR I.Issue IS NULL)

And another Fiddle demo

Frequently Asked Questions

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