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Why Doesn't My LEFT JOIN Query Return Organizations with a Count of Zero?

Patricia Arquette
Release: 2025-01-14 11:57:46
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Why Doesn't My LEFT JOIN Query Return Organizations with a Count of Zero?

PostgreSQL LEFT JOIN: Addressing Zero-Count Organization Issues

A common challenge when using LEFT JOIN in PostgreSQL involves retrieving counts of associated records. If the count is zero for a particular organization, that organization might be omitted from the results. This article addresses this problem.

The following query attempts to count used items per organization:

<code class="language-sql">SELECT o.name as organisation_name,
         coalesce(COUNT(exam_items.id)) as total_used
  FROM organisations o
  LEFT JOIN exam_items e on o.id = e.organisation_id
  WHERE e.item_template_id = #{sanitize(item_template_id)}
  and e.used = true
  GROUP BY o.name
  ORDER BY o.name</code>
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This query's flaw lies in the WHERE clause. Conditions placed within the WHERE clause filter after the join, effectively turning the LEFT JOIN into an INNER JOIN when a matching record isn't found in exam_items.

Here are improved versions:

Solution 1: Correcting the JOIN Condition

The WHERE clause filters out organizations with zero counts. Moving the item_template_id and used conditions into the ON clause of the LEFT JOIN solves this:

<code class="language-sql">SELECT o.name AS organisation_name, COUNT(e.id) AS total_used
FROM organisations o
LEFT JOIN exam_items e ON e.organisation_id = o.id AND e.item_template_id = #{sanitize(item_template_id)} AND e.used
GROUP BY o.name
ORDER BY o.name;</code>
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This revised query ensures all organizations are included, even those with no matching entries in exam_items.

Solution 2: Pre-Aggregation for Efficiency

For larger datasets, pre-aggregating exam_items can significantly improve performance:

<code class="language-sql">SELECT o.id, o.name AS organisation_name, COALESCE(e.total_used, 0) AS total_used
FROM organisations o
LEFT JOIN (
   SELECT organisation_id AS id, COUNT(*) AS total_used
   FROM exam_items
   WHERE item_template_id = #{sanitize(item_template_id)} AND used = TRUE
   GROUP BY 1
) e ON o.id = e.id
ORDER BY o.name, o.id;</code>
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This approach first aggregates the counts for each organization and then performs the LEFT JOIN, resulting in a more efficient query. Note the use of COALESCE to handle cases where total_used is NULL.

By implementing either of these solutions, you'll accurately retrieve the count of used items for all organizations, including those with a count of zero. Choose the solution that best suits your data size and performance requirements.

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