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    <p>The only additional piece of information in the 1st query that's missing from the second is</p> <pre><code>COUNT(ra.question_option_id) AS TotalAnswers </code></pre> <p>on the <code>dbo.form_response_answers ra</code> table.</p> <p>So, just add this count into your select:</p> <pre><code> (select count(*) from dbo.form_response_answers ra where qo.question_option_id = ra.question_option_id) as AS TotalAnswers </code></pre> <p>as in:</p> <pre><code>SELECT p.program_id, p.pre_survey_form_id, p.post_survey_form_id, fq.form_id, sq.question_id, sq.question_text, qo.question_option_id, qo.option_text, G.Total, (select count(*) from dbo.form_response_answers ra where qo.question_option_id = ra.question_option_id) as AS TotalAnswers FROM dbo.program p LEFT OUTER JOIN dbo.form_question fq ON p.pre_survey_form_id = fq.form_id OR p.post_survey_form_id = fq.form_id LEFT OUTER JOIN dbo.survey_question sq ON fq.question_id = sq.question_id LEFT OUTER JOIN dbo.question_option qo ON sq.question_id = qo.question_id LEFT OUTER JOIN ( SELECT ra.question_id, ra.question_option_id, COUNT(*) AS Total FROM dbo.form_response_answers ra GROUP BY ra.question_option_id, ra.question_id ) G ON G.question_id = sq.question_id AND G.question_option_id = qo.question_option_id ORDER BY p.program_id, fq.form_id, sq.question_id, qo.question_option_id </code></pre> <hr> <p>EDIT: You wanted the total # of Answers for each sq.question_id.</p> <p>So, I should have inserted:</p> <pre><code>(select count(ra2.question_option_id) from dbo.question_option qo2 LEFT OUTER JOIN dbo.form_response_answers ra2 ON qo2.question_option_id = ra2.question_option_id where qo2.question_id = sq.question_id) as TotalAnswers </code></pre> <p>Now, of course, that will be repeated multiple times because there are more rows in Query 2 than Query 1.</p>
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    1. POHow to perform a OUTER JOIN on two complex SQL queries?
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    1. USMike Ryan
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    1. POHow to perform a OUTER JOIN on two complex SQL queries?
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    1. COThis does not work because it loses the grouping that I have in place in the first query and therefore the count returns different results.
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    2. COThat is actually desirable that it is repeated because I will be using that column to calculate the percent for the number of times a particular response was chosen.
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