{"id":34503,"date":"2023-03-13T03:08:22","date_gmt":"2023-03-12T21:38:22","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-sql-querying-doctors-and-hospitals-closed\/"},"modified":"2023-03-13T03:08:22","modified_gmt":"2023-03-12T21:38:22","slug":"solved-sql-querying-doctors-and-hospitals-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-sql-querying-doctors-and-hospitals-closed\/","title":{"rendered":"[Solved] SQL querying &#8211; Doctors and Hospitals [closed]"},"content":{"rendered":"<p> [ad_1]<br \/>\n<\/p>\n<div id=\"answer-16570560\" class=\"answer js-answer accepted-answer js-accepted-answer\" data-answerid=\"16570560\" data-parentid=\"16569914\" data-score=\"1\" data-position-on-page=\"1\" data-highest-scored=\"1\" data-question-has-accepted-highest-score=\"1\" itemprop=\"acceptedAnswer\" itemscope itemtype=\"https:\/\/schema.org\/Answer\">\n<div class=\"post-layout\">\n<div class=\"votecell post-layout--left\"><\/div>\n<div class=\"answercell post-layout--right\">\n<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I&#8217;ll help you with your first question, and I&#8217;ll leave to you the second.<\/p>\n<blockquote>\n<ol>\n<li>Display doctorid, dname, total fees received by the doctor(s) who have treated more than one patient?<\/li>\n<\/ol>\n<\/blockquote>\n<p>Let&#8217;s split this problem in pieces:<\/p>\n<p>So you need first to know which doctors have treated more than one patient. That information is in the table <code>billing<\/code>. So:<\/p>\n<pre><code>select doctorId, count(patientId) as patientCount\nfrom (select distinct doctorId, patientId from billing) as a\ngroup by doctorId\nhaving count(patientId)&gt;1;\n<\/code><\/pre>\n<p>This query will return only the Ids of the doctors that have more than one patient. Notice that I&#8217;m using a subquery to deduplicate the doctor-patient tuple.<\/p>\n<p>Now let&#8217;s attack the other part of this question: The total fees of each doctor. Again, that info is in the table <code>billing<\/code>:<\/p>\n<pre><code>select doctorId, sum(fees) as totalFees\nfrom billing\ngroup by doctorId;\n<\/code><\/pre>\n<p>Finally, let&#8217;s put it all together, and include the doctor&#8217;s info, which is in the table <code>doctor<\/code>:<\/p>\n<pre><code>select\n    d.doctorId, d.doctorName, a.totalFees\nfrom\n    doctor as d\n    inner join (\n        select doctorId, sum(fees) as totalFees\n        from billing\n        group by doctorId\n    ) as a on d.doctorId = a.doctorId\n    inner join (\n        select doctorId, count(patientId) as patientCount\n        from (select distinct doctorId, patientId from billing) as a\n        group by doctorId\n        having count(patientId)&gt;1;\n    ) as b on d.doctorId = b.doctorId;\n<\/code><\/pre>\n<p>Hope this helps<\/p>\n<hr>\n<p>Things you need to study and (or) keep in mind:<\/p>\n<ol>\n<li>You need to understand how to relate data stored in different tables. Study how to use <code>INNER JOIN<\/code> (and also <code>LEFT JOIN<\/code> and <code>RIGHT JOIN<\/code>)<\/li>\n<li>You need to understand how does <code>GROUP BY<\/code> works, and how to use aggregate functions (<code>sum()<\/code>, <code>count()<\/code>, etcetera).<\/li>\n<li>You know how to write subqueries. Now try to use them not only for <code>where<\/code> conditions, but as data sources (including them in <code>from<\/code> statements)<\/li>\n<li>Keep a copy of the reference manual of your RDBMS at hand. Also a good book on SQL can help you (go to a bookstore or library and find one you like).<\/li>\n<\/ol><\/div>\n<div class=\"mt24\"><\/div>\n<\/div>\n<p>            <span class=\"d-none\" itemprop=\"commentCount\">6<\/span> <\/p><\/div>\n<\/div>\n<p>[ad_2]<\/p>\n<p>solved SQL querying &#8211; Doctors and Hospitals [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"<p>[ad_1] I&#8217;ll help you with your first question, and I&#8217;ll leave to you the second. Display doctorid, dname, total fees received by the doctor(s) who have treated more than one patient? Let&#8217;s split this problem in pieces: So you need first to know which doctors have treated more than one patient. That information is in &#8230; <a title=\"[Solved] SQL querying &#8211; Doctors and Hospitals [closed]\" class=\"read-more\" href=\"https:\/\/jassweb.com\/solved\/solved-sql-querying-doctors-and-hospitals-closed\/\" aria-label=\"More on [Solved] SQL querying &#8211; Doctors and Hospitals [closed]\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[341],"class_list":["post-34503","post","type-post","status-publish","format-standard","hentry","category-solved","tag-sql"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.5 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>[Solved] SQL querying - Doctors and Hospitals [closed] - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-sql-querying-doctors-and-hospitals-closed\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] SQL querying - Doctors and Hospitals [closed] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] I&#8217;ll help you with your first question, and I&#8217;ll leave to you the second. Display doctorid, dname, total fees received by the doctor(s) who have treated more than one patient? Let&#8217;s split this problem in pieces: So you need first to know which doctors have treated more than one patient. That information is in ... 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Display doctorid, dname, total fees received by the doctor(s) who have treated more than one patient? Let&#8217;s split this problem in pieces: So you need first to know which doctors have treated more than one patient. That information is in ... 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