[lab-mysql-select] Sanober Khoso

your-code/solutions.sql

@@ -0,0 +1,97 @@
+-- Challenge 1 -------------------------
+
+USE publications;
+
+SELECT 
+    authors.au_id AS `Author Id`,
+    au_lname AS `Last Name`,
+    au_fname AS `First Name`,
+    titles.title AS `Title`,
+    pub_name AS `Publisher`
+FROM
+    authors
+        INNER JOIN
+    titleauthor ON authors.au_id = titleauthor.au_id
+        INNER JOIN
+    titles ON titleauthor.title_id = titles.title_id
+        INNER JOIN
+    publishers ON titles.pub_id = publishers.pub_id
+
+
+-- Challenge 2 - Solution 1 ----------------------------
+USE publications;
+
+SELECT 
+    authors.au_id AS `Author Id`,
+    au_lname AS `Last Name`,
+    au_fname AS `First Name`,
+    pub_name AS `Publisher`,
+    COUNT(titles.title_id) AS `Title Count`
+FROM
+    authors
+        INNER JOIN
+    titleauthor ON authors.au_id = titleauthor.au_id
+        INNER JOIN
+    titles ON titleauthor.title_id = titles.title_id
+        INNER JOIN
+    publishers ON titles.pub_id = publishers.pub_id
+GROUP BY authors.au_id , au_lname , au_fname , pub_name
+
+
+-- Challenge 2 - Solution 2 using subquery --------
+USE publications;
+
+SELECT 
+    `Author Id`,
+    `Last Name`,
+    `First Name`,
+    `Publisher`,
+    COUNT(`Title`) AS `Title Count`
+FROM
+    (SELECT 
+        authors.au_id AS `Author Id`,
+            au_lname AS `Last Name`,
+            au_fname AS `First Name`,
+            titles.title AS `Title`,
+            pub_name AS `Publisher`
+    FROM
+        authors
+    INNER JOIN titleauthor ON authors.au_id = titleauthor.au_id
+    INNER JOIN titles ON titleauthor.title_id = titles.title_id
+    INNER JOIN publishers ON titles.pub_id = publishers.pub_id) t
+GROUP BY `Author Id` , `Last Name` , `First Name` , `Publisher`
+
+-- Challenge 3 -----------
+USE publications;
+
+SELECT 
+    authors.au_id AS `Author Id`,
+    au_lname AS `Last Name`,
+    au_fname AS `First Name`,
+    SUM(qty) AS `Total`
+FROM
+    authors
+        INNER JOIN
+    titleauthor ON authors.au_id = titleauthor.au_id
+        INNER JOIN
+    sales ON titleauthor.title_id = sales.title_id
+GROUP BY authors.au_id , au_lname , au_fname
+ORDER BY `Total` DESC
+LIMIT 3
+
+-- Challenge 4 -----------------
+USE publications;
+
+SELECT 
+    authors.au_id AS `Author Id`,
+    au_lname AS `Last Name`,
+    au_fname AS `First Name`,
+    IFNULL(SUM(qty), 0) AS `Total`
+FROM
+    authors
+        LEFT JOIN
+    titleauthor ON authors.au_id = titleauthor.au_id
+        LEFT JOIN
+    sales ON titleauthor.title_id = sales.title_id
+GROUP BY authors.au_id , au_lname , au_fname
+ORDER BY `Total` DESC