From e2621eea4dfc9771ee0a4f361a5c4391948b07cb Mon Sep 17 00:00:00 2001
From: atefehsafarkhah <50791492+atefehsafarkhah@users.noreply.github.com>
Date: Tue, 17 Dec 2024 10:11:18 +0000
Subject: [PATCH 1/2] joins

---
 .vscode/settings.json |  1 -
 Scripts/Query1.sql    |  0
 Scripts/query.sql     | 53 +++++++++++++++++++++++++++++++++++++++++++
 3 files changed, 53 insertions(+), 1 deletion(-)
 create mode 100644 Scripts/Query1.sql
 create mode 100644 Scripts/query.sql

diff --git a/.vscode/settings.json b/.vscode/settings.json
index 9bc98c8..20de972 100644
--- a/.vscode/settings.json
+++ b/.vscode/settings.json
@@ -18,7 +18,6 @@
   "screencastMode.onlyKeyboardShortcuts": true,
   "terminal.integrated.fontSize": 18,
   "window.zoomLevel": 3,
-  "workbench.activityBar.visible": true,
   "workbench.colorTheme": "Default Light+",
   "workbench.fontAliasing": "antialiased",
   "workbench.statusBar.visible": true,
diff --git a/Scripts/Query1.sql b/Scripts/Query1.sql
new file mode 100644
index 0000000..e69de29
diff --git a/Scripts/query.sql b/Scripts/query.sql
new file mode 100644
index 0000000..7fa1525
--- /dev/null
+++ b/Scripts/query.sql
@@ -0,0 +1,53 @@
+SELECT *
+FROM employee
+LIMIT 5;
+Go
+SELECT * 
+FROM model
+LIMIT 5;
+
+SELECT sql
+FROM sqlite_schema
+WHERE name= 'employee';
+
+# Make a list of employees and their immediate managers
+SELECT emp.firstName AS employee_first_name,emp.lastName AS employee_last_name,emp.title,
+ mng.firstName AS manager_first_name,mng.lastName AS manager_last_name
+FROM employee e
+INNER JOIN employee mng
+ON emp.managerId = mng.employeeId;
+
+#Find salespeople who have zero sales
+SELECT emp.firstName, emp.lastName
+FROM  employee emp
+LEFT JOIN  sales s ON emp.employeeId = sls.employeeId
+WHERE emp.title= 'Sales Person' AND salesAmount ISNULL;
+
+#List all customers and their sales, even if some data is gone
+SELECT cus.firstName, cus.lastName, sls.salesAmount
+FROM customer cus
+INNER JOIN sales sls
+    ON cus.customerId = sls.customerId
+UNION
+-- UNION WITH CUSTOMERS WHO HAVE NO SALES
+SELECT cus.firstName, cus.lastName,sls.salesAmount
+FROM customer cus
+LEFT JOIN sales sls
+    ON cus.customerId = sls.customerId
+WHERE sls.salesId IS NULL
+UNION
+-- UNION WITH SALES MISSING CUSTOMER DATA
+SELECT cus.firstName, cus.lastName, sls.salesAmount
+FROM sales sls
+LEFT JOIN customer cus
+    ON cus.customerId = sls.customerId
+WHERE cus.customerId IS NULL;
+
+#another optimized and cleaner way: 
+SELECT 
+    COALESCE(cus.firstName, 'N/A') AS firstName, 
+    COALESCE(cus.lastName, 'N/A') AS lastName, 
+    COALESCE(sls.salesAmount, 0) AS salesAmount
+FROM customer cus
+FULL OUTER JOIN sales sls
+    ON cus.customerId = sls.customerId;

From e22f9df138f781e62f1bd4642ed00dae90b9324e Mon Sep 17 00:00:00 2001
From: atefehsafarkhah <50791492+atefehsafarkhah@users.noreply.github.com>
Date: Thu, 19 Dec 2024 13:20:05 +0000
Subject: [PATCH 2/2] Grouping

---
 Scripts/Query1.sql | 22 ++++++++++++++++++++++
 1 file changed, 22 insertions(+)

diff --git a/Scripts/Query1.sql b/Scripts/Query1.sql
index e69de29..0855593 100644
--- a/Scripts/Query1.sql
+++ b/Scripts/Query1.sql
@@ -0,0 +1,22 @@
+#How many cars have been sold per employee?
+SELECT emp.employeeId ,emp.firstName,emp.lastName, COUNT(sls.salesAmount) totalSale
+FROM sales sls
+JOIN employee emp ON sls.employeeId = emp.employeeId
+GROUP BY emp.employeeId 
+ORDER BY 4 DESC ;
+#Find the least and most expensive car sold by each employee
+SELECT emp.employeeId ,emp.firstName,emp.lastName, MIN(sls.salesAmount) minSale, MAX(sls.salesAmount) maxSale
+FROM sales sls
+JOIN employee emp ON sls.employeeId = emp.employeeId
+GROUP BY emp.employeeId ;
+
+#Display a report for employees who have sold more than five cars
+SELECT emp.employeeId empId, emp.firstName empFirstName, emp.lastName empLastName, COUNT(sls.salesAmount) sales_count
+FROM employee emp
+JOIN sales sls ON emp.employeeId=sls.employeeId
+GROUP BY 1,2,3
+HAVING COUNT(sls.salesAmount)> 5
+ORDER BY 4 DESC;
+
+
+