rlrightleft
diff --git a/‎sqlzoo/9+_COVID19.md
Lines changed: 161 additions & 0 deletions b/‎sqlzoo/9+_COVID19.md
Lines changed: 161 additions & 0 deletions
diff --git a/‎sqlzoo/9+_COVID19.sql
Lines changed: 0 additions & 95 deletions b/‎sqlzoo/9+_COVID19.sql
Lines changed: 0 additions & 95 deletions
@@ -0,0 +1,161 @@
+# SQLZoo Solutions: COVID-19 Window Functions (LAG)
+
+This document contains my solutions to the SQLZoo ['Window LAG' (COVID-19 Data) section](https://sqlzoo.net/wiki/Window_LAG) using MySQL syntax, along with my personal learning notes and explanations.
+
+---
+
+## Problem 1
+Modify the query to show data from Spain (instead of Italy).
+
+**My Solution:**
+
+```sql
+SELECT 
+  name, DAY(whn), confirmed, deaths, recovered
+FROM covid
+WHERE name = 'Spain'
+  AND MONTH(whn) = 3 
+  AND YEAR(whn) = 2020
+ORDER BY whn;
+```
+
+---
+
+## Problem 2
+Show confirmed cases for the day before, using `LAG`.
+
+**My Solution:**
+
+```sql
+SELECT 
+  name, DAY(whn), confirmed,
+  LAG(confirmed, 1) OVER (PARTITION BY name ORDER BY whn)
+FROM covid
+WHERE name = 'Italy'
+  AND MONTH(whn) = 3 
+  AND YEAR(whn) = 2020
+ORDER BY whn;
+```
+
+---
+
+## Problem 3
+Show the number of new cases for each day in Italy, March 2020.
+
+**My Solution:**
+
+```sql
+SELECT 
+  name, DAY(whn), 
+  (
+  confirmed - LAG(confirmed, 1) OVER (PARTITION BY name ORDER BY whn)
+  ) AS new_cases
+FROM covid
+WHERE name = 'Italy'
+  AND MONTH(whn) = 3 
+  AND YEAR(whn) = 2020
+ORDER BY whn;
+```
+
+**My Notes:**  
+`LAG()` allows comparing to the previous row's confirmed cases.
+
+---
+
+## Problem 4
+Show the number of new cases in Italy for each week in 2020 (Monday only).
+
+**My Solution:**
+
+```sql
+SELECT 
+  name, 
+  DATE_FORMAT(whn,'%Y-%m-%d') AS date, 
+  (
+  confirmed - LAG(confirmed, 1) OVER (PARTITION BY name ORDER BY whn)
+  ) AS new_cases
+FROM covid
+WHERE name = 'Italy'
+  AND WEEKDAY(whn) = 0 
+  AND YEAR(whn) = 2020
+ORDER BY whn;
+```
+
+**My Notes:**  
+Filter with `WEEKDAY(whn) = 0` to get Monday's data.
+
+---
+
+## Problem 5
+Show the number of new cases in Italy for each week using `JOIN`.
+
+**My Solution:**
+
+```sql
+SELECT 
+  tw.name, 
+  DATE_FORMAT(tw.whn,'%Y-%m-%d') AS date, 
+ (tw.confirmed - lw.confirmed) AS new_cases_each_week
+FROM covid tw 
+  LEFT JOIN covid lw 
+    ON DATE_ADD(lw.whn, INTERVAL 1 WEEK) = tw.whn 
+    AND tw.name = lw.name
+WHERE tw.name = 'Italy'
+  AND WEEKDAY(tw.whn) = 0
+ORDER BY tw.whn;
+```
+
+**My Notes:**  
+Using `JOIN` with `DATE_ADD()` to compare weekly data.
+
+---
+
+## Problem 6
+Add a column to show the ranking for deaths due to COVID on 2020-04-20.
+
+**My Solution:**
+
+```sql
+SELECT 
+  name,
+  confirmed,
+  RANK() OVER (ORDER BY confirmed DESC) rc,
+  deaths,
+  RANK() OVER (ORDER BY deaths DESC) deaths_rank
+FROM covid
+WHERE whn = '2020-04-20'
+ORDER BY confirmed DESC;
+```
+
+---
+
+## Problem 7
+Show the infection rate ranking (cases per 100,000) for countries with population over 10 million.
+
+**My Solution:**
+
+```sql
+SELECT 
+   world.name,
+   ROUND(100000 * confirmed / population, 2) rate,
+   RANK() OVER (ORDER BY rate) rank
+FROM covid 
+JOIN world ON covid.name = world.name
+WHERE whn = '2020-04-20' 
+  AND population > 10000000
+ORDER BY population DESC;
+```
+
+**My Notes:**  
+Calculate infection rates using population data from `world` table.
+
+---
+
+## Problem 8
+For each country with at least 1000 new cases in a single day, show the date of the peak number of new cases.
+
+**My Solution:**  
+_(No solution given in original SQL. Placeholder for future completion.)_
+
+---
+