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chore: update sql problems
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solution/1000-1099/1083.Sales Analysis II/README.md

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@@ -100,7 +100,7 @@ FROM
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Sales
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JOIN Product USING (product_id)
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GROUP BY 1
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HAVING sum(product_name = 'S8') > 0 AND sum(product_name = 'iPhone') = 0;
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HAVING SUM(product_name = 'S8') > 0 AND SUM(product_name = 'iPhone') = 0;
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```
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solution/1000-1099/1083.Sales Analysis II/README_EN.md

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@@ -90,7 +90,7 @@ FROM
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Sales
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JOIN Product USING (product_id)
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GROUP BY 1
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HAVING sum(product_name = 'S8') > 0 AND sum(product_name = 'iPhone') = 0;
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HAVING SUM(product_name = 'S8') > 0 AND SUM(product_name = 'iPhone') = 0;
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```
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solution/1000-1099/1083.Sales Analysis II/Solution.sql

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Sales
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JOIN Product USING (product_id)
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GROUP BY 1
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HAVING sum(product_name = 'S8') > 0 AND sum(product_name = 'iPhone') = 0;
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HAVING SUM(product_name = 'S8') > 0 AND SUM(product_name = 'iPhone') = 0;

solution/1100-1199/1173.Immediate Food Delivery I/README.md

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```sql
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# Write your MySQL query statement below
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SELECT
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round(sum(order_date = customer_pref_delivery_date) / count(1) * 100, 2) AS immediate_percentage
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ROUND(SUM(order_date = customer_pref_delivery_date) / COUNT(1) * 100, 2) AS immediate_percentage
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FROM Delivery;
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```
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solution/1100-1199/1173.Immediate Food Delivery I/README_EN.md

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@@ -65,7 +65,7 @@ We can use the `sum` function to count the number of instant orders, and then di
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```sql
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# Write your MySQL query statement below
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SELECT
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round(sum(order_date = customer_pref_delivery_date) / count(1) * 100, 2) AS immediate_percentage
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ROUND(SUM(order_date = customer_pref_delivery_date) / COUNT(1) * 100, 2) AS immediate_percentage
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FROM Delivery;
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```
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# Write your MySQL query statement below
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SELECT
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round(sum(order_date = customer_pref_delivery_date) / count(1) * 100, 2) AS immediate_percentage
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ROUND(SUM(order_date = customer_pref_delivery_date) / COUNT(1) * 100, 2) AS immediate_percentage
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FROM Delivery;

solution/2000-2099/2072.The Winner University/README.md

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@@ -155,7 +155,7 @@ SELECT
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ELSE 'No Winner'
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END AS winner
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FROM
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(SELECT count(1) AS cnt FROM NewYork WHERE score >= 90) AS n1,
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(SELECT COUNT(1) AS cnt FROM NewYork WHERE score >= 90) AS n1,
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(SELECT COUNT(1) AS cnt FROM California WHERE score >= 90) AS n2;
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```
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solution/2000-2099/2072.The Winner University/README_EN.md

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ELSE 'No Winner'
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END AS winner
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FROM
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(SELECT count(1) AS cnt FROM NewYork WHERE score >= 90) AS n1,
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(SELECT COUNT(1) AS cnt FROM NewYork WHERE score >= 90) AS n1,
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(SELECT COUNT(1) AS cnt FROM California WHERE score >= 90) AS n2;
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```
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solution/2000-2099/2072.The Winner University/Solution.sql

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ELSE 'No Winner'
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END AS winner
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FROM
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(SELECT count(1) AS cnt FROM NewYork WHERE score >= 90) AS n1,
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(SELECT COUNT(1) AS cnt FROM NewYork WHERE score >= 90) AS n1,
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(SELECT COUNT(1) AS cnt FROM California WHERE score >= 90) AS n2;

solution/2300-2399/2329.Product Sales Analysis V/README.md

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```sql
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# Write your MySQL query statement below
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SELECT user_id, sum(quantity * price) AS spending
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SELECT user_id, SUM(quantity * price) AS spending
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FROM
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Sales
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JOIN Product USING (product_id)

solution/2300-2399/2329.Product Sales Analysis V/README_EN.md

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@@ -89,7 +89,7 @@ Users 102 and 103 spent the same amount and we break the tie by their ID while u
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```sql
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# Write your MySQL query statement below
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SELECT user_id, sum(quantity * price) AS spending
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SELECT user_id, SUM(quantity * price) AS spending
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FROM
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Sales
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JOIN Product USING (product_id)

solution/2300-2399/2329.Product Sales Analysis V/Solution.sql

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# Write your MySQL query statement below
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SELECT user_id, sum(quantity * price) AS spending
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SELECT user_id, SUM(quantity * price) AS spending
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FROM
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Sales
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JOIN Product USING (product_id)

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