Enhance readme with additional SQL queries and expected results for c…

readme.md

@@ -28,6 +28,13 @@ Assume that any `_id` columns are incremental, meaning that higher ids always oc
 
 Get the `id` values of the first 5 clients from `district_id` with a value equals to 1.
 
+```sql
+select c.client_id
+from client c 
+where district_id = 1
+limit 5
+```
+
 Expected result:
 
 ```shell
@@ -42,16 +49,31 @@ Expected result:
 
 In the `client` table, get an `id` value of the last client where the `district_id` equals to 72.
 
+```sql
+select c.client_id
+from client c 
+where district_id = 72
+order by client_id DESC
+limit 1
+```
+
 Expected result:
 
-```shell
+```sql
 13576
 ```
 
 ### Query 3
 
 Get the 3 lowest amounts in the `loan` table.
 
+```sql
+select l.amount
+from loan l 
+order by amount ASC
+limit 3
+```
+
 Expected result:
 
 ```shell
@@ -64,6 +86,12 @@ Expected result:
 
 What are the possible values for `status`, ordered alphabetically in ascending order in the `loan` table?
 
+```sql
+select l.status
+from loan l 
+group by status
+```
+
 Expected result:
 
 ```shell
@@ -77,6 +105,13 @@ D
 
 What is the `loan_id` of the highest payment received in the `loan` table?
 
+```sql
+select l.loan_id
+from loan l
+order by l.payments DESC
+limit 1
+```
+
 Expected result:
 
 ```shell
@@ -87,6 +122,13 @@ Expected result:
 
 What is the loan `amount` of the lowest 5 `account_id`s in the `loan` table? Show the `account_id` and the corresponding `amount`
 
+```sql
+SELECT l.account_id, l.amount
+from loan l 
+order by l.account_id ASC
+limit 5
+```
+
 Expected result:
 
 ```shell
@@ -102,6 +144,14 @@ Expected result:
 
 What are the `account_id`s with the lowest loan `amount` that have a loan `duration` of 60 in the `loan` table?
 
+```sql
+select l.account_id 
+from loan l 
+where l.duration = 60
+order by l.amount ASC
+limit 5
+```
+
 Expected result:
 
 ```shell
@@ -118,6 +168,12 @@ What are the unique values of `k_symbol` in the `order` table?
 
 Note: There shouldn't be a table name `order`, since `order` is reserved from the `ORDER BY` clause. You have to use backticks to escape the `order` table name.
 
+```sql
+select o.k_symbol
+from "order" o 
+group by o.k_symbol
+```
+
 Expected result:
 
 ```shell
@@ -131,6 +187,12 @@ UVER
 
 In the `order` table, what are the `order_id`s of the client with the `account_id` 34?
 
+```sql
+select o.order_id
+from "order" o 
+where o.account_id = 34
+```
+
 Expected result:
 
 ```shell
@@ -143,6 +205,12 @@ Expected result:
 
 In the `order` table, which `account_id`s were responsible for orders between `order_id` 29540 and `order_id` 29560 (inclusive)?
 
+```sql
+select distinct o.account_id 
+from "order" o 
+where o.order_id between 29540 and 29560
+```
+
 Expected result:
 
 ```shell
@@ -156,6 +224,12 @@ Expected result:
 
 In the `order` table, what are the individual amounts that were sent to (`account_to`) id 30067122?
 
+```sql
+select o.amount
+from "order" o 
+where o.account_to = 30067122
+```
+
 Expected result:
 
 ```shell
@@ -166,6 +240,14 @@ Expected result:
 
 In the `trans` table, show the `trans_id`, `date`, `type` and `amount` of the 10 first transactions from `account_id` 793 in chronological order, from newest to oldest.
 
+```sql
+select t.trans_id, t.date, t.type, t.amount
+from trans t 
+where t.account_id = 793
+order by t.date DESC
+limit 10
+```
+
 Expected result:
 
 ```shell
@@ -185,6 +267,14 @@ Expected result:
 
 In the `client` table, of all districts with a `district_id` lower than 10, how many clients are from each `district_id`? Show the results sorted by the `district_id` in ascending order.
 
+```sql
+select c.district_id, count(*)
+from client c 
+where c.district_id < 10
+group by c.district_id
+order by c.district_id ASC
+```
+
 Expected result:
 
 ```shell
@@ -203,6 +293,13 @@ Expected result:
 
 In the `card` table, how many cards exist for each `type`? Rank the result starting with the most frequent `type`.
 
+```sql
+select c.type, count(*) as count
+from card c
+group by type
+order by count DESC
+```
+
 Expected result:
 
 ```shell
@@ -215,6 +312,14 @@ gold	88
 
 Using the `loan` table, print the top 10 `account_id`s based on the sum of all of their loan amounts.
 
+```sql
+select l.account_id, SUM(l.amount) as total_amount
+from loan l
+group by l.account_id
+order by total_amount DESC
+limit 10;
+```
+
 Expected result:
 
 ```shell
@@ -234,6 +339,14 @@ Expected result:
 
 In the `loan` table, retrieve the number of loans issued for each day, before (excl) 930907, ordered by date in descending order.
 
+```sql
+select l.date, count(*)
+from loan l
+where l.date < 930907
+group by date
+order by l.date DESC
+```
+
 Expected result:
 
 ```
@@ -248,6 +361,14 @@ Expected result:
 
 In the `loan` table, for each day in December 1997, count the number of loans issued for each unique loan duration, ordered by date and duration, both in ascending order. You can ignore days without any loans in your output.
 
+```sql
+select l.date, l.duration, COUNT(*) as loan_count
+from loan l
+where l.date >= 971201 AND l.date <= 971231
+group by l.date, l.duration
+order by l.date ASC, l.duration ASC;
+```
+
 Expected result:
 
 ```shell
@@ -277,6 +398,14 @@ Expected result:
 
 In the `trans` table, for `account_id` 396, sum the amount of transactions for each type (`VYDAJ` = Outgoing, `PRIJEM` = Incoming). Your output should have the `account_id`, the `type` and the sum of amount, named as `total_amount`. Sort alphabetically by type.
 
+```sql
+select t.account_id, t.type, SUM(t.amount) AS total_amount
+from trans t
+where t.account_id = 396
+group by t.account_id, t.type
+order by t.type ASC;
+```
+
 Expected result:
 
 ```shell