From 1fce67fe474efedcd34118ac50f41ec60570f49b Mon Sep 17 00:00:00 2001
From: BENAYG <125310603+BENAYG@users.noreply.github.com>
Date: Tue, 26 May 2026 21:37:44 +0300
Subject: [PATCH 1/3] Add files via upload
---
solution_sql.sqbpro | 114 ++++++++++++++++++++++++++++++++++++++++++++
1 file changed, 114 insertions(+)
create mode 100644 solution_sql.sqbpro
diff --git a/solution_sql.sqbpro b/solution_sql.sqbpro
new file mode 100644
index 0000000..e16a6d3
--- /dev/null
+++ b/solution_sql.sqbpro
@@ -0,0 +1,114 @@
+-- Query 1
+-- Get the id values of the first 5 clients from district_id with a value equals to 1.
+SELECT client_id
+FROM client
+WHERE district_id = 1
+ORDER BY client_id
+LIMIT 5;
+--Query 2
+--In the client table, get an id value of the last client where the district_id equals to 72.
+SELECT max(client_id)
+FROM client
+WHERE district_id = 72;
+--Query 3
+--Get the 3 lowest amounts in the loan table.
+SELECT amount
+FROM loan
+ORDER BY amount
+LIMIT 3;
+--Query 4
+--What are the possible values for status, ordered alphabetically in ascending order in the loan table?
+SELECT distinct status
+FROM loan
+ORDER BY 1;
+--Query 5
+--What is the loan_id of the highest payment received in the loan table?
+SELECT loan_id
+FROM loan
+ORDER BY payments DESC
+LIMIT 1;
+--Query 6
+--What is the loan amount of the lowest 5 account_ids in the loan table? Show the account_id and the corresponding amount
+SELECT account_id, amount
+FROM loan
+ORDER BY account_id
+LIMIT 5;
+--Query 7
+--What are the account_ids with the lowest loan amount that have a loan duration of 60 in the loan table?
+SELECT account_id
+FROM loan
+WHERE amount = (SELECT MIN(amount) FROM loan WHERE duration =60);
+--Query 8
+--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.
+SELECT DISTINCT k_symbol
+FROM 'order';
+--Query 9
+--In the order table, what are the order_ids of the client with the account_id 34?
+SELECT order_id
+FROM 'order'
+WHERE account_id = 34;
+--Query 10
+--In the order table, which account_ids were responsible for orders between order_id 29540 and order_id 29560 (inclusive)?
+SELECT DISTINCT account_id
+FROM 'order'
+WHERE order_id between 29540 and 29560;
+--Query 11
+--In the order table, what are the individual amounts that were sent to (account_to) id 30067122?
+SELECT DISTINCT amount
+FROM 'order'
+WHERE account_to= 30067122;
+--Query 12
+--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.
+SELECT trans_id, date, type, amount
+FROM trans
+WHERE account_id = 793
+ORDER BY DATE DESC
+LIMIT 10;
+--Query 13
+--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.
+SELECT district_id, COUNT(DISTINCT client_id)
+FROM client
+WHERE district_id < 10
+GROUP BY district_id
+ORDER BY district_id ASC;
+--Query 14
+--In the card table, how many cards exist for each type? Rank the result starting with the most frequent type.
+SELECT type, COUNT(*)
+FROM card
+GROUP BY type
+ORDER BY 2 DESC;
+--Query 15
+--Using the loan table, print the top 10 account_ids based on the sum of all of their loan amounts.
+SELECT account_id, SUM(amount)
+FROM loan
+GROUP BY account_id
+ORDER BY 2 DESC
+LIMIT 10;
+--Query 16
+--In the loan table, retrieve the number of loans issued for each day, before (excl) 930907, ordered by date in descending order.
+SELECT date, COUNT(loan_id)
+FROM loan
+WHERE date < 930907
+GROUP BY date
+ORDER BY date DESC;
+--Query 17
+--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.
+SELECT date, duration, COUNT(loan_id)
+FROM loan
+WHERE date LIKE '9712%'
+GROUP BY duration
+HAVING COUNT(loan_id)>0
+ORDER BY date, duration;
+--Query 18
+--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.
+SELECT account_id,
+CASE WHEN type = 'VYDAJ' THEN 'Outgoing' WHEN type='PRIJEM' THEN 'Incoming' ELSE 'Others' END AS type,
+SUM(amount) AS total_amount
+FROM trans
+WHERE account_id = 396
+GROUP BY account_id,
+CASE WHEN type = 'VYDAJ' THEN 'Outgoing' WHEN type='PRIJEM' THEN 'Incoming' ELSE 'Others' END
+ORDER BY 2;
From 0a9120b9ce1831c556b6007d223a44caf9f2f768 Mon Sep 17 00:00:00 2001
From: BENAYG <125310603+BENAYG@users.noreply.github.com>
Date: Tue, 26 May 2026 21:43:24 +0300
Subject: [PATCH 2/3] Delete solution_sql.sqbpro
---
solution_sql.sqbpro | 114 --------------------------------------------
1 file changed, 114 deletions(-)
delete mode 100644 solution_sql.sqbpro
diff --git a/solution_sql.sqbpro b/solution_sql.sqbpro
deleted file mode 100644
index e16a6d3..0000000
--- a/solution_sql.sqbpro
+++ /dev/null
@@ -1,114 +0,0 @@
--- Query 1
--- Get the id values of the first 5 clients from district_id with a value equals to 1.
-SELECT client_id
-FROM client
-WHERE district_id = 1
-ORDER BY client_id
-LIMIT 5;
---Query 2
---In the client table, get an id value of the last client where the district_id equals to 72.
-SELECT max(client_id)
-FROM client
-WHERE district_id = 72;
---Query 3
---Get the 3 lowest amounts in the loan table.
-SELECT amount
-FROM loan
-ORDER BY amount
-LIMIT 3;
---Query 4
---What are the possible values for status, ordered alphabetically in ascending order in the loan table?
-SELECT distinct status
-FROM loan
-ORDER BY 1;
---Query 5
---What is the loan_id of the highest payment received in the loan table?
-SELECT loan_id
-FROM loan
-ORDER BY payments DESC
-LIMIT 1;
---Query 6
---What is the loan amount of the lowest 5 account_ids in the loan table? Show the account_id and the corresponding amount
-SELECT account_id, amount
-FROM loan
-ORDER BY account_id
-LIMIT 5;
---Query 7
---What are the account_ids with the lowest loan amount that have a loan duration of 60 in the loan table?
-SELECT account_id
-FROM loan
-WHERE amount = (SELECT MIN(amount) FROM loan WHERE duration =60);
---Query 8
---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.
-SELECT DISTINCT k_symbol
-FROM 'order';
---Query 9
---In the order table, what are the order_ids of the client with the account_id 34?
-SELECT order_id
-FROM 'order'
-WHERE account_id = 34;
---Query 10
---In the order table, which account_ids were responsible for orders between order_id 29540 and order_id 29560 (inclusive)?
-SELECT DISTINCT account_id
-FROM 'order'
-WHERE order_id between 29540 and 29560;
---Query 11
---In the order table, what are the individual amounts that were sent to (account_to) id 30067122?
-SELECT DISTINCT amount
-FROM 'order'
-WHERE account_to= 30067122;
---Query 12
---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.
-SELECT trans_id, date, type, amount
-FROM trans
-WHERE account_id = 793
-ORDER BY DATE DESC
-LIMIT 10;
---Query 13
---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.
-SELECT district_id, COUNT(DISTINCT client_id)
-FROM client
-WHERE district_id < 10
-GROUP BY district_id
-ORDER BY district_id ASC;
---Query 14
---In the card table, how many cards exist for each type? Rank the result starting with the most frequent type.
-SELECT type, COUNT(*)
-FROM card
-GROUP BY type
-ORDER BY 2 DESC;
---Query 15
---Using the loan table, print the top 10 account_ids based on the sum of all of their loan amounts.
-SELECT account_id, SUM(amount)
-FROM loan
-GROUP BY account_id
-ORDER BY 2 DESC
-LIMIT 10;
---Query 16
---In the loan table, retrieve the number of loans issued for each day, before (excl) 930907, ordered by date in descending order.
-SELECT date, COUNT(loan_id)
-FROM loan
-WHERE date < 930907
-GROUP BY date
-ORDER BY date DESC;
---Query 17
---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.
-SELECT date, duration, COUNT(loan_id)
-FROM loan
-WHERE date LIKE '9712%'
-GROUP BY duration
-HAVING COUNT(loan_id)>0
-ORDER BY date, duration;
---Query 18
---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.
-SELECT account_id,
-CASE WHEN type = 'VYDAJ' THEN 'Outgoing' WHEN type='PRIJEM' THEN 'Incoming' ELSE 'Others' END AS type,
-SUM(amount) AS total_amount
-FROM trans
-WHERE account_id = 396
-GROUP BY account_id,
-CASE WHEN type = 'VYDAJ' THEN 'Outgoing' WHEN type='PRIJEM' THEN 'Incoming' ELSE 'Others' END
-ORDER BY 2;
From 04e6ee0701328a24f14aa167f6b5fbafef35c301 Mon Sep 17 00:00:00 2001
From: BENAYG <125310603+BENAYG@users.noreply.github.com>
Date: Tue, 26 May 2026 21:43:45 +0300
Subject: [PATCH 3/3] Updated file upload
---
solution_sql.sqbpro | 93 +++++++++++++++++++++++++++++++++++++++++++++
1 file changed, 93 insertions(+)
create mode 100644 solution_sql.sqbpro
diff --git a/solution_sql.sqbpro b/solution_sql.sqbpro
new file mode 100644
index 0000000..fb0a5b1
--- /dev/null
+++ b/solution_sql.sqbpro
@@ -0,0 +1,93 @@
+-- Get the id values of the first 5 clients from district_id with a value equals to 1.
+SELECT client_id
+FROM client
+WHERE district_id = 1
+ORDER BY client_id
+LIMIT 5;
+--In the client table, get an id value of the last client where the district_id equals to 72.
+SELECT max(client_id)
+FROM client
+WHERE district_id = 72;
+--Get the 3 lowest amounts in the loan table.
+SELECT amount
+FROM loan
+ORDER BY amount
+LIMIT 3;
+--What are the possible values for status, ordered alphabetically in ascending order in the loan table?
+SELECT distinct status
+FROM loan
+ORDER BY 1;
+--What is the loan_id of the highest payment received in the loan table?
+SELECT loan_id
+FROM loan
+ORDER BY payments DESC
+LIMIT 1;
+--What is the loan amount of the lowest 5 account_ids in the loan table? Show the account_id and the corresponding amount
+SELECT account_id, amount
+FROM loan
+ORDER BY account_id
+LIMIT 5;
+--What are the account_ids with the lowest loan amount that have a loan duration of 60 in the loan table?
+SELECT account_id
+FROM loan
+WHERE amount = (SELECT MIN(amount) FROM loan WHERE duration =60);
+--What are the unique values of k_symbol in the order table?
+SELECT DISTINCT k_symbol
+FROM 'order';
+--In the order table, what are the order_ids of the client with the account_id 34?
+SELECT order_id
+FROM 'order'
+WHERE account_id = 34;
+--In the order table, which account_ids were responsible for orders between order_id 29540 and order_id 29560 (inclusive)?
+SELECT DISTINCT account_id
+FROM 'order'
+WHERE order_id between 29540 and 29560;
+--In the order table, what are the individual amounts that were sent to (account_to) id 30067122?
+SELECT DISTINCT amount
+FROM 'order'
+WHERE account_to= 30067122;
+--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.
+SELECT trans_id, date, type, amount
+FROM trans
+WHERE account_id = 793
+ORDER BY DATE DESC
+LIMIT 10;
+--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.
+SELECT district_id, COUNT(DISTINCT client_id)
+FROM client
+WHERE district_id < 10
+GROUP BY district_id
+ORDER BY district_id ASC;
+--In the card table, how many cards exist for each type? Rank the result starting with the most frequent type.
+SELECT type, COUNT(*)
+FROM card
+GROUP BY type
+ORDER BY 2 DESC;
+--Using the loan table, print the top 10 account_ids based on the sum of all of their loan amounts.
+SELECT account_id, SUM(amount)
+FROM loan
+GROUP BY account_id
+ORDER BY 2 DESC
+LIMIT 10;
+--in the loan table, retrieve the number of loans issued for each day, before (excl) 930907, ordered by date in descending order.
+SELECT date, COUNT(loan_id)
+FROM loan
+WHERE date < 930907
+GROUP BY date
+ORDER BY date DESC;
+--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.
+SELECT date, duration, COUNT(loan_id)
+FROM loan
+WHERE date LIKE '9712%'
+GROUP BY duration
+HAVING COUNT(loan_id)>0
+ORDER BY date, duration;
+--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.
+SELECT account_id,
+CASE WHEN type = 'VYDAJ' THEN 'Outgoing' WHEN type='PRIJEM' THEN 'Incoming' ELSE 'Others' END AS type,
+SUM(amount) AS total_amount
+FROM trans
+WHERE account_id = 396
+GROUP BY account_id,
+CASE WHEN type = 'VYDAJ' THEN 'Outgoing' WHEN type='PRIJEM' THEN 'Incoming' ELSE 'Others' END
+ORDER BY 2;