| 1 |
user_query |
[3.5 Sonnet] prompts.py:generate_prompt |
[3.5 Sonnet] prompts.py:generate_prompt_with_examples |
[3.5 Sonnet] prompts.py:generate_prompt_with_cot |
[3.5 Sonnet] prompts.py:generate_prompt_with_rag |
[3.5 Sonnet] prompts.py:generate_prompt |
[3 Haiku] prompts.py:generate_prompt |
[3.5 Sonnet] prompts.py:generate_prompt_with_examples |
[3 Haiku] prompts.py:generate_prompt_with_examples |
[3.5 Sonnet] prompts.py:generate_prompt_with_cot |
[3 Haiku] prompts.py:generate_prompt_with_cot |
[3.5 Sonnet] prompts.py:generate_prompt_with_rag |
[3 Haiku] prompts.py:generate_prompt_with_rag |
| 2 |
What are the names of all employees in the Engineering department? |
[PASS] (1.00)
<sql>
SELECT employees.name
FROM employees
JOIN departments ON employees.department_id = departments.id
WHERE departments.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<sql>
SELECT e.name
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
1. We need to join the employees and departments tables to get the department information for each employee.
2. We'll match employees.department_id with departments.id for the join.
3. We need to filter for employees in the Engineering department.
4. We only need to return the employee names.
5. We'll use a JOIN operation to combine the tables and a WHERE clause to filter for the Engineering department.
</thought_process>
<sql>
SELECT e.name
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
To answer this query, we need to:
1. Join the employees and departments tables since the department information is in a separate table.
2. Filter for employees in the Engineering department.
3. Select only the names of these employees.
The steps to create the SQL query:
1. Start with SELECT to choose the employee names.
2. FROM the employees table.
3. JOIN with the departments table to get department information.
4. Use WHERE to filter for the Engineering department.
5. The join condition will be on the department_id in employees and id in departments.
</thought_process>
<sql>
SELECT employees.name
FROM employees
JOIN departments ON employees.department_id = departments.id
WHERE departments.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<sql>
SELECT employees.name
FROM employees
JOIN departments ON employees.department_id = departments.id
WHERE departments.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<sql>
SELECT employees.name
FROM employees
JOIN departments ON employees.department_id = departments.id
WHERE departments.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<sql>
SELECT e.name
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<sql>
SELECT e.name
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
1. We need to join the employees and departments tables to get the department information for each employee.
2. We'll match employees.department_id with departments.id to connect the two tables.
3. We'll filter for the 'Engineering' department name.
4. We only need to return the employee names.
</thought_process>
<sql>
SELECT e.name
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
1. We need to join the employees and departments tables to get the department information for each employee.
2. We'll match employees.department_id with departments.id to connect the two tables.
3. We'll filter for the 'Engineering' department name.
4. We only need to return the employee names.
</thought_process>
<sql>
SELECT e.name
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
To solve this query, I will need to join the `employees` and `departments` tables on the `department_id` and `id` columns, and then select the `name` column from the `employees` table where the `name` column in the `departments` table is 'Engineering'.
The key steps are:
1. Join the `employees` and `departments` tables on the `department_id` and `id` columns.
2. Filter the results to only include rows where the `name` column in the `departments` table is 'Engineering'.
3. Select the `name` column from the `employees` table.
</thought_process>
<sql>
SELECT e.name
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
To solve this query, I will need to join the `employees` and `departments` tables on the `department_id` and `id` columns, and then select the `name` column from the `employees` table where the `name` column in the `departments` table is 'Engineering'.
The key steps are:
1. Join the `employees` and `departments` tables on the `department_id` and `id` columns.
2. Filter the results to only include rows where the `name` column in the `departments` table is 'Engineering'.
3. Select the `name` column from the `employees` table.
</thought_process>
<sql>
SELECT e.name
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
| 3 |
How many employees are in the Engineering department? |
[PASS] (1.00)
<sql>
SELECT COUNT(*)
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<sql>
SELECT COUNT(*) FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
1. We need to count employees in the Engineering department.
2. This requires joining the employees and departments tables.
3. We'll match employees.department_id with departments.id.
4. We need to filter for the Engineering department.
5. We'll use COUNT(*) to get the number of employees.
6. Since we're counting all employees in the department, we don't need to select any specific columns from the employees table.
</thought_process>
<sql>
SELECT COUNT(*)
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
To answer this query, we need to:
1. Join the employees and departments tables since the department name is in the departments table, but we need to count employees.
2. Filter for the Engineering department.
3. Count the number of employees in that department.
We'll use:
- A JOIN between employees and departments tables on department_id
- A WHERE clause to filter for the Engineering department
- COUNT(*) to count the number of employees
</thought_process>
<sql>
SELECT COUNT(*) AS employee_count
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<sql>
SELECT COUNT(*) AS num_employees
FROM employees
WHERE department_id = (
SELECT id
FROM departments
WHERE name = 'Engineering'
);
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<sql>
SELECT COUNT(*) AS num_employees
FROM employees
WHERE department_id = (
SELECT id
FROM departments
WHERE name = 'Engineering'
);
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<sql>
SELECT COUNT(*) FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<sql>
SELECT COUNT(*) FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
1. We need to join the employees and departments tables to get the department information for each employee.
2. We'll match employees.department_id with departments.id to connect the two tables.
3. We'll filter for the 'Engineering' department name.
4. We'll use the COUNT function to count the number of rows that match the filter.
</thought_process>
<sql>
SELECT COUNT(*) FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
1. We need to join the employees and departments tables to get the department information for each employee.
2. We'll match employees.department_id with departments.id to connect the two tables.
3. We'll filter for the 'Engineering' department name.
4. We'll use the COUNT function to count the number of rows that match the filter.
</thought_process>
<sql>
SELECT COUNT(*) FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
To solve this query, I will need to join the `employees` and `departments` tables on the `department_id` column in the `employees` table and the `id` column in the `departments` table. This will allow me to access the department name and count the number of employees in the "Engineering" department.
The key steps are:
1. Join the `employees` and `departments` tables on the appropriate columns.
2. Filter the results to only include rows where the department name is "Engineering".
3. Count the number of rows in the filtered result to get the total number of employees in the Engineering department.
</thought_process>
<sql>
SELECT COUNT(*) AS num_employees
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
To solve this query, I will need to join the `employees` and `departments` tables on the `department_id` column in the `employees` table and the `id` column in the `departments` table. This will allow me to access the department name and count the number of employees in the "Engineering" department.
The key steps are:
1. Join the `employees` and `departments` tables on the appropriate columns.
2. Filter the results to only include rows where the department name is "Engineering".
3. Count the number of rows in the filtered result to get the total number of employees in the Engineering department.
</thought_process>
<sql>
SELECT COUNT(*) AS num_employees
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering';
</sql>
Pass Reason: All assertions passed |
| 4 |
Give me the name, age, and salary of the oldest employee in the Engineering department. |
[PASS] (1.00)
<sql>
SELECT e.name, e.age, e.salary
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering'
ORDER BY e.age DESC
LIMIT 1
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<sql>
SELECT e.name, e.age, e.salary
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering'
ORDER BY e.age DESC
LIMIT 1;
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
1. We need to join the employees and departments tables to filter for the Engineering department.
2. We'll match employees.department_id with departments.id.
3. We need to filter for the Engineering department.
4. To find the oldest employee, we'll order by age in descending order and limit to 1 result.
5. We need to select the name, age, and salary columns from the employees table.
</thought_process>
<sql>
SELECT e.name, e.age, e.salary
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering'
ORDER BY e.age DESC
LIMIT 1;
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
To solve this query, I'll follow these steps:
1. We need to join the employees and departments tables to get the department information.
2. We need to filter for employees in the Engineering department.
3. We need to find the oldest employee, which means the maximum age.
4. We need to select the name, age, and salary of this employee.
5. We'll use a subquery to find the maximum age in the Engineering department.
6. Then we'll use this subquery in the main query to get the desired information.
</thought_process>
<sql>
SELECT e.name, e.age, e.salary
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering'
AND e.age = (
SELECT MAX(e2.age)
FROM employees e2
JOIN departments d2 ON e2.department_id = d2.id
WHERE d2.name = 'Engineering'
)
LIMIT 1;
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
Here is the SQL query to get the name, age, and salary of the oldest employee in the Engineering department:
<sql>
SELECT e.name, e.age, e.salary
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering'
ORDER BY e.age DESC
LIMIT 1;
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
Here is the SQL query to get the name, age, and salary of the oldest employee in the Engineering department:
<sql>
SELECT e.name, e.age, e.salary
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering'
ORDER BY e.age DESC
LIMIT 1;
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<sql>
SELECT e.name, e.age, e.salary
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering'
ORDER BY e.age DESC
LIMIT 1;
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<sql>
SELECT e.name, e.age, e.salary
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering'
ORDER BY e.age DESC
LIMIT 1;
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
1. We need to work with the employees and departments tables to get the information for the employees in the Engineering department.
2. We'll join the two tables on the department_id foreign key.
3. We'll filter for the Engineering department by checking the name column in the departments table.
4. To get the oldest employee, we'll order the results by the age column in descending order and take the first result.
5. We'll select the name, age, and salary columns from the employees table.
</thought_process>
<sql>
SELECT e.name, e.age, e.salary
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering'
ORDER BY e.age DESC
LIMIT 1;
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
1. We need to work with the employees and departments tables to get the information for the employees in the Engineering department.
2. We'll join the two tables on the department_id foreign key.
3. We'll filter for the Engineering department by checking the name column in the departments table.
4. To get the oldest employee, we'll order the results by the age column in descending order and take the first result.
5. We'll select the name, age, and salary columns from the employees table.
</thought_process>
<sql>
SELECT e.name, e.age, e.salary
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering'
ORDER BY e.age DESC
LIMIT 1;
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
To solve this query, I will need to follow these steps:
1. Identify the relevant tables and columns needed to answer the query.
- The relevant tables are 'employees' and 'departments'.
- The relevant columns are 'name', 'age', 'salary', and 'department_id' from the 'employees' table, and 'name' from the 'departments' table.
2. Filter the 'employees' table to only include employees from the 'Engineering' department.
3. Order the filtered results by the 'age' column in descending order to get the oldest employee.
4. Select the 'name', 'age', and 'salary' columns from the filtered and ordered results.
</thought_process>
<sql>
SELECT e.name, e.age, e.salary
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering'
ORDER BY e.age DESC
LIMIT 1;
</sql>
Pass Reason: All assertions passed |
[PASS] (1.00)
<thought_process>
To solve this query, I will need to follow these steps:
1. Identify the relevant tables and columns needed to answer the query.
- The relevant tables are 'employees' and 'departments'.
- The relevant columns are 'name', 'age', 'salary', and 'department_id' from the 'employees' table, and 'name' from the 'departments' table.
2. Filter the 'employees' table to only include employees from the 'Engineering' department.
3. Order the filtered results by the 'age' column in descending order to get the oldest employee.
4. Select the 'name', 'age', and 'salary' columns from the filtered and ordered results.
</thought_process>
<sql>
SELECT e.name, e.age, e.salary
FROM employees e
JOIN departments d ON e.department_id = d.id
WHERE d.name = 'Engineering'
ORDER BY e.age DESC
LIMIT 1;
</sql>
Pass Reason: All assertions passed |
Mahesh Murag