Parameters (IN, OUT, INOUT) in SQL

Introduction

Parameters let you send and receive values when running a stored procedure. They help you share information between your code and the database.

You want to give a procedure some input values to work with.

You want a procedure to send back results or status after running.

You want a procedure to both receive input and send back output.

You want to reuse a procedure with different inputs without rewriting it.

You want to keep your database logic organized and clear.

Syntax

SQL

CREATE PROCEDURE procedure_name (
  IN param1 datatype,
  OUT param2 datatype,
  INOUT param3 datatype
)
BEGIN
  -- procedure body
END;

IN means the parameter is for input only.

OUT means the parameter is for output only.

INOUT means the parameter can be used for both input and output.

Examples

This procedure takes two input numbers and returns their sum as output.

SQL

CREATE PROCEDURE AddNumbers(
  IN num1 INT,
  IN num2 INT,
  OUT sum INT
)
BEGIN
  SET sum = num1 + num2;
END;

This procedure takes a number, adds 10 to it, and returns the new value using the same parameter.

SQL

CREATE PROCEDURE UpdateAndReturn(
  INOUT value INT
)
BEGIN
  SET value = value + 10;
END;

Sample Program

This procedure calculates the area of a rectangle using width and height as input, and returns the area as output.

We call it with width=5 and height=4, then select the output variable to see the result.

SQL

DELIMITER $$
CREATE PROCEDURE CalculateArea(
  IN width INT,
  IN height INT,
  OUT area INT
)
BEGIN
  SET area = width * height;
END$$
DELIMITER ;

-- Call the procedure
CALL CalculateArea(5, 4, @result);
SELECT @result AS Area;

Important Notes

You must use session variables (like @result) to get OUT or INOUT values after calling a procedure.

IN parameters cannot be changed inside the procedure.

INOUT parameters start with a value and can be changed inside the procedure.

Summary

Parameters let procedures receive inputs and send outputs.

IN is input only, OUT is output only, INOUT is both.

Use session variables to capture OUT and INOUT results after calling procedures.

Practice

1. What does the IN parameter type in a SQL stored procedure do?

easy

A. It allows the procedure to send output values only.

B. It allows the procedure to receive input values only.

C. It allows the procedure to both receive input and send output.

D. It declares a variable inside the procedure.

Solution

1. Step 1: Understand parameter types in SQL procedures

   Parameters can be IN, OUT, or INOUT. IN means input only.

2. Step 2: Identify the role of IN parameters

   IN parameters pass values into the procedure but do not return values.

3. Final Answer:

   It allows the procedure to receive input values only. -> Option B

4. Quick Check:

   IN = input only [OK]

Hint: IN means input only, no output from procedure [OK]

Common Mistakes:

• Confusing IN with OUT or INOUT
• Thinking IN parameters can return values
• Assuming IN declares internal variables

2. Which of the following is the correct syntax to declare an OUT parameter named result of type INT in a MySQL stored procedure?

easy

A. CREATE PROCEDURE proc_name(IN result INT)

B. CREATE PROCEDURE proc_name(INOUT result INT)

C. CREATE PROCEDURE proc_name(OUT result INT)

D. CREATE PROCEDURE proc_name(RESULT OUT INT)

Solution

1. Step 1: Recall parameter declaration syntax in MySQL

   Parameters are declared as IN, OUT, or INOUT followed by name and type.

2. Step 2: Check each option's syntax

   CREATE PROCEDURE proc_name(OUT result INT) uses correct syntax: OUT result INT. CREATE PROCEDURE proc_name(INOUT result INT) uses INOUT, not OUT. CREATE PROCEDURE proc_name(IN result INT) uses IN, not OUT. CREATE PROCEDURE proc_name(RESULT OUT INT) has wrong order and keyword.

3. Final Answer:

   CREATE PROCEDURE proc_name(OUT result INT) -> Option C

4. Quick Check:

   OUT parameter syntax = OUT name type [OK]

Hint: OUT parameters use OUT before name and type [OK]

Common Mistakes:

• Using IN instead of OUT
• Swapping parameter order
• Using wrong keywords like RESULT OUT

3. Consider this MySQL procedure call:

CALL calculate_sum(5, 10, @total);
SELECT @total;

Given the procedure signature:

CREATE PROCEDURE calculate_sum(IN a INT, IN b INT, OUT sum INT)
BEGIN
  SET sum = a + b;
END;

What will be the output of SELECT @total;?

medium

A. 5

B. Syntax error

C. NULL

D. 15

Solution

1. Step 1: Understand procedure parameters and call

   The procedure takes two input numbers and outputs their sum via the OUT parameter sum. The call passes 5 and 10 as inputs, and @total as the OUT variable.

2. Step 2: Calculate the output value

   The procedure sets sum = 5 + 10 = 15. This value is assigned to the session variable @total.

3. Final Answer:

   15 -> Option D

4. Quick Check:

   5 + 10 = 15 [OK]

Hint: OUT parameters update session variables after CALL [OK]

Common Mistakes:

• Expecting NULL because of no SELECT inside procedure
• Confusing IN and OUT parameters
• Not using session variables for OUT

4. You have this procedure:

CREATE PROCEDURE update_value(INOUT val INT)
BEGIN
  SET val = val + 1;
END;

Which of the following calls will correctly update the session variable @x from 10 to 11?

medium

A. CALL update_value(@x);

B. CALL update_value(10);

C. CALL update_value(INOUT @x);

D. CALL update_value(IN @x);

Solution

1. Step 1: Understand INOUT parameter usage

   INOUT parameters require a variable to pass in and receive updated value. Session variables like @x can be used.

2. Step 2: Analyze each call

   CALL update_value(@x); passes @x correctly. CALL update_value(10); passes a literal 10, which is not a variable. CALL update_value(INOUT @x); uses invalid syntax. CALL update_value(IN @x); uses IN keyword, which is incorrect for INOUT.

3. Final Answer:

   CALL update_value(@x); -> Option A

4. Quick Check:

   INOUT needs variable, not literal [OK]

Hint: Use session variables for INOUT parameters [OK]

Common Mistakes:

• Passing literals instead of variables
• Using IN or INOUT keywords in CALL
• Expecting procedure to update literals

5. You want to create a procedure that swaps two integer values using INOUT parameters a and b. Which of the following procedure bodies correctly swaps the values?

hard

A. BEGIN DECLARE temp INT; SET temp = a; SET a = b; SET b = temp; END

B. BEGIN SET a = b; SET b = a; END

C. BEGIN SET a = a + b; SET b = a - b; SET a = a - b; END

D. BEGIN SET a = b; SET b = a + b; END

Solution

1. Step 1: Understand swapping logic with INOUT parameters

   Swapping requires a temporary variable to hold one value while assigning the other.

2. Step 2: Evaluate each option's correctness

   BEGIN DECLARE temp INT; SET temp = a; SET a = b; SET b = temp; END uses a temporary variable to swap correctly. BEGIN SET a = b; SET b = a; END overwrites a before assigning b. BEGIN SET a = a + b; SET b = a - b; SET a = a - b; END uses arithmetic swap but is valid only if no overflow and no NULLs. BEGIN SET a = b; SET b = a + b; END does not swap correctly.

3. Final Answer:

   BEGIN DECLARE temp INT; SET temp = a; SET a = b; SET b = temp; END -> Option A

4. Quick Check:

   Temporary variable swap = correct [OK]

Hint: Use temp variable to swap INOUT parameters safely [OK]

Common Mistakes:

• Overwriting variables without temp
• Using arithmetic swap without checks
• Ignoring NULL or overflow risks