How can the "SELECT" statement be used to determine the output value depending on the values of a column using "IF"?

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The "IF" statement is not used in a "SELECT" statement. Instead, the "CASE" statement can be used to determine the output value depending on the values of a column.

The syntax for using "CASE" statement in a "SELECT" statement is as follows:

SELECT column1, column2, 
  CASE 
    WHEN condition1 THEN output_value1 
    WHEN condition2 THEN output_value2 
    ELSE default_output_value 
  END as output_column 
FROM table_name;

Here, "column1" and "column2" are the columns in the table "tablename". "condition1" and "condition2" are the conditions that are evaluated for each row in the table. If "condition1" is true for a row, "outputvalue1" is returned for the "outputcolumn" for that row. Similarly, if "condition2" is true, "outputvalue2" is returned for the "outputcolumn". If none of the conditions are true, "defaultoutputvalue" is returned for the "outputcolumn".

For example, suppose we have a table "students" with columns "name" and "grade". Now, we want to create a column "result" which indicates whether the student has passed or failed based on their grade. We can use the following query:

SELECT name, grade, 
  CASE 
    WHEN grade >= 60 THEN 'Pass' 
    ELSE 'Fail' 
  END as result 
FROM students;

In this query, if the grade is greater than or equal to 60, "Pass" is returned for the "result" column, otherwise "Fail" is returned.