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SQL CHOOSE() Function: Complete Guide

SQL CHOOSE() function introduced in SQL Server 2012. This SQL Choose function returns the item at the specified index from the list of items.

SQL CHOOSE

SQL CHOOSE() is a built-in function used to return an item from the list of items based on a specified index. The CHOOSE() function returns the item from a list of items at a specified index.

Syntax

CHOOSE (Index, Value1, Value2, ...., ValueN);

Parameters

  1. Index: It is an expression integer that specifies the index of the element to be returned. Here, indexing is based on 1-based.
  2. Value1, Value2, …., ValueN: It denotes the list of items separated by commas.

Note

  1. Value1 will be returned if the index specified is 1, and Value2 will be returned if the index specified is 2 and so on.
  2. If the index is not in the integer format, then it will be type-casted to an integer.
  3. If the index is out of bounds, then it will return NULL.

Query 1

Select CHOOSE (1, ‘App’, ‘Dividend’, ‘Com’);

Output

App

Explanation

The index specified was 1, and the App was indexed at the first position. So, the function returned the App as the output.

Query 2

Select CHOOSE (0, ‘App’, ‘Dividend’, ‘Com’);

Output

NULL

Explanation

Here, the index specified was 0, and as it was out of bounds, so the NULL value was returned.

Query 3

Select CHOOSE (2.3, ‘App’, ‘Dividend’, ‘Com’);

Output

Dividend

Explanation

Here, an index that was specified at decimal representation was type-casted to an integer data type, which rounded off the value to 2. So, Dividend, which was indexed at 2nd position, was returned.

Query 4

Select CHOOSE (‘3’, ‘App’, ‘Dividend’, ‘Com’);

Output

Com

Explanation

Here, an index that was specified at varchar was type-casted to the integer data type. So, Com, which was indexed at 3rd position, was returned.

That’s it for this tutorial.

See also

SQL SESSION_USER

SQL SYSTEM_USER

SQL USER_NAME

SQL CURRENT_USER

SQL ISNULL

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