FIND_IN_SET() vs. IN() in MySQL: When Should I Use Each?

Find_in_set() vs IN(): Understanding the Difference

In MySQL, the FIND_IN_SET() and IN() operators play crucial roles in querying data from tables. While both can be used to test for membership in a set of values, they differ in how they handle specific scenarios.

Consider the following query, which aims to retrieve the names of companies associated with an order:

SELECT name
FROM orders, company
WHERE orderID = 1 AND FIND_IN_SET(companyID, attachedCompanyIDs)

This query successfully yields all three company names: Company 1, Another Company, and StackOverflow. However, if we replace FIND_IN_SET() with IN(), the result is unexpected:

SELECT name
FROM orders, company
WHERE orderID = 1 AND companyID IN (attachedCompanyIDs)

In this case, only the first company name, Company 1, is returned. To understand why, it's essential to grasp the underlying mechanics of these operators.

FIND_IN_SET()

FIND_IN_SET() performs a literal search for a specific value within a comma-separated string. In this instance, attachedCompanyIDs contains a string like '1,2,3'. FIND_IN_SET() returns the position of the target value (1, 2, or 3) within this string, or 0 if not found.

IN()

IN(), on the other hand, expects a list of values to compare against the target value. However, in MySQL, attachedCompanyIDs is a scalar value, not an array or list. Thus, MySQL attempts to cast this scalar string into an integer (the type of companyID) for the IN() comparison.

During casting, MySQL interprets the first comma in the string ('1,2,3') as the end of the numeric part, resulting in an integer value of 1. In essence, the IN() condition effectively becomes:

companyID IN (1)

This explains why only the first company name, Company 1, is retrieved. The IN() condition is true only for the first value in the comma-separated string.

Solutions

If the assumption holds that the number of values in the comma-separated lists is limited, one can employ alternative approaches to handle such scenarios effectively.

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