ILIKE ANY¶

Syntax¶

<subject> ILIKE ANY (<pattern1> [, <pattern2> ... ] ) [ ESCAPE <escape_char> ]

Arguments¶

Required:

subject

The string to compare to the pattern(s).

pattern#

The pattern(s) that the string is to be compared to. You must specify at least one pattern.

Optional:

escape_char

Character(s) inserted in front of a wildcard character to indicate that the wildcard should be interpreted as a regular character rather than as a wildcard.

Returns¶

The data type of the returned value is VARCHAR.

Usage Notes¶

• SQL wildcards are supported in pattern:

  • An underscore (_) matches any single character.

  • A percent sign (%) matches any sequence of zero or more characters.

• Wildcards in pattern include newline characters (\n) in subject as matches.

• The pattern is considered a match if the pattern matches the entire input string (subject). To match a sequence anywhere within a string, start and end the pattern with %, for example ‘%something%’.

• NULL does not match NULL. In other words, if the subject is NULL and one of the patterns is NULL, that is not considered a match.

• If the function is used with a subquery, the subquery should return a single row.

  For example, the following should be used only if the subquery returns a single row:

  SELECT ...
      WHERE x ILIKE ANY (SELECT ...)
  

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Examples¶

This shows how to use ILIKE ANY:

Create a table that contains some strings:

CREATE OR REPLACE TABLE ilike_example(subject varchar(20));
INSERT INTO ilike_example VALUES
    ('jane doe'),
    ('Jane Doe'),
    ('JANE DOE'),
    ('John Doe'),
    ('John Smith');

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SELECT * 
  FROM ilike_example 
  WHERE subject ILIKE ANY ('jane%', '%SMITH')
  ORDER BY subject;
+------------+
| SUBJECT    |
|------------|
| JANE DOE   |
| Jane Doe   |
| John Smith |
| jane doe   |
+------------+

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For examples of how to use wildcard characters, see the documentation of the related function LIKE ANY.