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javaormjpajpql

JPQL (JPA) search substring

Im facing simple problem with searching entities by some (sub)string, which they might contain.

E.g. I have users user1, usr2, useeeer3, user4 and I will enter to search window "use" and I expect to return user1, useeer3, user4.

Im sure you know what I mean now. Is there any construction in JPA (JQPL)? It would be nice to search using WHERE somehow in named queries. Something like "SELECT u FROM User u WHERE u.nickname contains :substring"

Solution

  • Use a LIKE expression. Here is a quote from the section 4.6.9 Like Expression of the JPA 1.0 spec (JSR 220):

    The syntax for the use of the comparison operator [NOT] LIKE in a conditional expression is as follows:

    string_expression [NOT] LIKE pattern_value [ESCAPE escape_character]

    The string_expression must have a string value. The pattern_value is a string literal or a string-valued input parameter in which an underscore (_) stands for any single character, a percent (%) character stands for any sequence of characters (including the empty sequence), and all other characters stand for themselves. The optional escape_character is a single-character string literal or a character-valued input parameter (i.e., char or Character) and is used to escape the special meaning of the underscore and percent characters in pattern_value.

    Examples are:

    • address.phone LIKE ‘12%3’ is true for ‘123’ ‘12993’ and false for ‘1234’
    • asentence.word LIKE ‘l_se’ is true for ‘lose’ and false for ‘loose’
    • aword.underscored LIKE ‘\_%’ ESCAPE ‘\’ is true for ‘_foo’ and false for ‘bar’
    • address.phone NOT LIKE ‘12%3’ is false for ‘123’ and ‘12993’ and true for ‘1234’

    If the value of the string_expression or pattern_value is NULL or unknown, the value of the LIKE expression is unknown. If the escape_character is specified and is NULL, the value of the LIKE expression is unknown.