java - JPQL (JPA) 搜索子字符串

Question

我面临着一个简单的问题，即通过实体可能包含的某些(子)字符串搜索实体。

例如我有用户 user1、usr2、useeeer3、user4，我将进入搜索窗口“use”，我希望返回 user1、useeer3、user4。

我相信你现在明白我的意思了。 JPA(JQPL)中有构造吗？在命名查询中以某种方式使用 WHERE 进行搜索会很好。类似于“SELECT u FROM User u WHERE u.nickname contains :substring”

Answer 1

使用 LIKE 表达式。以下是 JPA 1.0 规范 (JSR 220) 的 4.6.9 Like Expression 部分的引述:

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.