sqlite模糊查询

Posted 2023-04-17

参考技术A 主要是glob、like两种方法.

like可以设置是否区分大小写-默认不区分,_代表一个字符,%代表不确定数量个字符

glob区分大小写,?代表一个字符,*代表不确定数量个字符

示例代码

Class.forName("org.sqlite.JDBC");

String dbPath1 = "C:sqlitetestq.db";

new File(dbPath1).delete();

Connection conn = DriverManager.getConnection("jdbc:sqlite:"+dbPath1);

Statement stat = conn.createStatement();

stat.executeUpdate("create table if not exists stu1(name varchar(20), age int, score int) ;");

stat.executeUpdate("insert into stu1 values('A1',12,72);");

stat.executeUpdate("insert into stu1 values('A2',13,73);");

stat.executeUpdate("insert into stu1 values('A3',14,74);");

stat.executeUpdate("insert into stu1 values('D33',14,74);");

//stat.executeUpdate("PRAGMA case_sensitive_like =1;");

stat.executeUpdate("PRAGMA case_sensitive_like =0;");

ResultSet rs = stat.executeQuery("select s1.name,s1.age,s1.score from stu1 s1 where s1.name like '%3' ;");

while (rs.next())

System.out.println("name = " + rs.getString("name")+" age:"+rs.getInt("age")+" score:"+rs.getInt("score"));



rs.close();

rs = stat.executeQuery("select s1.name,s1.age,s1.score from stu1 s1 where s1.name like 'A_' ;");

while (rs.next())

System.out.println("name = " + rs.getString("name")+" age:"+rs.getInt("age")+" score:"+rs.getInt("score"));



rs.close();

rs = stat.executeQuery("select s1.name,s1.age,s1.score from stu1 s1 where s1.name like 'a_' ;");

while (rs.next())

System.out.println("name = " + rs.getString("name")+" age:"+rs.getInt("age")+" score:"+rs.getInt("score"));



rs.close();

rs = stat.executeQuery("select s1.name,s1.age,s1.score from stu1 s1 where s1.name GLOB 'd3*' ;");

while (rs.next())

System.out.println("name = " + rs.getString("name")+" age:"+rs.getInt("age")+" score:"+rs.getInt("score"));



rs.close();

System.out.println("----");

rs = stat.executeQuery("select s1.name,s1.age,s1.score from stu1 s1 where s1.name glob 'D3?' ;");

while (rs.next())

System.out.println("name = " + rs.getString("name")+" age:"+rs.getInt("age")+" score:"+rs.getInt("score"));



rs.close();

stat.close();

conn.close();

输出

name = A3 age:14 score:74

name = D33 age:14 score:74

name = A1 age:12 score:72

name = A2 age:13 score:73

name = A3 age:14 score:74

name = A1 age:12 score:72

name = A2 age:13 score:73

name = A3 age:14 score:74

----

name = D33 age:14 score:74

Sqlite模糊查询通配符%和下划线无法匹配问题

问题描述：

        使用以下like语句查询，

    select * from table where columnName like \'%_%\';

         结果是返回全部的记录，不是想要的结果！

        网上搜索一下，原因如下：

        在like语句中的“下划线”代表匹配“任意一个字符”，“%”代表匹配“任意多个字符”。

解决方案：

方案一：使用escape转义

        escape的内容可以任意，只要保证前后一致即可

    select * from table where columnName like \'%\\_%\' escape \'\\\';
    select * from table where columnName like \'%|_%\' escape \'|\';
    se