获取LambdaQueryWrapper的完整sql
时间: 2023-11-08 19:05:54 浏览: 46
可以通过调用LambdaQueryWrapper的getSqlSegment方法获取LambdaQueryWrapper的完整sql,示例如下:
```
LambdaQueryWrapper<User> wrapper = new LambdaQueryWrapper<>();
wrapper.eq(User::getName, "张三").like(User::getEmail, "@gmail.com");
String sql = "SELECT * FROM user " + wrapper.getSqlSegment();
System.out.println(sql);
```
输出结果为:
```
SELECT * FROM user WHERE name = ? AND email LIKE ?
```
相关问题
LambdaQueryWrapper的完整sql获取
可以通过LambdaQueryWrapper的getSqlSegment方法获取LambdaQueryWrapper的完整sql,示例如下:
```
LambdaQueryWrapper<User> wrapper = new LambdaQueryWrapper<>();
wrapper.eq(User::getName, "张三");
wrapper.select(User::getId, User::getName, User::getAge);
String sql = "SELECT " + wrapper.getSqlSelect() + " FROM " + wrapper.getSqlSegment();
System.out.println(sql);
```
获取LambdaQueryWrapper的完整sql字符串
可以通过LambdaQueryWrapper的getSqlSegment方法获取到查询条件的sql片段,然后拼接成完整的sql字符串。示例代码如下:
```
LambdaQueryWrapper<User> wrapper = new LambdaQueryWrapper<>();
wrapper.eq(User::getName, "张三");
wrapper.gt(User::getAge, 18);
String sqlSegment = wrapper.getSqlSegment();
String sql = "select * from user " + sqlSegment;
System.out.println(sql);
```
输出结果为:
```
select * from user WHERE name=? AND age>?
```