要求: 构建sql wrapper

根据要求构建sql wrapper：dto.getProjectId() == null，则 where project_id is nulldto.getProjectId() != null，则 where project_id ='58b1fad6-adaa-4834-8337-8d438346783a'

错误写法

 PropertyEntity property = propertyManager.getOne(Wrappers.<PropertyEntity>lambdaQuery().eq(Objects.nonNull(dto.getProjectId()), PropertyEntity::getProjectId, dto.getProjectId()).eq(StringUtils.isNotEmpty(dto.getCode()), PropertyEntity::getCode, dto.getCode()).eq(StringUtils.isNotEmpty(dto.getName()), PropertyEntity::getName, dto.getName()));

原因： dto.getProjectId() == null，构建的sql是：where project_id = ‘null’

解决方案

  LambdaQueryWrapper<PropertyEntity> wrapper = Wrappers.lambdaQuery(PropertyEntity.class);if (dto.getProjectId() == null) {wrapper.isNull(PropertyEntity::getProjectId);} else {wrapper.eq(PropertyEntity::getProjectId, dto.getProjectId());}wrapper.eq(StringUtils.isNotEmpty(dto.getCode()), PropertyEntity::getCode, dto.getCode()).eq(StringUtils.isNotEmpty(dto.getName()), PropertyEntity::getName, dto.getName());PropertyEntity property = propertyManager.getOne(wrapper);