1. 复现问题

使用如下SQL查询数据：

SELECT
  id as id,
	`user`.login_name AS user_mobile,
	apply_status,
	( SELECT `value` FROM data_dict WHERE CODE = apply_status ) AS apply_status_value,
	apply_no,
	application_name,
	belong_org_code,
	belong_org_data_dict.`value` AS belong_org_code_value,
	business_contact_name,
	business_contact_mobile,
	auth_way,
	( SELECT `value` FROM data_dict WHERE CODE = auth_way ) AS auth_way_value,
	REPLACE ( REPLACE ( apply_service_type, ",  ), ", " ) apply_service_type,
	apply_need_desc,
	apply_time,
	audit_time,
	audit_suggest 
FROM
	application_apply 
LEFT JOIN user ON user.id = application_apply.apply_user_id
LEFT JOIN data_dict belong_org_data_dict ON belong_org_data_dict.`code` = application_apply.belong_org_code

却报出如下错误：

即Column id in field list is ambiguous

2. 分析问题

我们在解决问题之前，首先要分析问题。

做到知其然，知其所以然，这样才能有所成长，进而避坑。

将Column id in field list is ambiguous翻译成中文就是字段列表中的列id不明确。

为什么不明确这个id呢？

通过如上的·mysql语句可得，application_apply表关联user表，但 application_apply表中存在id字段，而user表中也存在id字段。

但如上mysql语句，并没有说明id字段是哪张表中的，因而mysql认为这个id字段是不明确的。

3. 解决问题

既然知道问题的原因，我们便可如下修改SQL语句：

SELECT
  	application_apply.id as id,
	`user`.login_name AS user_mobile,
	apply_status,
	( SELECT `value` FROM data_dict WHERE CODE = apply_status ) AS apply_status_value,
	apply_no,
	application_name,
	belong_org_code,
	belong_org_data_dict.`value` AS belong_org_code_value,
	business_contact_name,
	business_contact_mobile,
	auth_way,
	( SELECT `value` FROM data_dict WHERE CODE = auth_way ) AS auth_way_value,
	REPLACE ( REPLACE ( apply_service_type, ",  ), ", " ) apply_service_type,
	apply_need_desc,
	apply_time,
	audit_time,
	audit_suggest 
FROM
	application_apply 
LEFT JOIN user ON user.id = application_apply.apply_user_id
LEFT JOIN data_dict belong_org_data_dict ON belong_org_data_dict.`code` = application_apply.belong_org_code

即在id前加上application_apply.

查询结果如下图所示：

4. 问题总结

Column ‘xxx‘ in field list is ambiguous这种错误，往往存在于多表查询语句中。

多表存在相同的字段，但select语句中并没有指名该字段属于哪张表，便报出了这种错误。