完美解决Column ‘xxx‘ in field list is ambiguous问题

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

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。

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：

•