完美解决Column ‘xxx‘ in field list is ambiguous问题-Toy模板网

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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, '"', '' ), '&quot;', '"' ) 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, '"', '' ), '&quot;', '"' ) 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