完美解决Column ‘xxx‘ in field list is ambiguous问题-蒲公英云

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

女爷i 2024-03-26 15:12 191阅读 0赞

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