复现问题
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使用如下SQL查询数据：
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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_idLEFT 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：
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即
Column 'id' in field list is ambiguous
分析问题
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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。
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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。
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为什么不明确这个id呢？
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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
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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_idLEFT 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： -
来源地址：https://blog.csdn.net/Andrew_Chenwq/article/details/129253748
