我一直收到错误代码：1064 sintaxys mysql workbench

我一直在看这个错误，但什么也没发现

代码：

错误：

选择sv.id，sv.service\u state\u id作为sstId，sha.id，sha.address，sha.user\u guide作为guide，sha.users\u id，sst.id从deliver\u db\u v2.service作为sv internal join deliver\u db\u v2.service\u state作为sst on sv.id=sha.users\u id on sv.sstId=sst.id错误代码：1064。您的SQL语法有错误；请查看与MySQL服务器版本对应的手册，以获取第6行0.094秒处“on sv.sstId=sst.id”附近要使用的正确语法

您将第一个内部联接的ON子句放错了位置

你可以做：

SELECT
  sv.id, sv.service_state_id as sstId, sha.id, sha.address,
  sha.user_guide as guide, sha.users_id, sst.id
FROM deliver_db_v2.service as sv
inner join deliver_db_v2.service_has_address as sha on sv.id = sha.users_id
inner join deliver_db_v2.service_state as sst on sv.sstId = sst.id;

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每个内部联接都必须有自己的ON子句。

您的标题上说您使用的是MySQL Workbench，但您使用的是phpMyAdmin标记，您使用的是哪一个？

SELECT
  sv.id, sv.service_state_id as sstId, sha.id, sha.address,
  sha.user_guide as guide, sha.users_id, sst.id
FROM deliver_db_v2.service as sv
inner join deliver_db_v2.service_has_address as sha on sv.id = sha.users_id
inner join deliver_db_v2.service_state as sst on sv.sstId = sst.id;