error: column reference "id" is ambiguous 文心快码BaiduComate 当你遇到错误 error: column reference "id" is ambiguous 时,这通常意味着在你的SQL查询中,id 列在多个表中存在,而你没有明确指定你想要引用的是哪个表的 id 列。为了解决这个问题,你可以按照以下步骤操作: 确定查询中涉及的表格及其别名: 首先...
join lot l on l.id = a.lot_id and l.numero ='20230620V327'; The result is displayed You try to edit and modify a value, and the you save it. It returns this message SQL Error [42702]: ERROR: column reference "id" is ambiguous Position : 323 Script-8.sql /Prod-rw/Scripts lin...
Hi, the bug 'column reference "id" is ambiguous' is resolved. But now we have a new problem when pk name is different from default (id) and the autocomplete field is a foreign key. The problem is in "SELECT (CASE WHEN the_pk_name='877' THEN 0 END) AS "ordering", "autht..."...
设置表别名 a,b 若返回对象只需要取a的id,select a.id即可,如图.XXMapper.xml 同理 PSQLException: ERROR: column reference "XX" is ambiguous
This message indicates that the two tables to be joined contain the same column, but the owner of the column is not specified when the command is executed.For example, ta
https://stackoverflow.com/questions/37792662/column-reference-is-ambiguous-sequelize-error 我有study和study_user 两张表,它们都有一个列叫user_id, 我用study去include study_user,以下是我的查询语句: let studies = await Study.findAll({ where: { department_id: { [Op.in]: upsDepartments, }, '...
错误Column 'id' in order clause is ambiguous 易出现在 p=Product.find(:all,:include=>:kind,:order=>"id desc") 类似的语句上,用include包含了一个left join对象,所以id可能存在于多个表内,就引起了二义性 修改为 p=Product.find(:all,:include=>:kind,:order=>"products.id desc") ...
https://sentry.gitlap.com/gitlab/gitlabcom/issues/218059/ PG::AmbiguousColumn: ERROR: column reference "name" is ambiguous
awe have round faces 我们有围绕面孔[translate] aFAILED: Error in semantic analysis: Line 22:1 Ambiguous table alias or column reference c 不合格: 错误在语义分析: 排行22:1模棱两可的桌或别名专栏参考c[translate]
time=1698685733318","message":"An exception occurred while executing a query: SQLSTATE[42702]: Ambiguous column: 7 ERROR: column reference \"poll_id\" is ambiguous\nLINE 1: ...\"c1\".\"user_id\") WHERE \"p\".\"poll_id\" = $1 GROUP BY \"poll_id\",...\n ^","userAgent":...