有一个查询,如下所示
q = 'select distinct t.uuid, t.tags
from tags as t
inner join tag_item ti on t.uuid = ti.item_id
inner join tag_words tw on ti.tag_id = tw.id
where tw.tag in (?) and ti.item_id in (?)'如果我使用err := s.db.QueryContext(ctx, &resp, q, p1, p2),它工作得很好
但我并不总是有p1和p2,所以我希望能够很好地将它们分开。我尝试了这样的东西:
var resp []*models.TagRecord
q := s.db.ModelContext(ctx, &resp).ColumnExpr("distinct tags.uuid, tags.tags")
q = q.Join("inner join tag_item as ti").JoinOn("ti.item_id=tags.uuid")
if len(filters.UUIDs) != 0 {
q = q.JoinOn("ti.item_id IN (?)", pg.In(filters.UUIDs))
}
q = q.Join("inner join tag_words as tw").JoinOn("tw.id=ti.tag_id")
if len(filters.Tags) != 0 {
q = q.JoinOn("tw.tag IN (?)", pg.In(filters.Tags))
}
err := q.Select()但是我得到了这个错误:
错误#42P01对表“tags”的FROM-子句条目的引用无效)
我做错了什么?我如何才能让它工作?
发布于 2021-04-29 15:08:25
我有一个聪明的想法来记录查询在普通sql中的样子,tx到这个线程的答案convert go-pg query into plain sql做到了,并弄清楚了如何让上面的代码工作,它被证明是相当简单的:
var resp []*models.TagRecord
q := s.db.ModelContext(ctx, &resp).ColumnExpr("distinct tag_record.uuid, tag_record.tags")
q = q.Join("inner join tag_item as ti").JoinOn("tag_record.uuid=ti.item_id")
if len(filters.UUIDs) != 0 {
q = q.JoinOn("ti.item_id IN (?)", pg.In(filters.UUIDs))
}
q = q.Join("inner join tag_words as tw").JoinOn("ti.tag_id=tw.id")
if len(filters.Tags) != 0 {
q = q.JoinOn("tw.tag IN (?)", pg.In(filters.Tags))
}
err := q.Select()表标记别名为tag_record,因此更改了对该别名的调用,一切就绪)
https://stackoverflow.com/questions/67307137
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