2016-11-16 36 views

回答

0

假設你的問題是關係到另一個(Sqlalchemy dynamically create joins),在這裏你去(我伸出接受的答案):

s_stmt = [ 
    col.label(col.name + '_' + ti.name) 
    for ti in t for col in ti.columns 
] 

t_joined = t[0] 
for ti in t[1:]: 
    t_joined = t_joined.join(ti, ti.c.id == t[0].c.id, full=False) 


result = select(s_stmt).select_from(t_joined) 
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