問(wèn)題描述

我必須用一對(duì)多的關(guān)系建模，我試圖用它來(lái)區(qū)分我的記錄類(lèi)型.假設(shè)第一個(gè)模型專(zhuān)用于 Book 信息，第二個(gè)模型是一些類(lèi)型，例如 A、B、C，并且 Type 表與 Book 之間存在間接關(guān)系，因此每本書(shū)可以是 A、B 或 C 或任何可能的組合類(lèi)型.我想使用串聯(lián)(或注釋中的任何其他可能的函數(shù)來(lái)收集字段中的所有類(lèi)型).

I have to models with one to many relation with which I try to distinguish the type of my records. Let's say First model is dedicated to Book information and second model is some types such as A, B , C and there is an indirect relation from the Type table to Book, so each book could be A, B or C or any possible combination of Types. I want to use concatenation (or any other possible function in annotation to gather all the types in a field).

Book.objects.all(
).annotate(
    Types = F('TableRelation__Type__Name')
).annotate(
    CombinedTypes = Concat('Types')
)

拋出錯(cuò)誤，因?yàn)橹挥幸粋€(gè)參數(shù)被傳遞給連接.我正在尋找的結(jié)果是一個(gè)針對(duì) Book 的任何唯一 id 填充了ABAB"的 CombinedTypes 字段，這表明該記錄是AB"(或 A、B 和 C 的任何其他組合).

which throws an error since only one argument is passed to be Concatenated. The result I am looking for is a CombinedTypes field filled with "ABAB" for any unique id of Book which shows that that record is an "AB" (or any other combination of A,B and C).

我怎樣才能做到這一點(diǎn)?

How can I achieve this?

推薦答案

我最終使用了 this 中解釋的解決方案@Ahmad 在評(píng)論中介紹的帖子中的堆棧溢出答案.

I ended up using the solution explained in this Stack-overflow answer in the post introduced by @Ahmad in comments.

這篇關(guān)于連接與 Django 中查詢(xún)集中記錄的多行相關(guān)的字段的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！