具有多个条件的左访问联接 [英] ACCESS LEFT JOIN with multiple criteria

问题描述

我想做这样的事情:

从t1左联接t2上选择t1.*，t2.* t1.id = t2.id AND t1.field1 = false

因此，我不想从 t2 表中看到来自 t1 表中那些记录的数据，其中 t1.field1 = false .在ORACLE中有可能，但是在MS ACCESS中有可能吗?

So, I don't want to see data from t2 table for those records from t1 table where t1.field1=false. It's possible in ORACLE, but is this possible in MS ACCESS?

我在JOIN操作中出现语法错误，并且 JOIN表达式不受支持和函数的无效参数.

I got Syntax error in JOIN operation and JOIN expression not supported and Invalid argument to function.

Edit2:为防止进一步的误解和放入 WHERE 子句"的注释.

To prevent further misunderstanding and "put in the WHERE clause" comments.

选择t1.*，t2.*从t1左联接t2到t1.id = t2.id

如果您在ORACLE中编写我的第一个 SELECT ，您将看到此信息(我必须在MS中执行此操作).我想查看 t1 表中的所有记录，但是我不想联接每条记录，而只想联接 t1.f1 = false 的地方.您可以看到 id = 2 和 id = 5 .

If you write my very first SELECT in ORACLE, you will see this (and I have to do it in MS). I want to see ALL records from t1 table, but I don't want to join every record, but only where t1.f1=false. You can see that where id=2 and id=5.

推荐答案

我通过多个 SELECT 设法获得了期望的结果，如果您有一个长且多个 SELECT的选项，那真是令人作呕首先.

I managed to get the desired result with multiple SELECT, which is disgusting, if you have a long and multiple SELECT in the first place.

SELECT t1.*，a.c1从t1左联接(选择t2.*从t1左联接t2打开t1.id = t2.id其中t1.f1 = false)作为开启t1.id = a.id

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