在MySQL中为多个属性添加列

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我真的不确定如何解释这一点,因为今天我才开始尝试学习如何以更高级的方式使用MySQL查询.这是我所拥有的：

SELECT pa.displayvalue as Brand
FROM product p
JOIN categoryheader ch ON ch.templatetype = 0 AND p.categoryid=ch.categoryid AND ch.isactive=1
JOIN categorydisplayattributes cda ON ch.headerid = cda.headerid AND ch.categoryid = cda.categoryid AND cda.templatetype = ch.templatetype AND cda.isactive=1
JOIN productattribute pa ON p.productid = pa.productid AND cda.attributeid = pa.attributeid AND pa.localeid = 1 AND pa.isactive =1
JOIN attributenames an ON pa.attributeid = an.attributeid AND an.localeid = 1
WHERE p.isactive = 1 AND p.categoryid = 4871
AND an.attributeid = 113319;

因此,我有了该查询,从字面上看,我要做的就是为“ And an.attributeid = 113319”添加另一列

我只想添加另一列,而不是113319,我希望它将例如1762的值拉到其旁边的列中.

因此,第1列将具有“和an.attributeid = 113319”提取的所有值,而第2列将具有“和an.attributeid = 1762”提取的所有值.

我得到的这段代码可以工作：

SELECT pa.displayvalue as Brand
FROM product p
JOIN categoryheader ch ON ch.templatetype = 0 AND p.categoryid=ch.categoryid AND ch.isactive=1
JOIN categorydisplayattributes cda ON ch.headerid = cda.headerid AND ch.categoryid = cda.categoryid AND cda.templatetype = ch.templatetype AND cda.isactive=1
JOIN productattribute pa ON p.productid = pa.productid AND cda.attributeid = pa.attributeid AND pa.localeid = 1 AND pa.isactive =1
JOIN attributenames an ON pa.attributeid = an.attributeid AND an.localeid = 1
WHERE p.isactive = 1 AND p.categoryid = 4871
AND an.attributeid = 113319;

SELECT pa.displayvalue as Type
FROM product p
JOIN categoryheader ch ON ch.templatetype = 0 AND p.categoryid=ch.categoryid AND ch.isactive=1
JOIN categorydisplayattributes cda ON ch.headerid = cda.headerid AND ch.categoryid = cda.categoryid AND cda.templatetype = ch.templatetype AND cda.isactive=1
JOIN productattribute pa ON p.productid = pa.productid AND cda.attributeid = pa.attributeid AND pa.localeid = 1 AND pa.isactive =1
JOIN attributenames an ON pa.attributeid = an.attributeid AND an.localeid = 1
WHERE p.isactive = 1 AND p.categoryid = 4871
AND an.attributeid = 1762;

但这在navicat中作为两个单独的结果返回.我希望这两个结果在一个结果中但作为两列,而不是两个结果各为一列.

如您所见,除了最后一行,所有这些Select中的所有行都是相同的.

我还觉得,有一种更好的方法,那就是不必为获得an.attributeid的另一列而两次使用整个代码.无论如何,我会尽力而为.

如果您有任何建议,我将非常感谢.

解决方法:

如果您确实要这样做,则出于格式化或报告目的.这是您将要使用的一种技巧.

使用LEFT JOIN将返回左表中的所有行,并在右表中返回匹配的行.如果没有匹配项,则结果在右侧为NULL.

SELECT p.productid, t2.Type, t1.Brand, t3.Resolution 
FROM product p 
JOIN 
( 
    SELECT p.productid, pa.displayvalue as Brand 
    FROM product p 
    JOIN categoryheader ch ON ch.templatetype = 0 AND p.categoryid=ch.categoryid AND ch.isactive=1 
    JOIN categorydisplayattributes cda ON ch.headerid = cda.headerid AND ch.categoryid = cda.categoryid AND cda.templatetype = ch.templatetype AND cda.isactive=1 
    JOIN productattribute pa ON p.productid = pa.productid AND cda.attributeid = pa.attributeid AND pa.localeid = 1 AND pa.isactive =1 
    JOIN attributenames an ON pa.attributeid = an.attributeid AND an.localeid = 1 
    WHERE p.isactive = 1 AND p.categoryid = 4871 
    AND an.attributeid = 113319 
) t1 ON t1.productid = p.productid 
LEFT OUTER JOIN 
( 
    SELECT p.productid, pa.displayvalue as Type 
    FROM product p 
    JOIN categoryheader ch ON ch.templatetype = 0 AND p.categoryid=ch.categoryid AND ch.isactive=1 
    JOIN categorydisplayattributes cda ON ch.headerid = cda.headerid AND ch.categoryid = cda.categoryid AND cda.templatetype = ch.templatetype AND cda.isactive=1 
    JOIN productattribute pa ON p.productid = pa.productid AND cda.attributeid = pa.attributeid AND pa.localeid = 1 AND pa.isactive =1 
    JOIN attributenames an ON pa.attributeid = an.attributeid AND an.localeid = 1 
    WHERE p.isactive = 1 AND p.categoryid = 4871 
    AND an.attributeid = 1100 
) t2 ON t2.productid = p.productid 
LEFT OUTER JOIN 
( 
    SELECT p.productid, pa.displayvalue as Resolution 
    FROM product p 
    JOIN categoryheader ch ON ch.templatetype = 0 AND p.categoryid=ch.categoryid AND ch.isactive=1 
    JOIN categorydisplayattributes cda ON ch.headerid = cda.headerid AND ch.categoryid = cda.categoryid AND cda.templatetype = ch.templatetype AND cda.isactive=1 
    JOIN productattribute pa ON p.productid = pa.productid AND cda.attributeid = pa.attributeid AND pa.localeid = 1 AND pa.isactive =1 
    JOIN attributenames an ON pa.attributeid = an.attributeid AND an.localeid = 1 
    WHERE p.isactive = 1 AND p.categoryid = 4871 
    AND an.attributeid = 1762 
) t3 ON t3.productid = p.productid

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