我是一名图形设计师，尽我所能来理解表别名，但是它不起作用。这是我到目前为止的内容：
SELECT colours.colourid AS colourid1, combinations.manufacturercolourid AS colourmanid1, colours.colourname AS colourname1, colours.colourhex AS colourhex1, combinations.qecolourid2 AS colouridqe2, colours.colourid AS colourid2, colours.colourname AS colourname2, colours.colourhex AS colourhex2, colours.colourid AS colourid3, combinations.qecolourid3 AS colouridqe3, colours.colourname AS colourname3, colours.colourhex AS colourhex3, colours.colourid AS colourid4, combinations.qecolourid4 AS colouridqe4, colours.colourname AS colourname4, colours.colourhex AS colourhex4, combinations.coloursupplierid FROM combinations INNER JOIN colours ON colours.colourid = combinations.manufacturercolourid;
现在，想法是在颜色查找表中，id将从查找表中提取颜色代码，十六进制和名称，以便我可以提取要查找的4种颜色的颜色代码，十六进制和名称。我可以使用它，但是它只会产生名字，代码和十六进制，而我只是看不到我做错了什么。
您的问题是您只链接了colors表中的单个记录，因为您的SQL中只有一个JOIN。该记录将与Manufacturer_colour_id指定的颜色匹配。
您可能还会遇到另一个问题，即您的组合表似乎没有采用正确的标准格式（尽管我可能是错的，因为您不知道要表示的数据的实际性质）。
如果我正确理解了您的问题，那么解决方案（使用您当前的表结构）将更像是：
SELECT C1.colourid AS colourid1, CMB.manufacturercolourid AS colourmanid1, C1.colourname AS colourname1, C1.colourhex AS colourhex1, CMB.qecolourid2 AS colouridqe2, C2.colourid AS colourid2, C2.colourname AS colourname2, C2.colourhex AS colourhex2, C3.colourid AS colourid3, CMB.qecolourid3 AS colouridqe3, C3.colourname AS colourname3, C3.colourhex AS colourhex3, C4.colourid AS colourid4, CMB.qecolourid4 AS colouridqe4, C4.colourname AS colourname4, C4.colourhex AS colourhex4, CMB.coloursupplierid FROM combinations CMB LEFT OUTER JOIN colours C1 ON C1.colourid = CMB.manufacturercolourid LEFT OUTER JOIN colours C2 ON C2.colourid = CMB.qecolourid2 LEFT OUTER JOIN colours C3 ON C3.colourid = CMB.qecolourid3 LEFT OUTER JOIN colours C4 ON C4.colourid = CMB.qecolourid4
这里发生的是，我将颜色表链接了四次，对于组合表中的每个colour_id字段一次。为此，我每次都需要对表名使用别名，以便我知道在返回的列的列表中使用四种可能的颜色实例中的哪一种。另外，如果一个或多个colour_id列可能为空，我将使用OUTER JOIN。如果使用INNER JOINs发生这种情况，则整个行将从结果集中删除。
