Question

我想这样做：

SELECT * FROM `langCategories` ORDER BY `name` ASC WHERE ORDER BY `amount` DESC LIMIT 8,0

但这不可能。

我的网站中有2个部分：的 1。前8个人口类别

(SELECT * FROM `langCategories` ORDER BY `langCategories`.`amount` DESC LIMIT 0,8)

2。类别的其余部分按名称排序（不包括前8名） （???）

我用php做到了：

$var = db_multiselect("SELECT * FROM `langCategories` ORDER BY `amount` DESC LIMIT 0,8");
$var2 = db_multiselect("SELECT * FROM `langCategories` ORDER BY `name` ASC");

$i=-1;
while ($row = mysqli_fetch_array($var)) // Save the top8 categories
{
   ++$i;
   $top8[$i] = $row["name"];
}

$i=0;
while ($row2 = mysqli_fetch_array($var2))
{
   if (!(in_array($row2["name"], $top8))) // compare the rest of categories excluding top 8
   {
      ...
   }
}

但我想用MySQL查询做到这一点，怎么做呢？

Answer 1

子查询！

SELECT * FROM (
  SELECT * FROM `langCategories` 
  ORDER BY `langCategories`.`amount` DESC 
  LIMIT 8,2000000000
) AS baseview 
ORDER BY name ASC

按字母顺序为您提供NOT前8个类别

Answer 2

这应该适用于MySQL（我希望）：

SELECT *
FROM (
  SELECT lc.*, @rownum:=@rownum+1 AS rownum
  FROM `langCategories` lc, (SELECT @rownum:=0) r
  ORDER BY `amount` DESC
)
ORDER BY CASE WHEN rownum <= 8 THEN rownum ELSE 999 END, name

在与MySQL相同的表中排除第二个查询的第一个查询的结果

2 个答案: