Question

我有一个运行2个相当慢的SQL查询的脚本几乎完全相同。从数据库中选择应该在页面上“特色”的记录，即它们首先出现。另一个从数据库中选择常规记录。每个查询大约需要0.3秒，并且由于两个查询同时运行，因此我们会丢失0.6秒。所以我想知道是否可以组合这些查询。这是他们的样子：

SELECT * FROM records WHERE [other where conditions] AND featured>=$timestamp ORDER BY featured DESC

SELECT * FROM records WHERE [other where conditions] AND featured<$timestamp ORDER BY created DESC

在两种情况下，我标记的[其他条件]相同。所以我基本上需要做的是：

SELECT * FROM records WHERE [other where conditions] ORDER BY featured DESC UNLESS featured < $timestamp, created DESC

除了我不知道如何告诉MySQL“ORDER BY ... UNLESS”。

Answer 1

您可以单独使用order by执行此操作：

SELECT *
FROM records
WHERE [other where conditions]
ORDER BY (case when features >= $timestamp then featured else $timestamp end) DESC,
         created desc;

Answer 2

您可以使用IF()。

SELECT * FROM records
WHERE [other where conditions]
ORDER BY IF(featured >= $timestamp, featured, created) DESC

Answer 3

是的，你可以这样做，代码是这样的：

SELECT * FROM records 
    WHERE [other where conditions] 
ORDER BY 
    IF(featured < $timestamp, created, featured) DESC

php版本将是这样的：

$timestamp = now();
$SQLCond = "id = 4";
$SQLLimit = "LIMIT 0, 100";
$SQLMask = "SELECT * FROM records where " . $SQLCond . " ORDER BY IF(featured < " . mysql_real_escape_string($timestamp) . ", created, featured) DESC " . $SQLLimit;

Answer 4

你可以在返回的行中创建一个特色标志，如下所示：

SELECT CASE WHEN featured>=$timestamp THEN 1 ELSE 0 END AS `featuredRecord`, *
FROM records
WHERE [other where conditions]
ORDER BY featured DESC

第一条记录应该是特色记录，其余记录是常规记录。

除非value1＆lt; 0，然后按value2排序

4 个答案: