mySQL UNION 错误

Question

我在涉及联合的 sql 连接方面遇到问题。我试图从 2 个表中提取 COUNT 和一个字段，但出现错误。

查询：

$sql_result7 = mysql_query("(SELECT COUNT (*) as alertcount, date as alertdate FROM alerts WHERE to_id='$id' AND date > '$lastcheck') UNION (SELECT COUNT (*) as mailcount, date maildate FROM mobmail WHERE to_id='$id' AND to_del=0 AND seen = '0')", $db); 

$rs7 = mysql_fetch_array($sql_result7);
$alerts = $rs7[alertcount]; 
$mails = $rs7[mailcount]; 
$last_alert = $rs7[alertdate]; 
$last_mail = $rs7[maildate];

这与 date as alertdate 部分有关吗？

我得到的错误是：

警告：mysql_fetch_array()：提供的参数不是有效的 MySQL 结果资源

Answer 1

除了 COUNT 和 (*) 之间的空格之外，还有另一个问题。您不能在 PHP 代码中使用 $rs7[mailcount] 或 $rs7[maildate]，因为您的查询相当于：

SELECT 
      COUNT(*) as alertcount
    , date as alertdate 
FROM alerts 
WHERE to_id = '$id' 
  AND date > '$lastcheck' 
UNION 
SELECT 
      COUNT(*)                      --- No "as mailcount" here
    , date                          --- No "as maildate" either 
FROM mobmail 
WHERE to_id = '$id' 
  AND to_del = 0 
  AND seen = '0'

并且将返回两行且仅 2 列：

alertcount | alertdate
-----------|------------
 24        | 2012-01-04
 73        | 2011-11-11

---

解决此问题的两种方法：

要么保留查询（将 UNION 更改为 UNION ALL 以确保始终获得 2 行），然后更改 PHP 以使用 2 行行。

或者将查询更改为：

SELECT alertcount, alertdate, mailcount, maildate 
FROM
      ( SELECT 
              COUNT(*) AS alertcount
            , date     AS alertdate 
        FROM alerts 
        WHERE to_id = '$id' 
          AND date > '$lastcheck' 
      ) AS a 
  CROSS JOIN
      ( SELECT 
              COUNT(*) AS mailcount
            , date     AS maildate 
        FROM mobmail 
        WHERE to_id = '$id' 
          AND to_del = 0 
          AND seen = '0'
      ) AS b