mySQL UNION故障

Question

我在使用涉及联合的sql连接时遇到了问题。我试图从两个表中拉出一个计数和一个字段，但是得到了一个错误。

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];

是不是和date as alertdate部件有关？

我得到的错误是：

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

Answer 1

除了COUNT和(*)之间的空间之外，还有另一个问题。您不能在PHP代码中使用$rs7[mailcount]或$rs7[maildate]，因为您的查询等同于：

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'

并将返回两行且仅返回两列：

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

有两种方法可以解决这个问题：

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以使用这2行。

或将查询更改为：

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