mySQL UNION错误

Question

我遇到了涉及工会的sql连接问题。我试图从2个表中拉出一个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]; 

有什么事情做的date as alertdate一部分？

错误即时得到的是：

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

Answer 1

除了COUNT和(*)之间的空间，还有另外一个问题。你不能在你的PHP代码中使用$rs7[mailcount]也不$rs7[maildate]，因为你的查询等效于：

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' 

，并返回两行，只有2列：

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

---

两种方法可以解决这个问题：

要么保持查询（更改UNION为UNION ALL以确保您总是获得2行）并将PHP转换为u在2排。

或更改查询：

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