mySQL语法有什么问题？

我正在使用以下代码进行计数，并对数据库中的值进行求和。mySQL语法有什么问题？

$query = "SELECT 
      COUNT(n.*) AS cnt_news, 
      COUNT(a.*) AS cnt_adv, 
      COUNT(c.*) AS cnt_comm, 
      SUM(CASE WHEN c.approve = '1' AND c.spam = '0' THEN 1 ELSE 0 END) AS cnt_approved, 
      SUM(CASE WHEN c.approve = '0' AND c.spam = '0' THEN 1 ELSE 0 END) AS cnt_unapproved, 
      SUM(CASE WHEN c.spam = '0' THEN 1 ELSE 0 END) AS cnt_spam, 
      SUM(a.amount) AS t_amnt, 
      SUM(a.cashpaid) AS t_cpaid, 
      SUM(a.balance) AS t_bal 
      FROM 
      news n, advertisements a, comments c"; 
      $result = mysql_query($query) or die(mysql_error()); 
      $row = mysql_fetch_array($result);

下面的代码给我一个错误，错误的是

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '*) AS cnt_news, COUNT(a.*) AS cnt_adv, COUNT(c.*) AS cnt_c' at line 2

如果我删除前三行的选择查询的，它并没有显示错误，而不是它打印错误的价值观。

这是错误的我的代码。 ??

以下代码对我来说工作得非常好。

$query = "SELECT COUNT(*) as cnt_news FROM news"; 
$result = mysql_query($query); 
$row = mysql_fetch_array($result); 


$query = "SELECT COUNT(*) as cnt_adv FROM advertisements"; 
$result = mysql_query($query); 
$row = mysql_fetch_array($result); 

$query = "SELECT COUNT(*) as cnt_comm FROM comments"; 
$result = mysql_query($query); 
$row = mysql_fetch_array($result); 


$query = "SELECT SUM(CASE WHEN c.approve = '1' AND c.spam = '0' THEN 1 ELSE 0 END) AS cnt_approved, 
      SUM(CASE WHEN c.approve = '0' AND c.spam = '0' THEN 1 ELSE 0 END) AS cnt_unapproved, 
      SUM(CASE WHEN c.spam = '1' THEN 1 ELSE 0 END) AS cnt_spam 
      FROM COMMENTS c"; 
$result = mysql_query($query); 
$row = mysql_fetch_array($result); 


$query = "SELECT SUM(a.amount) as t_amnt, 
      SUM(a.cashpaid) as t_cpaid, 
      SUM(a.balance) as t_bal 
      FROM advertisements a"; 
$result = mysql_query($query); 
$row = mysql_fetch_array($result);

我在哪里出错了？

来源

2010-09-13 Ibrahim Azhar Armar

那么我放弃了使我的查询成为一个单一的想法，并作为建议由Col.Shrapnel构建，我为它做了一个自定义函数，并且我发现用这种方法维护代码非常容易。谢谢Col.Sharpnel，我发布了他提出的答案。

这是我创建的用户定义函数。

function dbgetvar($query) { 
      $res = mysql_query($query); 
     if(!$res) { 
      trigger_error("dbget: ". mysql_error(). " in " .$query); 
      return false; 
      } 
      $row = mysql_fetch_array($res); 
      if(!$row) return ""; 
      return $row; 
      }

然后我使用这段代码调用了我的函数。

 $news = dbgetvar("SELECT COUNT(*) as count FROM news"); 
$comments = dbgetvar("SELECT SUM(CASE WHEN c.approve = '1' AND c.spam = '0' THEN 1 ELSE 0 END) AS approved, 
         SUM(CASE WHEN c.approve = '0' AND c.spam = '0' THEN 1 ELSE 0 END) AS pending, 
         SUM(CASE WHEN c.spam = '1' THEN 1 ELSE 0 END) AS spam, 
         COUNT(*) AS count 
         FROM COMMENTS c"); 
$advertise = dbgetvar("SELECT SUM(a.amount) AS amount, 
         SUM(a.cashpaid) AS cashpaid, 
         SUM(a.balance) AS balance, 
         COUNT(*) AS count 
         FROM advertisements a");

上面的代码工作对我来说完全没有问题。

来源

2010-09-13 13:29:43

它看起来像Mysql不喜欢那条线。将COUNT(n.*)更改为COUNT(n.id)或该表的主键字段的名称。对a和c做同样的事情。

来源

2010-09-13 08:49:47 Fanis

我试过你的方式错误消失，但为什么地狱它打印错误的值？我有两个数据库表中的行数，它显示30代替。 :( – 2010-09-13 08:53:23

你正在做一个完整的连接反对三个表新闻n，广告a，评论c ...尝试用COUNT（DISTINCT n.id）替换COUNT（n.id）（和您的每个表的等效项） – 2010-09-13 08:56:31

@Mark Baker很好，我甚至没有注意到查询的其余部分。这3个表格是否加入了任何列？我认为新闻有评论和广告。 – Fanis 2010-09-13 09:04:24

不能使用count(tablename.*)，请尝试使用count(tablename.columnname)

来源

2010-09-13 08:52:32 JohnoBoy

我试图更改为COUNT（n.id）作为cnt_news，ut工作，但打印出错误的值，..任何想法为什么？ – 2010-09-13 08:54:59

如果您指定的列具有NULL值，则count命令将忽略它们，尝试使用表的id列上的计数或非空列 – JohnoBoy 2010-09-13 10:05:36

您可以尝试

SELECT（SELECT COUNT（）从新闻）AS cnt_news，（SELECT COUNT（）FROM广告）作为cnt_adv ， ...

来源

2010-09-13 09:38:21 viriathus

mySQL语法有什么问题？

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