Question

我正在使用以下代码进行计数，并对数据库中的值求和。

$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);

以下代码给出了错误，错误是

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

如果我删除了选择查询的前三行，它不显示错误，而是打印错误的值。

我的代码错了。 ??

以下代码对我来说非常合适。

$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);

我哪里错了？

Answer 1

我放弃了将查询放入单个查询的想法，并且正如Col.Shrapnel所建议的那样，我为它做了一个自定义函数，我发现以这种方式维护代码非常容易。谢谢Col.Sharpnel我发布了他建议的答案。

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;
             }

然后我使用此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");

上面的代码对我来说非常好。

Answer 2

看起来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执行相同操作。

Answer 3

您无法使用count(tablename.*)，请尝试使用count(tablename.columnname)

Answer 4

你可以尝试

SELECT（SELECT COUNT（）FROM news）AS cnt_news，（SELECT COUNT（）FROM FROM advertisements）为cnt_adv， ...

mySQL语法有什么问题？

4 个答案: