mySQL語法有什麼問題？

我正在使用以下代碼進行計數，並對數據庫中的值進行求和。mySQL語法有什麼問題？

$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);

下面的代碼給我一個錯誤，錯誤的是

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

如果我刪除前三行的選擇查詢的，它並沒有顯示錯誤，而不是它打印錯誤的價值觀。

這是錯誤的我的代碼。 ??

以下代碼對我來說工作得非常好。

$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);

我在哪裏出錯了？

來源

2010-09-13 Ibrahim Azhar Armar

那麼我放棄了使我的查詢成爲一個單一的想法，並作爲建議由Col.Shrapnel構建，我爲它做了一個自定義函數，並且我發現用這種方法維護代碼非常容易。謝謝Col.Sharpnel，我發佈了他提出的答案。

這是我創建的用戶定義函數。

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; 
      }

然後我使用這段代碼調用了我的函數。

 $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");

上面的代碼工作對我來說完全沒有問題。

來源

2010-09-13 13:29:43

它看起來像Mysql不喜歡那條線。將COUNT(n.*)更改爲COUNT(n.id)或該表的主鍵字段的名稱。對a和c做同樣的事情。

來源

2010-09-13 08:49:47 Fanis

我試過你的方式錯誤消失，但爲什麼地獄它打印錯誤的值？我有兩個數據庫表中的行數，它顯示30代替。 :( – 2010-09-13 08:53:23

你正在做一個完整的連接反對三個表新聞n，廣告a，評論c ...嘗試用COUNT（DISTINCT n.id）替換COUNT（n.id）（和您的每個表的等效項） – 2010-09-13 08:56:31

@Mark Baker很好，我甚至沒有注意到查詢的其餘部分。這3個表格是否加入了任何列？我認爲新聞有評論和廣告。 – Fanis 2010-09-13 09:04:24

不能使用count(tablename.*)，請嘗試使用count(tablename.columnname)

來源

2010-09-13 08:52:32 JohnoBoy

我試圖更改爲COUNT（n.id）作爲cnt_news，ut工作，但打印出錯誤的值，..任何想法爲什麼？ – 2010-09-13 08:54:59

如果您指定的列具有NULL值，則count命令將忽略它們，嘗試使用表的id列上的計數或非空列 – JohnoBoy 2010-09-13 10:05:36

您可以嘗試

SELECT（SELECT COUNT（）從新聞）AS cnt_news，（SELECT COUNT（）FROM廣告）作爲cnt_adv ， ...

來源

2010-09-13 09:38:21 viriathus

mySQL語法有什麼問題？

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