SELECT查詢輸出每個單個結果六次

Question

此查詢輸出每個結果6次：因此，對於一列以「亞青」的主體，它會被輸出爲：亞青
亞青
亞青
亞青
阿約
亞青

function get_comments($page_type, $spot) { 
     $query = mysql_query(" 
           SELECT 
           comments.user_id AS user_id, 
           comments.page_type AS page_type, 
           comments.spot AS spot, 
           comments.comment AS comment, 
           comments.timestamp AS timestamp, 
           users.first_name AS first_name, 
           users.last_name AS last_name 
           FROM comments, users 
           WHERE page_type = '$page_type' AND spot = '$spot' AND (comments.user_id = users.user_id OR comments.user_id = '0') 
           ORDER BY timestamp DESC 
          "); 

     while($fetch = mysql_fetch_assoc($query)) { 

      $comment = $fetch['comment']; 
      $timestamp = $fetch['timestamp']; 
      $timestamp_date = date('M d y\'', $timestamp); 
      $timestamp_time = date('g:m', $timestamp); 
      $comments_user_id = $fetch['user_id']; 

       if(date('H', $timestamp) > 12) { 
        $am_pm = 'pm'; 
        } else { 
         $am_pm = 'am'; 
         } 

      $first_name = $fetch['first_name']; 
      $last_name = $fetch['last_name']; 

      ?> 

      <div id="replies"> 

       <?php if($comments_user_id == '0') {echo 'Guest';} else { echo $first_name.' '.$last_name; } ?><br /> 
       <?php echo $timestamp_date; ?><br /> 
       <?php echo $comment; ?> 

      </div> 

      <?php 

      } 
     } 

在這裏$ PAGE_TYPE和$發現是指：http://localhost/ $ PAGE_TYPE/$現貨

Answer 1

你得到重複，因爲你的隱式連接條件包含此：

OR comments.user_id = '0' 

，你有多個行與comments.user_id = 0。這些user_id = 0行中的每一行都會顯示匹配您的連接條件，並且會導致重複。解決方案是修正您的數據，以便每條評論有一個有效的user_id，其中「有效」意味着「有一個條目users」，並且放棄comments.user_id = 0連接條件的一部分。當你在它的時候，切換到InnoDB表並添加外鍵來強制執行這個參照完整性約束。

順便說一句，如果user_id是一個整數，那麼你真的不應該引用它;你正在使數據庫做不必要的工作，並建立一個壞習慣，如果你每個人都使用另一個數據庫，會導致你一些痛苦和痛苦。你也應該切換到一個明確的加盟：

from comments join users on comments.user_id = users.id 
where page_type = '$page_type' and spot = '$spot' 

Answer 2

$query = mysql_query(" 
         SELECT 
         comments.user_id AS user_id, 
         comments.page_type AS page_type, 
         comments.spot AS spot, 
         comments.comment AS comment, 
         comments.timestamp AS timestamp, 
         users.first_name AS first_name, 
         users.last_name AS last_name 
         FROM comments, users 
         WHERE page_type = '$page_type' AND spot = '$spot' AND (comments.user_id = users.user_id OR comments.user_id = '0') 
         GROUP BY comments.user_id 
         ORDER BY timestamp DESC 
        "); 

Answer 3

我不像我熟悉MS T-SQL那樣熟悉MySQL。但它看起來像你正在交叉連接你的桌子。用戶和評論之間可能應該有JOIN。否則，每條評論將被加入到每個用戶。

這是一個指向MySQL中JOIN細節的鏈接。

我可能錯過這裏的語法部分的MySQL細節，我可以; t檢驗這一點，但這裏有雲：

這將返回這些記錄爲用戶提供相關的註釋：

SELECT 
    comments.user_id AS user_id, 
    comments.page_type AS page_type, 
    comments.spot AS spot, 
    comments.comment AS comment, 
    comments.timestamp AS timestamp, 
    users.first_name AS first_name, 
    users.last_name AS last_name 
FROM 
    comments INNER JOIN users 
    ON comments.user_id = users.user_id 
WHERE page_type = '$page_type' 
AND spot = '$spot' 
AND (comments.user_id = users.user_id OR comments.user_id = '0') 
ORDER BY timestamp DESC 

這將返回所有用戶的記錄，並會匹配他們存在的註釋記錄： 選擇 comments.user_id AS USER_ID， comments.page_type AS PAGE_TYPE， comments.spot AS現貨， comments.comment AS評論， comments.timestamp時間戳， users.first_name AS FIRST_NAME， users.last_name AS姓氏 FROM 評論INNER JOIN用戶 ON comments.user_id = users.user_id WHERE PAGE_TYPE = '$ PAGE_TYPE' 和現貨=「$點」 AND（comments.user_id = users.user_id OR comments.user_id = '0'） ORDER BY時間戳DESC

Answer 4

因爲你(comments.user_id = users.user_id OR comments.user_id = '0')條款，因爲有記錄users每一位客人的評論會出現多次。顯然，這是六個。

編輯補充：

我想你想是這樣的：

SELECT comments.user_id AS user_id, 
     comments.page_type AS page_type, 
     comments.spot AS spot, 
     comments.comment AS comment, 
     comments.timestamp AS timestamp, 
     COALESCE(users.first_name, 'Guest') AS first_name, 
     COALESCE(users.last_name, 'User') AS last_name 
    FROM comments 
    LEFT 
OUTER 
    JOIN users 
    ON users.user_id = comments.user_id 
WHERE comments.page_type = '$page_type' 
    AND comments.spot = '$spot' 
ORDER BY comments.timestamp DESC 
; 

，但我有「萬畝太短」，你應該重新評估表的設計一致。使用comments.user_id = 0來表示「客人評論」，當沒有users.user_id = 0的記錄時，不是很好，因爲它給你留下了「幾乎外鍵」的關係。完整的外鍵關係將使您的表結構更易於理解和推理。

Answer 5

while($fetch = mysql_fetch_array($query)) 

變化而語法上面

和改變這種代碼的語法

WHERE page_type = '$page_type' AND spot = '$spot' AND (comments.user_id = users.user_id OR comments.user_id = '0') 

到

WHERE page_type = '$page_type' AND spot = '$spot' AND (comments.user_id = users.user_id)