MySQL不顯示數據

Question

我有一個網站充滿了文章，我希望能夠通過標籤來訂購它們。每篇文章都有一些關聯的標籤。

我用3個表：

媒體（展位的文章）

+----+--------+-----+ 
| id | title | ... | 
+----+--------+-----+ 
| 1 | hello | ... | 
| 2 | hi  | ... | 
+----+--------+-----+ 

標籤

+----+------+-----+ 
| id | name | ... | 
+----+------+-----+ 
| 1 | red | ... | 
| 2 |square| ... | 
+----+------+-----+ 

medias_tags（爲了鏈接標籤&媒體一起）

+----+----------+--------+ 
| id | media_id | tag_id | 
+----+----------+--------+ 
| 1 |  1 |  1 | 
| 2 |  1 |  2 | 
| 3 |  2 |  2 | 
+----+----------+--------+ 

到目前爲止，我使用的這個MySQL/PHP代碼命令由標籤的物品，這是工作（我故意刪除更好地瞭解分頁條件）：

$ BDD

try{ 
    $bdd = new PDO('mysql:host=localhost;dbname=DB;charset=utf8', 'username', 'pass');}catch(Exception $e) { die('Erreur : '.$e->getMessage()); 
} 



if(isset($_GET['tag'])){ 
    $tag_name_1 = htmlspecialchars($_GET['tag']); 
    $req_tag = $bdd->prepare('SELECT * FROM tags WHERE name = ?'); 
    $req_tag->execute(array($tag_name_1)); 
    $data_tag = $req_tag->fetch(); 
    $ttag_id = $data_tag['id'];  

    $q = 'SELECT * 
      FROM medias 
      JOIN medias_tags 
      ON medias.id=medias_tags.media_id 
      WHERE medias_tags.tag_id ='.$ttag_id.' 
      ORDER BY date DESC 
      LIMIT '.$start.','.$limit;  
    } 

    $req_mda_list = $bdd->query($q); 

問題是當我想顯示與文章相關的所有標籤時。 $ mda_tags_list爲空。

while ($data_mda = $req_mda_list->fetch()){ 

    $mda_id = $data_mda['id']; 
    $mda_title = $data_mda['title']; 

    #Get media tags 
    $req_mda_tags = $bdd->prepare('SELECT * FROM medias_tags WHERE media_id = ?'); 
    $req_mda_tags->execute(array($mda_id));  

    $mda_tags_list = null; 

    while ($data_mda_tags = $req_mda_tags->fetch()){ 
     $tag_id = $data_mda_tags['tag_id']; 

     $req_tag_name = $bdd->prepare('SELECT * FROM tags WHERE id = ?'); 
     $req_tag_name->execute(array($tag_id)); 
     $data_tag_name = $req_tag_name->fetch(); 

     $mda_tags_list .= '<a href="http://www.website.com/tags/'.$data_tag_name['name'].'">'.$data_tag_name['name'].'</a> '; 
} 

    echo $mda_title.' - '.$mda_tags_list; 

} 

我居然得到：

Order by tag -> Square 

Title1 - Tag : 
Title5 - Tag : 
Title6 - Tag : 

我試圖讓：

Order by tag -> Square 

Title1 - Tag : Square, Red, Thank 
Title5 - Tag : Yellow, Square, You 
Title6 - Tag : Square, Blue, Stackoverflow 

Answer 1

我做了什麼，簡化你的代碼，我已經減少你的SQL到一個聲明，我已經添加了一些非常基本的錯誤檢查：

$tagName = htmlspecialchars($_GET['tag']); 

$sql = "SELECT 
     medias.*, 
     GROUP_CONCAT(tags2.name) AS allTags 
    FROM 
     medias 
     -- Get the media tags 
     JOIN medias_tags ON medias.id = medias_tags.media_id 
     -- Join on the tags so we can add a where condition 
     JOIN tags ON medias_tags.tag_id = tags.id 
     -- Rejoin tags again, this won't be used as a condition hence the left 
     LEFT JOIN tags AS tags2 ON medias_tags.tag_id = tags2.id 
    WHERE 
     -- Only get medias where it has a matching tag name 
     tag.name = :name 
    GROUP BY 
     medias.id 
    ORDER BY 
     date DESC 
    LIMIT 
     :offset, :limit"; 

$sth = $bdd->prepare($sql); 

// Did it prepare ok? 
if (!$sth) { 
    echo "\nPDO::errorInfo():\n"; 
    print_r($dbh->errorInfo()); 
    exit; 
} 

$err = $sth->execute(array(':name' => $tagName, ':offset' => $start, ':limit' => $limit)); 

// Did it execute ok? 
if (!$sth) { 
    echo "\nPDO::errorInfo():\n"; 
    print_r($dbh->errorInfo()); 
    exit; 
} 

foreach ($sth->fetchAll() as $row) { 
    $tags = explode(',', $row['allTags']); 

    echo $row['title'] . ' - '; 

    foreach ($tags as $tag) { 
     echo '<a href="http://www.website.com/tags/' . $tag . '">' . $tag . '</a> '; 
    } 
} 

這可能是我的SQL註釋需要刪除，如果您的標籤包含逗號，這也無法正常工作。如果需要避免這種情況，您可以指定一個不同的分隔符。

基本上，這只是爲了減少可能失敗的區域並捕捉可能的區域。我只是試圖根據你的例子輸出結果，看起來你的代碼看起來像什麼，希望這會給你一些能夠實現你的目標的東西。

Answer 2

對於媒體獲得的所有標籤，你可以使用下面的查詢：

SELECT medias.id,medias.title,GROUP_CONCAT(tags.name) as tag 
FROM medias 
JOIN medias_tags ON medias.id = medias_tags.media_id 
JOIN tags ON medias_tags.tag_id = tags.id 
GROUP BY medias.id 

要獲得所有媒體與標籤，如果avai拉布勒其他空白，可以使用下面的查詢：

SELECT medias.id,medias.title,GROUP_CONCAT(tags.name) as tag 
FROM medias 
LEFT JOIN medias_tags ON medias.id = medias_tags.media_id 
LEFT JOIN tags ON medias_tags.tag_id = tags.id 
GROUP BY medias.id