結合兩個查詢與加入

Question

在許多提示後，我試着將它結合以下兩個查詢。我認爲代碼是自我解釋的。

表：

評論：ID（INT），用戶名（VARCHAR），autorpost（文本），評論（文本）， id_post（INT），時間（TIME）

帖子（TEXT）， date（DATE），longitude（FLOAT），latitude（FLOAT），title（TEXT）， text（TEXT），town（TEXT），time（） TIME）

$hostname='localhost'; 
      $user='root'; 
      $password=''; 
      $useron = $_COOKIE['username']; 
        try { 
          $dbh = new PDO("mysql:host=$hostname;dbname=searchfood",$user,$password); 

          $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); // <== add this line 
          $sql = "SELECT id_post 
    FROM comments 
    WHERE username = '$useron' 
    ORDER BY id DESC"; // oder (longitude between $loo and $lo or latitude between $laa and $la) versuchen 
     if ($own = $dbh->query($sql)) {// need to add this line in your code 
      // then after fetchColumn 
     $ownco = $own->fetchAll(PDO::FETCH_COLUMN, 0);  
     }       
        } 
        catch(PDOException $e) 
        { 
          echo $e->getMessage(); 
        } 


        try { 
          $dbh = new PDO("mysql:host=$hostname;dbname=searchfood",$user,$password); 

          $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); // <== add this line 
          $sql = "SELECT id, autorid, autor, date, longitude, latitude, title, text, town, time 
FROM posts 
WHERE id in (" . implode(",",$ownco) . ") 
    ORDER BY id DESC"; // oder (longitude between $loo and $lo or latitude between $laa and $la) versuchen 
     if ($resco = $dbh->query($sql)) {// need to add this line in your code 
      // then after fetchColumn 
     $resultcom = $resco->fetchAll();   
     }       
        } 
        catch(PDOException $e) 
        { 
          echo $e->getMessage(); 
        }  

我自己嘗試：

"SELECT posts.id, posts.autorid, posts.autor, posts.date, posts.longitude, posts.latitude, posts.title, posts.text, posts.town, posts.time, comments.id_post 
FROM posts 
INNER JOIN 
comments 
WHERE id = post_id 
ORDER BY id DESC" 

Answer 1

嘗試此查詢的$ ownco

"SELECT p.*, c.id_post 
FROM posts p 
JOIN comments c ON(c.id_post = p.id) 
ORDER BY p.id DESC"