Question

非常抱歉，因为这将是一个很长的帖子。我是MYSQL的初学者，我刚开始学习。我试图建立用户评论回复系统。我完全卡住了。

我有一个带有表名的数据库＆＃39; post＆＃39;用以下结构的评论

＆＃13;

 TABLE `post` (
  `id_posta` int(4) NOT NULL,
  `tekst_posta` text NOT NULL,
  `name` text NOT NULL,
  `slika_posta` blob NOT NULL,
  `type` text NOT NULL,
  `vreme_posta` datetime NOT NULL,
  `id_autora` int(4) NOT NULL,
  `ime` varchar(50) NOT NULL,
  `prezime` varchar(100) NOT NULL
   ) ENGINE=InnoDB DEFAULT CHARSET=latin1;

＆＃13;

和表名＆＃39; komentar＆＃39;，其中书面回复评论

＆＃13;

TABLE `komentar` (
  `id_komentara` int(11) NOT NULL,
  `tekst_komentara` text NOT NULL,
  `id_aut_komentara` int(4) NOT NULL,
  `id_autora` int(4) NOT NULL,
  `vreme_komentara` datetime NOT NULL,
  `id_posta` int(4) NOT NULL,
  `imek` varchar(50) NOT NULL,
  `prezimek` varchar(100) NOT NULL
   ) ENGINE=InnoDB DEFAULT CHARSET=latin

＆＃13;

当我写下所有评论时，一切都很完美。

＆＃13;

$upit = "SELECT id_posta, slika_posta, vreme_posta, tekst_posta, id_autora, ime, prezime FROM post ORDER BY `post`.`id_posta` DESC;";
		$postovi = $con->query($upit);
	
	if($postovi->num_rows > 0)
	{
		
		while($red = $postovi->fetch_assoc())
		{
			$idp = $red["id_posta"];
			echo "<table border-collapse: separate; empty-cells: hide; class='my-special-table'>";
			echo "<tr><td>";
			echo $red["ime"]. " " . $red["prezime"]. ':' ;
			echo "</td></tr>";
			echo "<tr><td class='prvatd'>";
			echo $red["tekst_posta"];
			echo "</td></tr>";
			if( $red["slika_posta"] != '' ){
			echo '<tr><td class="w_imgg"><center><img src="data:image/png;base64,'.base64_encode($red["slika_posta"]).'" class="w_img"></center></td></tr>';
			}
			echo "<br>";
			echo "<tr><td style= 'color:#FF0000; font-size:12px;'>";
			echo 'Objavljeno' . ' ' . $red["vreme_posta"];
			echo "</td></tr>";
			echo "<tr><td>";
			echo "<form method='post' action = '' >
           <textarea id = 'styled' name='tekst_komentara' placeholder='Prokomentarisite'></textarea><br>
		   <input type='hidden' name='id_pos' value= $idp/>
		   <input type='submit' name='unesi_komentar' value='Objavi komentar'/>
	        </form>";
			echo "</td></tr>";
			echo "<tr><td>";
			echo REPLY ?????
			echo "<td></td>";
			echo "<tr><td>Lajk<td></td>";
			echo "</table>";
			echo "<br><br>";
			
		}
	}
	 
   $con->close();

＆＃13;

当我尝试在回复td中回复相应评论时，问题就出现了。

我试试这个......

＆＃13;

$con = new mysqli ("localhost", "root", "", "introduce");

$upit = "SELECT id_posta, slika_posta, vreme_posta, tekst_posta, id_autora, ime, prezime FROM post ORDER BY `post`.`id_posta` DESC;";
$postovi = mysqli_query($con,$upit);
if($postovi->num_rows > 0)
{
  while($red = $postovi->fetch_assoc()) {
    
            			
			echo "<table>";
            echo "<tr><td>";
			echo $red["tekst_posta"];
			echo "</td></tr>";
			echo "</table>";	
			  

    $upit1 = "SELECT * FROM komentar INNER JOIN post ON post.id_posta=komentar.id_posta ;"; 
    $komentari = mysqli_query($con,$upit1);
	while($red = $komentari->fetch_assoc()) {
     
	  echo "<table>";
            echo "<tr><td>";
			echo $red["tekst_komentara"];
			echo "</td></tr>";
			echo "</table>";
    }
  }
  
}

＆＃13;

，结果是

＆＃13;

2                     2  (second post)
1-1                  2-1 (first reply)
1-2                  2-2 (second reply)
2-1
1    and should be    1   (first post)
1-1                  1-1  etc,etc....
1-2                  1-2   
2-1

＆＃13;

然后我试试这个

＆＃13;

 $con = new mysqli("localhost", "root", "", "introduce");
	
   $sql = "SELECT * FROM komentar INNER JOIN post ON post.id_posta=komentar.id_posta ; ";
  
   $komentari = $con->query($sql);
   if($komentari->num_rows > 0)
   {
	   while($red = $komentari->fetch_assoc())
	   {
		  
		   echo $red["tekst_posta"];
		   echo "<br>";
		   echo $red["tekst_komentara"];
		  
	   }
   }

＆＃13;

结果是

＆＃13;

1
  1-1
   1
  1-2
   2
  2-1 duplicate post on every replay....
   2
  2-2
   2
  2-3

＆＃13;

接下来我试试......

＆＃13;

 $sql = "SELECT id_posta, slika_posta, vreme_posta, tekst_posta, id_autora, ime, prezime FROM post ORDER BY `post`.`id_posta` DESC;";
   $sql .= "SELECT * FROM post JOIN SELECT * FROM komentar WHERE komentar.id_posta = 'post.id_posta'";

＆＃13;

我尝试GROUP BY并尝试，并尝试.....

我正在考虑阵列....所以，我请求你的建议，因为我将永远无法做到这一点。

使用PHP和MYSQL的用户评论回复系统

0 个答案: