將mysql結果顯示到表中

Question

我想在我的頁面中將我的mysql表顯示到表中。一切正常，但順序是相反的，第一行是20，然後是19,18,17等等。有人能幫我嗎？

 <?php 
$id = mysql_connect("localhost","root","") or die('Could not connect: ' . mysql_error()); 
mysql_select_db("angajati", $id) or die('Could not select db: ' . mysql_error()); 
$query1 = "SELECT * FROM angajati "; 
$result = mysql_query($query1) or die('Error querying database.'); 

echo "<table summary='text' cellpadding='0' cellspacing='0'> 
<thead> 
<tr> 
<th>ID</th> 
<th>Nume</th> 
<th>Prenume</th> 
</tr> 
</thead> 
<tbody> 
"; 

while($row = mysql_fetch_array($result)) 
{ 
echo "<tr class='dark'>"; 
echo "<td>" . $row['ID'] . "</td>"; 
echo "<td>" . $row['Nume'] . "</td>"; 
echo "<td>" . $row['Prenume'] . "</td>"; 
} 
echo "</tbody> </table>"; 
mysql_close(); 
?> 

Answer 1

1）由ID ASC

`SELECT * FROM angajati order by ID asc` 

2）缺少TR接近

警告的mysql_query，mysql_fetch_array，的mysql_connect等擴展是在PHP 5.5.0棄用使用順序，並在PHP 7.0.0中被刪除。 應該使用MySQLi或PDO_MySQL擴展。

嘗試使用mysqli_ *

  <?php 
     $id = mysqli_connect("localhost","root","") or die('Could not connect: ' . mysqli_connect_error()); 
     mysqli_select_db($id,"angajati") or die('Could not select db: ' . mysqli_error()); 

     $query1 = "SELECT * FROM angajati order by ID ASC"; 

     $stmt = $id->prepare($query1); 
     $stmt->execute(); 
     $result = $stmt->get_result(); 
     $count = $result->num_rows; 

     echo "<table summary='text' cellpadding='0' cellspacing='0'> 
     <thead> 
     <tr> 
     <th>ID</th> 
     <th>Nume</th> 
     <th>Prenume</th> 
     </tr> 
     </thead> 
     <tbody> 
     "; 
     if($count>0){ 
      while($row = $result->fetch_assoc()) 
      { 
      echo "<tr class='dark'>"; 
      echo "<td>" . $row['ID'] . "</td>"; 
      echo "<td>" . $row['Nume'] . "</td>"; 
      echo "<td>" . $row['Prenume'] . "</td>"; 
      echo "</tr>"; 
      } 
     } 
     echo "</tbody> </table>"; 
     mysqli_close($id); 
     ?>