php - 从表中全选并显示结果mysql

<?php
   $query_announcement = mysql_query("SELECT * FROM announcement");
   $get_announcement = mysql_fetch_assoc($query_announcement);
?>

  <?php
        if(mysql_num_rows($query_announcement) > 0){
            $row = mysql_num_rows($query_announcement);

               for($x = 1; $x<= $row; $x++){

               $html = '<tr>';
               $html .= '<td>'.$get_announcement['id'].'</td>';
               $html .= '<td>'.$get_announcement['announce_name'].'</td>';
               $html .= '<td>'.$get_announcement['announce_description'].'</td>';
               $html .= '<td>'.$get_announcement['announce_location'].'</td>';
               $html .= '<td>'.date('M d,Y', strtotime($get_announcement['date_start'])).'</td>';
               $html .= '<td>'.date('M d,Y', strtotime($get_announcement['date_end'])).'</td>';
               $html .= '<td>'.date('M d,Y', strtotime($get_announcement['date_added'])).'</td>';
               $html .= '<td width="15%">

              <button type="button" class="btn btn-default" title="Edit"><span class="fa fa-pencil"></span></button>
              <button type="button" class="btn btn-default" title="Delete" ><span class="fa fa-trash-o"></span></button>
              </td>';
              $html .= '</tr>';

              echo $html;
              }
           }
               else{
                echo '<tr><td colspan="8" align="center"><h3>No Announcement</h3></td></tr>';
              }
    ?>

/这是我的代码，但是当我运行它时，我只会得到具有相同ID的公告，而不是我创建的所有公告。请帮助谢谢

最佳答案

您一次只能提取数据

for($x = 1; $x<= $row; $x++){
$get_announcement= mysql_fetch_assoc($query_announcement);

关于php - 从表中全选并显示结果mysql，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/30172636/