单击搜索按钮后如何只显示一个表？

Question

当我搜索会员名称时，我有2个这样的表。 链接：

我希望搜索结果仅在搜索后显示。

我的代码：

            <div id = "subtitle">
            View Members
            </div>
            <div id = "searchbox">
            <form method="post">
<center><input type="text"  maxlength="100"  required placeholder="Enter Full Name" name ="search" autocomplete="off" value="">
<input type="submit" name="btn" value="SEARCH NOW!"></p></center>

            </form>
            </div>

            <?php

     if(isset($_POST["btn"]))

     {
         $search = $_POST["search"]; 
         $sql = "select * from member where Member_Name like '$search%' "; 
         $result = mysqli_query($conn,$sql);
         $rowcount = mysqli_num_rows($result);
         if($rowcount==0)
             echo "Sorry ,no records found!"; 
         else
         {
            ?>
        <center><table class="table table-bordered">
                <thead>
                    <tr>
                    <th>#</th>
                      <th>Member ID</th>
                      <th>Member Name</th>
                      <th>Actions</th>
                    </tr>
    <?php
            while($row=mysqli_fetch_assoc($result))  //display
            {
    ?>          <tr>
                    <td></td>
                      <td><?php echo $row["Member_ID"]?></td>
                      <td><?php echo $row["Member_Name"]?></td>
                      <td><a href=Admin_MemberDetails.php?id=".$row["Member_ID"]."><img src=../Images/ViewFile.png height=37px title=View></a>
                      <a href=Admin_EditMember.php?id=".$row["Member_ID"]."><img src=../Images/edit.png height=37px title=Edit></a>
                  </td>
                </tr>
                </table>
                </center>
    <?php
            }           
         }
     }


    ?>

            <center><table class="table table-bordered">
                <thead>
                    <tr>
                      <th>#</th>
                      <th>Member ID</th>
                      <th>Member Name</th>
                      <th>Actions</th>
                    </tr>

      <?php

      $sql = "select * from legoclub_guesthouse.member";
      $result = mysqli_query($conn,$sql);
        $rowcount= mysqli_num_rows($result);

      while($row=mysqli_fetch_assoc($result))
      {
        echo "<tr>";
        echo "<td></td>";
        echo "<td>$row[Member_ID]</td>";
        echo "<td>$row[Member_Name]</td>";
        echo "<td><a href=Admin_MemberDetails.php?id=".$row["Member_ID"]."><img src=../Images/ViewFile.png height=37px title=View></a>
                  <a href=Admin_EditMember.php?id=".$row["Member_ID"]."><img src=../Images/edit.png height=37px title=Edit></a>
              </td>";
        echo "</tr>";
      }


      ?>

      </table><center>

            </div>   

请为我包含php文件！

Answer 1

要以最简单的方式执行此操作，请尝试使用此代码：

<div id = "subtitle">
            View Members
            </div>
            <div id = "searchbox">
            <form method="post">
<center><input type="text"  maxlength="100"  required placeholder="Enter Full Name" name ="search" autocomplete="off" value="">
<input type="submit" name="btn" value="SEARCH NOW!"></p></center>

            </form>
            </div>

            <?php

     if(isset($_POST["btn"]))

     {
         $search = $_POST["search"]; 
         $sql = "select * from member where Member_Name like '$search%' "; 
         $result = mysqli_query($conn,$sql);
         $rowcount = mysqli_num_rows($result);
         if($rowcount==0)
             echo "Sorry ,no records found!"; 
         else
         {
            ?>
        <center><table class="table table-bordered">
                <thead>
                    <tr>
                    <th>#</th>
                      <th>Member ID</th>
                      <th>Member Name</th>
                      <th>Actions</th>
                    </tr>
    <?php
            while($row=mysqli_fetch_assoc($result))  //display
            {
    ?>          <tr>
                    <td></td>
                      <td><?php echo $row["Member_ID"]?></td>
                      <td><?php echo $row["Member_Name"]?></td>
                      <td><a href=Admin_MemberDetails.php?id=".$row["Member_ID"]."><img src=../Images/ViewFile.png height=37px title=View></a>
                      <a href=Admin_EditMember.php?id=".$row["Member_ID"]."><img src=../Images/edit.png height=37px title=Edit></a>
                  </td>
                </tr>
    <?php
            }
            $sql = "select * from legoclub_guesthouse.member";
            $result = mysqli_query($conn,$sql);
            $rowcount= mysqli_num_rows($result);
            while($row=mysqli_fetch_assoc($result))
            {
              echo "<tr>";
              echo "<td></td>";
              echo "<td>$row[Member_ID]</td>";
              echo "<td>$row[Member_Name]</td>";
              echo "<td><a href=Admin_MemberDetails.php?id=".$row["Member_ID"]."><img src=../Images/ViewFile.png height=37px title=View></a>
                        <a href=Admin_EditMember.php?id=".$row["Member_ID"]."><img src=../Images/edit.png height=37px title=Edit></a>
                    </td>";
              echo "</tr>";
            }
        ?>
                </table>
                </center>
<?php
         }
     }


    ?>

            </div>   

但我建议您使用SQL UNION Operator.

这将使您的生活轻松解决这些类型的问题