Question

如何展示与该主题相匹配的唯一内容？
此代码将显示所有'主题'和'内容' 请帮帮我

这是嗡嗡声（就像一个论坛）

    <thead>
    <tr align="center">
        <th>Topic</th>
    </tr>
   </thead>

   <tbody>
    <tr align="center">
        <?php

            $sql = "SELECT buzz_topic FROM buzz";
            $result = $con->query($sql);

                while($row = mysqli_fetch_assoc($result)) {
                    echo"<tr>
                    <td><a href='buzzContent.php'>{$row['buzz_topic']}</a></td>
                    </tr>";
                }
            ?>
    </tr>

    <td><input type="button" value="Add Topic" onClick="document.location.href='addBuzz.php'"></td>

  </tbody>
  </table>

   </form>

这是将主题和细节添加到buzz数据库中

            <tr>
                <td>Student ID</td>
                <td> 
                    <textarea rows="1" cols="50" type='text' name='student_ID' id='student_ID'></textarea> 
                </td>
            </tr>

            <tr>
                <td>Topic</td>
                <td>
                    <textarea rows="1" cols="50" type='text' name='buzz_topic' id='buzz_topic'></textarea>
                </td>
            </tr>

            <tr>
                <td>Content</td>
                <td>
                      <textarea rows="5" cols="50" type='text' name='buzz_content' id='buzz_content'></textarea>
                </td>
            </tr>

        </table>

            <td><input type="submit" value="Submit" ></td>
            <td><input type="button" onClick="history.go(0)" value="Cancel"></td>

        </form>

这是为了显示人们点击的嗡嗡声的内容，但它会显示所有内容和主题，我该如何解决？

    <h1>
        <?php
            $sql = "SELECT buzz_topic FROM buzz";
            $result = $con->query($sql);

            while($row = mysqli_fetch_assoc($result)) {
            echo $row['buzz_topic'];
            }
        ?>
    </h1>

    <p>
        <?php   

        $sql = "SELECT buzz_content FROM buzz";
        $result = $con->query($sql);

        while($row = mysqli_fetch_assoc($result)) {
        echo $row['buzz_content'];
        }
        ?>
    </p>

    <input type="button" onClick="document.location.href='buzz.php'" value="Back">

    </form>

Answer 1

使用id作为主键。并在用户点击时传递id。

<?php
            $sql = "SELECT buzz_topic,buzz_content FROM buzz WHERE id=".$id."";
            $result = $con->query($sql);

            while($row = mysqli_fetch_assoc($result)) {
            echo "<h>".$row['buzz_topic']."</h>";
            echo "<p>".$row['buzz_content']."</p>";
            }
        ?>

Answer 2

您的SQL语句需要一个LIKE语句

SELECT buzz_content FROM buzz WHERE buzz_content LIKE '%searchterm%'

仅显示与主题匹配的内容

2 个答案: