Question

我正在开发一个聊天网络应用程序。问题是每当每次在数据库中插入空消息时加载页面。

代码如下

          <!--   Insert MySQL datbase into HTML     -->
<?php
$connection = mysqli_connect("localhost", "root", "", "tru");
$query = "SELECT * FROM shouts ORDER BY id Desc LIMIT 8";
$shouts = mysqli_query($connection, $query);
?>



                  <!--   Insert MySQL datbase into HTML     -->
     <?php while ($row = mysqli_fetch_assoc($shouts)) : ?>
      <li> <?php echo $row['shout']; ?> <b>&nbspSent at&nbsp</b><?php echo $row['Time']; ?></li>
          <?php endwhile; ?>

    </ul>
  </div>
  <footer>
      <form action="index.php" method="post">
      <label>Shout Text: </label>
      <input type="text" name="shout" placeholder="Enter your message here">
      <input type="submit" id="submit" value="SHOUT!" >
    </form>
      <?php 
                  <!--   Insert  into MySQL datbase     -->

$link = mysqli_connect("localhost", "root", "", "tru");
$sql = "INSERT INTO shouts (name,shout) VALUES ('$_POST[name]','$_POST[shout]')";
if(mysqli_query($link, $sql)){
echo "Records added successfully.";
} else
{
  echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

// close connection
mysqli_close($link);
?>   
</footer>
</div>
</body>
</html>

Answer 1

def create

requestor = current_user
@meeting_with_params = meeting_params
@meeting = Meeting.new(@meeting_with_params)

respond_to do |format|
  if @meeting.save
    format.html { redirect_to home_url, notice: 'Your lex was successfully requested! Click Plan Meeting ' }
    format.json { render :show, status: :created, location: @meeting }
  else
    format.html { render :new }
    format.json { render json: @meeting.errors, status: :unprocessable_entity }
  end
end
end

有了if sentece，你就解决了。如果设置了POST名称变量，它只会添加一条记录。

Answer 2

你无条件insert一行，所以这是预期的。那么，条件应该是什么？当然，在页面加载时，您不希望仅在POST时insert喊叫。所以你需要测试请求是否是帖子：

if ($_SERVER['REQUEST_METHOD'] === 'POST') {
    // insert the row
}

但即使是POST，也需要测试它是否有效，所以检查姓名和喊叫：

if ($_SERVER['REQUEST_METHOD'] === 'POST') {
    if ((isset($_POST["name"])) && (isset($_POST["shout"])) && ($_POST["name"]) && ($_POST["shout"])) {
        //insert the row
    }
}

请注意SQL injection。

每次都会添加额外的Sql记录

2 个答案: