Question

我在Php做一个聊天应用程序。我面临的问题是，只有当我发送第二条消息时才显示第一条消息。例如，如果我必须看到第4个发送的消息，那么我必须发送第5个消息或刷新页面。请帮助找出问题所在。简化代码如下：

<!DOCTYPE html>
<html>
<head>
<title> Chatting project</title>
<link rel="stylesheet" href="css/style.css">

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js">    </script>
<script src="js/script.js"></script>
</head>
<body>
<div id="container">
  <header>
  <h1><span>Chatting project</span></h1>
  </header>
  <div id="shouts">
      <ul>

    <!--   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['name']; ?>: <?php echo $row['shout']; ?> [<?php echo $row['date']; ?>]</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 
$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>

Answer 1

问题出现在您的代码中，在INSERTING之后您是SELECTING。那些线：

<?php 
$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);
?>

应该在SELECT行之前：

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

所以它会像：

<?php
$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);
}
$query = "SELECT * FROM shouts ORDER BY id Desc LIMIT 8";
$shouts = mysqli_query($link, $query);

// close connection
mysqli_close($link);
?>

在您的情况下，您在将新数据添加到数据库之前获取数据，所以在您从数据库SELECT开始的时候，还没有新行。

仅当我在聊天应用程序中发送新消息或刷新页面时，才会显示旧消息

1 个答案: