Question

Iam尝试创建一个私人消息系统，其中用户将消息发送给另一个用户并将该内容插入到数据库中。我使用一个名为hash的随机数来识别两个人之间的对话..可以选择＆＃ 34; message_group＆＃34;保存消息的表格是＆＃34;消息＆＃34; ..这里出现问题..
当我在文本区域中键入内容并单击sendmessage按钮时，它会将数据插入到消息数据库中。但是如果再次输入内容并尝试发送它，则数据不会进入数据库。其他人只得到这个数据第一条消息。请帮助我解决这个问题......那里有代码

<html>
<head>
<title>new convo</title>
</head>
<body>
<?php include 'connect.php';?>
<?php include 'message_title_bar.php';?>
<?php include 'functions.php';?>

<div>
<?php 
if(isset($_GET['user']) && !empty($_GET['user'])){ 
?>


<form method='post'>

<?php
if(isset($_POST['message']) && !empty($_POST['message'])){


        $my_id=$_SESSION['user_id'];

    $user=$_GET['user'];
    $random_number=rand();
    $message=$_POST['message'];


    $connect = mysqli_connect('localhost','root','','php_mysql_login_system');
    if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

  $query_string = "SELECT `hash` FROM `message_group` WHERE (`user_one`='$my_id' AND `user_two`='$user') OR (`user_one`='$user' AND `user_two`='$my_id')";

  $check_con=mysqli_query($connect,$query_string) or die(mysqli_error($connect));
    if(mysqli_num_rows($check_con)==1){
        echo "<p>Conversation already Started</p>";
    }else{
        $connect = mysqli_connect('localhost','root','','php_mysql_login_system');

    if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
        mysqli_query($connect,"INSERT INTO message_group VALUES('$my_id' , '$user' , '$random_number')");
    mysqli_query($connect,"INSERT INTO messages VALUES ('','$random_number','$my_id','$message')");
    echo "<p>Conversation started</p>";
    }
}
?>

Enter message:<br />
<textarea name='message' rows='7' cols='60'></textarea>
<br />
<br />
<input type='submit' name="submit" value="sendmessage" />
</form>


<?php


}

else{
    echo "<b>Select User</b>";
$connect = mysqli_connect('localhost','root','','php_mysql_login_system');
   $user_list=mysqli_query($connect,"SELECT `id`,`username` FROM `users`");
    while($run_user=mysqli_fetch_array($user_list)){
       $user = $run_user['id'];
        $username = $run_user['username'];
        echo "<p><a href='send.php?user=$user'>$username</a></p>";
        }
}
?>
</div>
</body>
</html>

感谢任何帮助。

Answer 1

代码兄弟有很多错误首先阅读基本...而你没有使用echo函数来打印旧聊天。

回复旧聊天。在发布功能之前。

steps:
1. Check if users already chatting.
2. If they are chatting echo chat which they already chatted.
3. If not chatting then start new chat.
4. if they are not chatting you dont need to echo anything just echo new message.

您现在的代码只会向您显示我认为的最后一条消息。因为你的页面正在重新加载。加载后，只需在我的视图中打印您的最后一条消息..您还需要打印旧聊天。

Answer 2

<html><head><title>new convo</title></head><body>
<?php include 'connect.php'; ?>
<?php include 'message_title_bar.php'; ?>
<?php include 'functions.php'; ?>
<?php $connect = mysqli_connect('localhost', 'root', '', 'php_mysql_login_system'); if (mysqli_connect_errno()) echo "Failed to connect to MySQL: " . mysqli_connect_error(); ?>
<div>
<?php
if (isset($_GET['user']) && !empty($_GET['user'])) {
    ?>
    <form method='post'>
        <?php
        if (isset($_POST['message']) && !empty($_POST['message'])) {
            $my_id = $_SESSION['user_id'];
            $user = $_GET['user'];
            $message = $_POST['message'];
            $query_string = "SELECT `hash` FROM `message_group` WHERE (`user_one`='$my_id' AND `user_two`='$user') OR (`user_one`='$user' AND `user_two`='$my_id')";
            $check_con = mysqli_query($connect, $query_string);
            if (mysqli_num_rows($check_con)) {
                $f_array = mysqli_fetch_array($check_con);
                $hash = $f_array['hash'];
                echo "<p>Conversation already Started</p>";
            } else {
                $hash = rand();
                mysqli_query($connect, "INSERT INTO message_group VALUES('$my_id' , '$user' , '$hash')");
                echo "<p>Conversation started</p>";
            }
            mysqli_query($connect, "INSERT INTO messages VALUES ('', '$hash', '$my_id', '$message')");
        }
        ?>
        <label for="message">Enter message:</label>
        <textarea name='message' id="message" rows='7' cols='60'></textarea>
        <br/>
        <br/>
        <input type='submit' name="submit" value="sendmessage"/>
    </form>
<?php
} else {
    echo "<b>Select User</b>";
    $user_list = mysqli_query($connect, "SELECT `id`,`username` FROM `users`");
    while ($run_user = mysqli_fetch_array($user_list)) {
        $user = $run_user['id'];
        $username = $run_user['username'];
        echo "<p><a href='send.php?user=$user'>$username</a></p>";
    }
}
?>
</div>
</body>
</html>

数据库中的数据仅首次输入

2 个答案: