从我的数组中获取错误，我无法获取我的网址ID

Question

我正在做的是获取url id来分割聊天室内容，但现在我有错误

这是我的

chatMessage.php

include '../config.php';
include 'login.php';

$username = $_SESSION['id'];
$chatroomID =$_GET["chatroomID"];

$sql="SELECT * FROM chatroom_chat where chatroom_id ='$chatroomID'";

while ($row = mysqli_fetch_array($sql)) {
    $chat = $row['chat_ID'];
    $getChatData = "SELECT * FROM chat where id = $chat";
}

$sql1= "SELECT * FROM (
            SELECT * FROM chat ORDER BY id DESC LIMIT 0,40
            ) sub
            ORDER BY id ASC ";

$result =  mysqli_query($connection, $sql1);

    while($extract = mysqli_fetch_array($result)){
        $color = ($extract['id'] == $username) ? '#FFFFFF' : '#66FFFF';
        $position = ($extract['id'] == $username) ? 'right' : 'left';
        $border = ($extract['id'] == $username) ? ' 1px solid black ' : ' none ';

        echo "<div class='msg-dateandtime' style='text-align:$position; float:$position;'> <div class='left-username' style='color:blue;'>" . $extract['id'] ."</div>"
                . "<div class='space'></div>"
                . "<div class='right-date'>  ". $extract['date'] ." </div></div>"
                . "<div class='wrap-message' style='background-color:$color; border:$border; float:$position;'>"
                . "<p style 'text-align=$position; margin:0; padding:0; text-align:left;'> ".$extract['chat']."</p></div>";
    }

我不知道哪里出错了

Answer 1

请使用如下：

$chatroomID = isset($_GET["chatroomID"]) ? $_GET["chatroomID"] : '';

原因：当您为变量$ chatroomID分配值并且GET请求没有 chatroomID 索引可用时，会出现错误。在分配给变量之前，您应首先检查此索引上是否有值。

警告：防止在SQL查询中直接使用用户输入变量。而是使用准备好的陈述。

参考：http://php.net/manual/en/mysqli-stmt.bind-param.php