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PHP 发布 导致插入错误

转载 作者:行者123 更新时间:2023-11-29 13:13:41 24 4
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我正在编写一个 php 文件,该文件从表单中获取值并将它们发送到 mysql 数据库。其中一个表字段是一个按钮链接,指向单击时将播放的视频。如果我进入数据库并手动添加链接,效果会很好。但是我的 PHP 插入导致错误。请看一下这段代码:

$fileName = "video_".$id.".html";
$link = "<a href=\"javascript: void(0)\" onclick=\"MM_openBrWindow('videos/".$fileName."','','toolbar=no,location=no,status=no,menubar=no,scrollbars=no,resizable=yes,width=420,height=390')\"><button class=\"count\">Watch Video</button></a>";
$con=mysqli_connect("localhost","videomanager","password","my_database");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="INSERT INTO video_list (date, title, description, link) VALUES('$_POST[date]','".mysqli_real_escape_string($_POST['sermon'])."','".mysqli_real_escape_string($_POST['description'])."','$link' )";
if (!mysqli_query($sql,$con))
{
die('Error: ' . mysqli_error());
}
echo "This video has been successfully added to the video database.";
mysqli_close($con);

如果我通过执行以下操作来查看 $link:echo $link;死();它生成一个带有按钮的页面,并且按钮中的代码看起来不错。这是我试图插入它的方式吗?感谢您的帮助!

最佳答案

你有很多错误

<?php
$fileName = "video_".$id.".html";
$link = "<a href=\"javascript: void(0)\" onclick=\"MM_openBrWindow('videos/".$fileName."','','toolbar=no,location=no,status=no,menubar=no,scrollbars=no,resizable=yes,width=420,height=390')\"><button class=\"count\">Watch Video</button></a>";
$con = mysqli_connect("localhost","videomanager","password","my_database");

// Check connection
if ( mysqli_connect_errno() )
die('Failed to connect to MySQL: ' . mysqli_connect_error() );

// Check param is set
if( !isset($_POST['date'], $_POST['sermon'], $_POST['description']) )
die('Param Error');

// SQL Request
$sql = sprintf("INSERT INTO video_list (date, title, description, link) VALUES('%s','%s','%s','%s')",
mysqli_real_escape_string($con, $_POST['date']),
mysqli_real_escape_string($con, $_POST['sermon']),
mysqli_real_escape_string($con, $_POST['description']),
mysqli_real_escape_string($con, $link)
);

// SQL execute
$result = mysqli_query($con, $sql) or die('Error: ' . mysqli_error($con));

// Free result
mysqli_free_result($result);

// Close connection
mysqli_close($con);

echo "This video has been successfully added to the Kim Watt videos.";

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