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我目前正在为我的课程项目做一个项目。我目前正在尝试更新到数据库中，但我沿途收到一些错误，基本上它是一个单选按钮，用于设置链接到更新页面。任何帮助和见解将不胜感激！MySQL和PHP更新
<html>
<head>
<title>asdf</title>
<link rel="stylesheet" type="text/css" href="Background.css">
</head>
<?php
session_start();
if(!isset($_SESSION["login"]))
header("location:admin.php");
?>
<body>
<h1 style="color:white"><u><center></center></u></h1>
<div id="BG"></div>
<form action = "update1.php" method = "GET">
<table border = 0>
<tr>
<td>Image: <input type = "text" name = "image" id = "image"></td>
<br/>
<td>Hero Name: <input type = "text" name = "heroes" id = "heroes"></td>
<br/>
<td>Role: <input type = "text" name = "roles" id = "roles"></td>
<br/>
<td>Attribute: <input type = "text" name = "attribute" id = "attribute"></td>
<br/>
<td>Description: <input type = "text" name = "description" id = "description"></td>
<br/>
<td>General: <input type = "text" name = "general" id = "general"></td>
<br/>
</tr>
</table>
</br>
<input type = "submit" name="update" value = "Update">
</form>
</center>
</html>
<?php
ini_set('display_errors', 1); ini_set('display_startup_errors', 1); error_reporting(E_ALL);
define("DB_USER","*****");
define("DB_PASSWORD","****");
define("DB_HOST","*****");
define("DB_NAME","*****");
$dbc=mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD,DB_NAME);
if(isset($_GET['update']))
{
$image = $_GET['image'];
$heroes = $_GET['heroes'];
$roles = $_GET['roles'];
$attribute = $_GET['attribute'];
$description = $_GET['description'];
$general = $_GET['general'];
$sql = "update `Dota 2 select` set (`image` = '$image',`heroes` = '$heroes') WHERE (heroes= '$heroes', image = '$image')";
// $sql = "Update `Dota 2 select` SET (`image`= [$image]) = WHERE `image`)";
// $sql = "Update `Dota 2 select` SET (`image`= [$image],`heroes` =[$heroes],`roles` =[$roles],`attribute`=[$attribute],`description`=[$description],`general`=[$general]) = WHERE `heroes`='$heroes')";
// $sql = "Update `Dota 2 select` SET (`image`= [$image],`heroes`,`roles`,`attribute`,`description`,`general`) = WHERE (`image`,`heroes`,`roles`,`attribute`,`description`,`general`) = ('$image','$heroes','$roles','$attribute','$description','$general')";
if(!mysqli_query($dbc, $sql))
{
echo(mysqli_error($dbc));
}
else
{
echo 'Data successfully updated!';
}
mysqli_close($dbc);
}
?>
这是 此页“ 您的SQL语法错误的错误;检查对应于你的MySQL服务器版本使用附近“（image =‘正确的语法手册一’ ，heroes = 'A'），其中（英雄= 'A'，图像= 'A'）”位于第1行
“
'A' ='这里了'。你的更新语法无效。 –
表格名称的名称中不能有空格。 – Rahi
@Rahi *“表名不能有名字空格” - 噢，为什么不呢？他们已经逃过了表名。 –