写一个php修改的代码，数据库包含a.id,bookname,categoryName,salePrice,picture,publishingName from books AS a INNER JOIN books_category AS b ON a.categoryId=b.id

以下是一个简单的 PHP 代码示例，用于更新 `books` 表中的数据：

<?php
// 连接数据库
$conn = mysqli_connect("localhost", "username", "password", "database_name");

// 检查连接是否成功
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

// 获取要更新的数据
$id = $_POST['id'];
$bookname = $_POST['bookname'];
$categoryName = $_POST['categoryName'];
$salePrice = $_POST['salePrice'];
$picture = $_POST['picture'];
$publishingName = $_POST['publishingName'];

// 构造 SQL 语句
$sql = "UPDATE books a 
        INNER JOIN books_category b ON a.categoryId = b.id 
        SET a.bookname='$bookname', a.categoryName='$categoryName', a.salePrice='$salePrice', 
        a.picture='$picture', a.publishingName='$publishingName' 
        WHERE a.id='$id'";

// 执行 SQL 语句
if (mysqli_query($conn, $sql)) {
    echo "Record updated successfully";
} else {
    echo "Error updating record: " . mysqli_error($conn);
}

// 关闭数据库连接
mysqli_close($conn);
?>

请注意，此示例代码仅供参考，并且需要根据您的实际情况进行适当修改和调整。