Question

public $id;
public $filename;
public $type;
public $size;
public $description;
public $title;

这就是我现在正在使用的，这很糟糕，

$sql = "INSERT INTO photographgallery 
              (filename,type,size,description,title)
    VALUES ('$sanitized_filename', '$sanitized_type', '$sanitized_size', '$sanitized_description', '$sanitized_title')";

我想知道我是否可以为此编写一份准备好的声明，我将如何处理它，我是堆栈。帮助

Answer 1

在SQL中，使用问号占位符（?）替换变量。通过将查询传递给mysqli_stmt来创建mysqli::prepare，然后通过调用mysqli_stmt::bind_param将变量绑定到占位符。调用mysqli_stmt::execute执行插入。它看起来像这样：

$sql = "INSERT INTO photographgallery 
            (filename, type, size, description, title)
          VALUES
            (?, ?, ?, ?, ?)";
$stmt = $conn->prepare($sql);
// The string 'ssiss' in the following means 'string, string, int, string, string'
//  and describes the types of the parameters.
$stmt->bind_param('ssiss', $filename, $type, $size, $description, $title);
$stmt->execute();
$stmt->close();  // always clean up after yourself

Answer 2

// Your variables, however you get them.    
$filename = "name1";
$type = "type1";
$size = 100;
$desc = "test desc";
$title = "title"

if($stmt = $mysqli->prepare("INSERT INTO photographgallery (filename, type, size, description, title) VALUES (?, ?, ?, ?, ?)") {
    $stmt->bind_param('ssiss', $filename, $type, $size, $desc, $title); //Assuming the variables are string, string, int, string, string respectively
    $stmt->execute();
    $stmt->close();
}

在代码周围使用if可确保仅在prepare语句没有错误时运行。如果有错误，则prepare语句返回false。

我将如何编写准备好的sql语句而不是使用real_escape PHP，MySQL

2 个答案: