Question

从数据库获取文件路径。我将文件路径存储到变量中，但是当我尝试将该变量解析为fopen函数时，它会抛出一个错误，说明文件名不能为空。

的login.html

<form action="a.php" method="POST">
Username: <input type="text" name="username"></br>
Password: <input type="text" name="password"></br>
Email-Address: <input type="text" name="email_address"></br>
<input type="submit">
</form>

a.php只会

$con=mysqli_connect("localhost","abc","abc","mysql");

if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$username= $_POST['username'];
$password= $_POST['password'];
$email=   $_POST['email_address'];
$r= mysqli_query($con,"SELECT * FROM Assignment where email_address='$email'");
$row = mysqli_fetch_array($r);
$filename = mysqli_real_escape_string($con, $row['home_directory']);

if(isset($_POST['text1'])){
$file_open = fopen($filename,"w+"); 
fwrite($file_open, $_POST['text1']);
fclose($file_open);
}
?>

<form action="" method="POST">
<textarea  class="text_edit" name="text1"  id="my_text" ></textarea></br>

<input type="submit" name="button">
    </form>

我想知道这里的错误是什么。为什么fopen不能使用数据库中的retreived文件路径

Answer 1

使用var_dump函数检查$ filename是否为空。试试这段代码：

$filename = $row['home_directory'];
var_dump($filename);

如何从数据库中获取数据并将数据解析为fopen函数

1 个答案: