Question

我在数据库中有一个表，我在其中的wav列中存储longblob个文件。我正在尝试创建一个文件夹并将给定用户的所有wav都存储在服务器上。到目前为止，我所拥有的是查询数据库和表以及到目前为止仅存储一个wav的简单示例。我对PHP不太熟悉，因此需要一点帮助。

$con=mysqli_connect("localhost","root","xxxx","testDB");

if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$sql="SELECT recording FROM users WHERE accountID = '1234'";
$result=mysqli_query($con,$sql);

while ($result = mysqli_fetch_array($row)) {

    $data = $result["recording"];

    //mkdir('/var/spool/sounds/', 0755, true);

    $wavFileName = "/var/spool/sounds/wavfile.wav";
    $fh = fopen($wavFileName, 'w'); 
    fwrite($fh, $data);
    fclose($fh);
}

上面的摘要是在该目录中创建一个wav。

如果用户有5个保存的wav文件，如何在内部创建名称为userID和file1.wav，file2.wav等的目录？

Answer 1

只需将路径和文件名与用户ID和已创建的文件数串联起来即可。然后，检查该路径是否存在，否则创建它。

$directory = "/var/spool/sounds/" . $userID;
$fileExtension = ".wav";
$fileNumber = 1;
while ($result = mysqli_fetch_array($row))
{
    $data = $result["recording"];
    if (!is_dir($directory))
        mkdir($directory, 0755, true);

    $wavFileName = $directory . "/file" . $fileNumber;
    /*
     *  if you need to check if the file already exists,
     *  you can rename the new file "/var/spool/sounds/userId/file1_copy.wav"
     *  or "/var/spool/sounds/userId/file1_copy_copy.wav" if a copy already exists and so on.
     */
    while (file_exists($wavFileName . $fileExtension))
        $wavFileName .= "_copy";
    $fh = fopen($wavFileName . $fileExtension, 'w'); 
    fwrite($fh, $data);
    fclose($fh);
    $fileNumber++;
}

Answer 2

使用以下代码，将可以工作：

$account_id = 1234;
mkdir('/var/spool/sounds/'.$account_id, 0755, true);
$sql="SELECT recording FROM users WHERE accountID = ".$account_id ;
$result=mysqli_query($con,$sql);
$i=0;
while ($result = mysqli_fetch_array($row)) {
    $i++;
    $data = $result["recording"];

    $wavFileName = "/var/spool/sounds/".$account_id."/file".$i.".wav";
    $fh = fopen($wavFileName, 'w'); 
    fwrite($fh, $data);
    fclose($fh);
}

为数据库中的每个记录创建WAV文件

2 个答案: