Question

我只是想从数据库表中获取用户名并将文件上传到以username命名的文件夹中，但它给了我一个错误。这是错误：Parse error: syntax error, unexpected T_STRING in /home/u381071273/public_html/upload/upload.php on line 18但我无法找到它。我想从数据库表中获取用户名并将其显示在我放置&username

的位置

代码：

require("models/db-settings.php");
$mysqli = new mysqli($db_host, $db_user, $db_pass, $db_name);

$result = $mysqli->query("SELECT user_name FROM upl_users");

// This will move the internal pointer and skip the first row, we don't want that.
//$row = mysql_fetch_assoc($result);
//echo $row['user_name'];

while ( $row = $result->fetch_assoc() ) {
echo $row['user_name'];}
$dir = "uploads/".$row['user_name']."/";

if (file_exists($UploadedDirectory)) {

mkdir('uploads/".$row['user_name']."/', 0777, true);

}
if(isset($_FILES["FileInput"]) && $_FILES["FileInput"]["error"]== UPLOAD_ERR_OK)
{
############ Edit settings ##############
$UploadDirectory    = 'uplaods/".$row['user_name']."/'; //specify upload    directory ends with / (slash)
##########################################

/*
Note : You will run into errors or blank page if "memory_limit" or "upload_max_filesize" is set to low in "php.ini". 
Open "php.ini" file, and search for "memory_limit" or "upload_max_filesize" limit 
and set them adequately, also check "post_max_size".
*/

//check if this is an ajax request
if (!isset($_SERVER['HTTP_X_REQUESTED_WITH'])){
    die();
}


//Is file size is less than allowed size.
if ($_FILES["FileInput"]["size"] > 5242880) {
    die("File size is too big!");
}

//allowed file type Server side check
switch(strtolower($_FILES['FileInput']['type']))
    {
        //allowed file types
        case 'image/png': 
        case 'image/gif': 
        case 'image/jpeg': 
        case 'image/pjpeg':
        case 'text/plain':
        case 'text/html': //html file
        case 'application/x-zip-compressed':
        case 'application/pdf':
        case 'application/msword':
        case 'application/vnd.ms-excel':
        case 'video/mp4':
        case 'audio/mp3';
            break;
        default:
            die('Unsupported File!'); //output error
}

$File_Name          = strtolower($_FILES['FileInput']['name']);
$File_Ext           = substr($File_Name, strrpos($File_Name, '.')); //get file    extention
$Random_Number      = uniqid(); //Random number to be added to name.
$NewFileName        = $Random_Number.$File_Ext; //new file name

if(move_uploaded_file($_FILES['FileInput']['tmp_name'],    $UploadDirectory.$NewFileName ))
   {
    die('Success! File Uploaded.');
}else{
    die('error uploading File!');
}

}
else
{
die('Something wrong with upload! Is "upload_max_filesize" set correctly?');
}

Answer 1

mkdir('uploads/".$row['user_name']."/', 0777, true);

应该是：

mkdir('uploads/'.$row['user_name'].'/', 0777, true);

（您在进行连接时使用了不同的字符串锚点 - 使用'启动了字符串但尝试使用"将其关闭）

Answer 2

使用此行:(您可以使用单引号或双引号不要混合）

mkdir("uploads/".$row['user_name']."/", 0777, true);

无法从数据库表中选择用户名并显示它

2 个答案: