2016-01-24 25 views
1

我想在表單提交後創建一個目錄,然後在提交表單之後我想讓該目錄中有一個文件塞入其中。我已經嘗試了PHP複製功能,並使用if語句來查看它是否成功複製,但它不是。請看看我的代碼,看看發生了什麼。當它工作時,只會提交一次「1」輸出。沒有實際的文件被移動到文件夾內。如何在提交後將文件複製到另一個目錄

if($_POST['submit']=='Register') 
{ 
// If the Register form has been submitted 
$root = "/serves/registered.php"; 
$err = array(); 
$folder = mkdir($_POST['username']); 
$reg = "registered.php"; 
mkdir($_POST['username']); 
copy($root,$folder); 
if (!copy($root, $folder)) { 
     echo "failed to copy $root...\n"; 
} else { 
    echo "Account was successfuly created."; 
} 

謝謝

+0

爲什麼在地球上,你做的拷貝兩次? – MikeCAT

+0

要移動文件/文件夾,您可以使用「重命名」功能。但我不明白你想做什麼。爲什麼你'mkdir($ _ POST ['username'])'和'copy($ root,$ folder)'兩次? – HtmHell

+0

@HtmHell右邊是用戶正在創建的用戶名,然後用他選擇的用戶名創建一個新目錄。現在我們需要點擊提交按鈕上該目錄內的文件。 – aidangig

回答

1

這可以做到這一點:

$root = "serves/registered.php"; 
$folder = mkdir($_POST['username']); 
if($folder) { 
    $reg = "registered.php"; 
    if (!copy($root, $_POST['username']."/".$reg)) { 
     echo "failed to copy $root...\n"; 
    } else { 
     echo "Account was successfuly created."; 
    } 
} 
else { 
    echo "Could not create folder"; 
} 
+0

非常感謝工作:) – aidangig

0
if($_POST['submit']=='Register') 
{ 
    // If the Register form has been submitted 
    $root = "contact.php"; 
    $folder = $_POST['username'] . '/' . $root; 
    mkdir($_POST['username']); 
    if (!copy($root, $folder)) { 
     echo "failed to copy $root...\n"; 
    } else { 
    echo "Account was successfuly created."; 
    } 
} 
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