如何使用Foreach为每个用户以不同的名称上传相同的文件。

Posted 2023-01-20

Foreach($_POST['...]...).其他查询通过foreach循环工作，但文件上传为数组的第1个索引，这不是多文件上传，我想为每个用户以不同的名称上传相同的文件。

    foreach($_POST['groupmem'] as $user)
    //Some Queries

    $filename2 = str_replace(" ", "_","$user.$_FILES['proposal']['name']");
    $destination2 = '../img/proposal/' . $filename2;
    $extension2 = pathinfo($filename2, PATHINFO_EXTENSION);
    $file2 = $_FILES['proposal']['tmp_name'];
    $size2 = $_FILES['proposal']['size'];
    if (!in_array($extension2, ['zip', 'pdf', 'docx'])) 
        echo "You file extension must be .zip, .pdf or .docx";
     elseif ($_FILES['proposal']['size'] > 200000000)  // file shouldn't be larger than 200Megabyte
        echo "File too large!";
     else 
        if (move_uploaded_file($file2, $destination2)) 
            $sql = "UPDATE project SET proposal_name='$filename2' WHERE u_id='$user' ";
            if (mysqli_query($conn, $sql)) 
                echo "File uploaded successfully";
            
         else 
            echo "Failed to upload file.";
        
    
    

答案

你不能在循环中进行move_uploaded_file的操作。

$user1 = $_POST['groupmem'][0];
$filename1 = str_replace(" ", "_","$user1.$_FILES['proposal']['name']");
$destination1 = '../img/proposal/' . $filename1;
$extension1 = pathinfo($filename1, PATHINFO_EXTENSION);
$file1 = $_FILES['proposal']['tmp_name'];
$size1 = $_FILES['proposal']['size'];
if (!in_array($extension1, ['zip', 'pdf', 'docx'])) 
    echo "You file extension must be .zip, .pdf or .docx";
 elseif ($_FILES['proposal']['size'] > 200000000)  // file shouldn't be larger than 200Megabyte
    echo "File too large!";
 else 
    if (move_uploaded_file($file1, $destination1)) 

    foreach($_POST['groupmem'] as $user)
        $filename2 = str_replace(" ", "_","$user.$_FILES['proposal']['name']");
        $destination2 = '../img/proposal/' . $filename2;
        if ($user <> $user1) 
            if (!copy($destination1, $destination2))  echo "failed to copy $file...\n";
        

        $sql = "UPDATE project SET proposal_name='$filename2' WHERE u_id='$user' ";
        if (mysqli_query($conn, $sql)) 
        echo "File uploaded successfully";
        
    

     else 
        echo "Failed to upload file.";