Question

这是我将图像上传到FTP服务器及其名称到Mysql db：

的代码

session_start();

require 'config2.php';
require_once 'user.class.php';

 $target = "sent/"; 
 $target1 = "sent/";
 $target2 = "sent/";
 $target3 = "sent/";
  $target4 = "sent/";
 $target5 = "sent/";

 $target = $target . basename( $_FILES['photo']['name']); 
$target1 = $target1 . basename( $_FILES['photo1']['name']); 
$target2 = $target2 . basename( $_FILES['photo2']['name']);
$target3 = $target3 . basename( $_FILES['photo3']['name']);
$target4 = $target4 . basename( $_FILES['photo4']['name']);
$target5 = $target5 . basename( $_FILES['photo5']['name']);

 //This gets all the other information from the form 
$login = $_SESSION['login'];    
  $name=$_POST['name']; 
$name1 =$_POST['name1'];
$name2 =$_POST['name2'];
$name3 =$_POST['name3'];
$name4 =$_POST['name4'];
$name5 =$_POST['name5'];
  $pic=($_FILES['photo']['name']); 
 $pic1=($_FILES['photo1']['name']); 
  $pic2=($_FILES['photo2']['name']); 
  $pic3=($_FILES['photo3']['name']); 
   $pic4=($_FILES['photo4']['name']); 
 $pic5=($_FILES['photo5']['name']); 

  $id=$_SESSION['id'];
 // Connects to your Database 



   //$op = mysql_query("select id from users where id = '$id' ");
 //Writes the information to the database 
//mysql_query("INSERT INTO users (name, photo, photo1, name1) VALUES ('$name', '$pic'  ,'$pic1', '$name1')") ; 


  mysql_query("UPDATE users SET name='$name', name1='$name1', name2='$name2', name3='$name3', name4='$name4', name5='$name5', photo = '$pic', photo1 = '$pic1', photo2 = '$pic2', photo3 = '$pic3', photo4 = '$pic4', photo5 = '$pic5' WHERE  login = '$login' ");


$allowed = array('jpg', 'jpeg', 'gif', 'png');  

 //Writes the photo to the server 
 if(move_uploaded_file($_FILES['photo']['tmp_name'] ,$target) )
 { 




 //Tells you if its all ok 
 echo "Plik ". basename( $_FILES['photo']['name'] ). " uploaded successfuly"; 
echo implode(', ', $allowed);
 } 


if(move_uploaded_file($_FILES['photo1']['tmp_name'] ,$target1) )
 { 

 //Tells you if its all ok 
 echo "Plik ". basename( $_FILES['photo1']['name'] ). " uploaded successfuly"; 
 echo implode(', ', $allowed);
 } 

 if(move_uploaded_file($_FILES['photo2']['tmp_name'] ,$target2) )
 { 

 //Tells you if its all ok 
  echo "Plik ". basename( $_FILES['photo2']['name'] ). " uploaded successfuly"; 
 echo implode(', ', $allowed);
 } 

 if(move_uploaded_file($_FILES['photo3']['tmp_name'] ,$target3) )
 { 

 //Tells you if its all ok 
 echo "Plik ". basename( $_FILES['photo3']['name'] ). " uploaded successfuly"; 
 echo implode(', ', $allowed);
 } 

if(move_uploaded_file($_FILES['photo4']['tmp_name'] ,$target4) )
   { 

 //Tells you if its all ok 
  echo "Plik ". basename( $_FILES['photo4']['name'] ). " uploaded successfuly"; 
 echo implode(', ', $allowed);
 } 

 if(move_uploaded_file($_FILES['photo5']['tmp_name'] ,$target5) )
 { 

 //Tells you if its all ok 
 echo "Plik ". basename( $_FILES['photo5']['name'] ). " uploaded successfuly"; 
 echo implode(', ', $allowed);

工作正常。但是现在我想添加用于分配新名称的代码，如果有任何用户上传了具有相同名称的图像。因为当两个或多个图像的名称相同时，脚本将覆盖较旧的图像。那么，你能看到这个问题的任何解决方案吗？

Answer 1

为文件名添加时间戳。

$target = $target . time().basename( $_FILES['photo']['name']); 
$target1 = $target1 . time().basename( $_FILES['photo1']['name']); 
$target2 = $target2 . time().basename( $_FILES['photo2']['name']);
$target3 = $target3 . time().basename( $_FILES['photo3']['name']);
$target4 = $target4 . time().basename( $_FILES['photo4']['name']);
$target5 = $target5 . time().basename( $_FILES['photo5']['name']);

Answer 2

只需更改目标文件的名称即可。这是您设置这些名称的地方：

$target = $target . basename( $_FILES['photo']['name']); 
$target1 = $target1 . basename( $_FILES['photo1']['name']); 
$target2 = $target2 . basename( $_FILES['photo2']['name']);
$target3 = $target3 . basename( $_FILES['photo3']['name']);
$target4 = $target4 . basename( $_FILES['photo4']['name']);
$target5 = $target5 . basename( $_FILES['photo5']['name']);

您可以将这些变量设置为您想要的任何字符串。例如，如果要创建全局唯一的文件名，则可以使用GUID或某些unique ID作为整个文件名或附加到文件名。 您想要的“唯一性”的衡量取决于您。但最终您的代码显式设置目标文件名，因此您可以显式将该值设置为您想要的任何值。

Answer 3

我已经尝试过，这成功了。 $target = time().( $_FILES['photo']['name']);

例如。 // Example of accessing data for a newly uploaded file $fileName = time().$_FILES["pic"]["name"]; $fileTmpLoc = $_FILES["pic"]["tmp_name"];

time()此函数将重命名具有相同名称的文件，并保留用户上传的另一个相同文件。希望这会有所帮助。感谢

当用户在ftp服务器上传同名文件（图像）时，如何更改文件名

3 个答案: