PHP：如何處理上傳2個文件和輸出

，我會說我剛開始學習PHP的，所以把它容易對我做起......

基本上，我有一個表單，允許用戶上傳4張圖片（與完成其他字段一起）。提交時，表單調用一個PHP文件（下面的代碼），它基本上將細節添加到數據庫並將圖像上傳到文件服務器。這對我正在努力完成的工作很好。我遇到麻煩的地方是確認信息。

由於我有2個獨立的上傳領域，我基本上有2如果語句，確認兩個文件已被正確上傳。我想簡化這一點，所以我只需要顯示1條確認消息。

有關如何簡化此操作的任何想法？我在考慮如果兩個文件都成功上傳，代碼將會顯示，如果是這樣的話echo「x」，否則echo「y」。我對move_uploaded_file函數不太熟悉，所以我不確定是否可以在那裏使用用戶語句...任何想法都將非常感激。

//This is the directory where images will be saved 
$target = "path/"; 
$target = $target . basename($_FILES[controlcreative][name]); 
$target2 = "path/"; 
$target2 = $target2 . basename($_FILES[winnercreative][name]); 

$pic=($_FILES['controlcreative']['name']); 
$pic2=($_FILES['winnercreative']['name']); 

$con=mysqli_connect(); 
// Check connection 
if (mysqli_connect_errno()) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 

$sql="INSERT INTO experiments (vertical, pagetype, pagename, primarykpitype, primarykpilift, primarysignificant, testobjective, takeawayone, optimizationtype, controlcreative, winnercreative) 
VALUES 
('$_POST[vertical]','$_POST[pagetype]','$_POST[pagename]','$_POST[primarykpitype]','$_POST[primarykpilift]','$_POST[primarysignificant]','$_POST[testobjective]','$_POST[takeawayone]','$_POST[optimizationtype]','$pic','$pic2')"; 

if (!mysqli_query($con,$sql)) 
    { 
    die('Error: ' . mysqli_error($con)); 
    } 
echo "1 record added"; 


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

//Tells you if its all ok 
echo "The file ". basename($_FILES[controlcreative][name]). " has been uploaded, and your information has been added to the directory"; 
} 
else { 

//Gives and error if its not 
echo "Sorry, there was a problem uploading your file."; 
} 

//Writes the photo to the server 
if(move_uploaded_file($_FILES[winnercreative][tmp_name], $target2)) 
{ 

//Tells you if its all ok 
echo "The file ". basename($_FILES[winnercreative][name]). " has been uploaded, and your information has been added to the directory"; 
} 
else { 

//Gives and error if its not 
echo "Sorry, there was a problem uploading your file."; 
} 


mysqli_close($con);

來源

2013-09-29 user2828701

我通過$_FILES就不斷循環，這可能會爲您節省一些代碼，並設置一個錯誤標誌

$error = false; 
foreach($_FILES as $name => $file) 
{ 
    $target = '/path/to/destination/' . $file['name']; 
    if(!move_uploaded_file($file['tmp_name'], $target)) $error = true; 
}

然後

if(!$error) echo 'All files uploaded';

來源

2013-09-29 16:24:59 Mina

我的實際工作問題的答案...感謝儘管協助。 if（（move_uploaded_file（$ _ FILES [controlcreative] [tmp_name]，$ target））&& move_uploaded_file（$ _ FILES [winnercreative] [tmp_name]，$ target2））' – user2828701

PHP：如何處理上傳2個文件和輸出

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