PHP文件上傳計數集文件

Question

我想統計設置的文件上傳。這是我使用的代碼。有沒有更好的方法來做到這一點。謝謝。

<form action="index.php" method="post" enctype="multipart/form-data"> 
    <input name="new_image[]" type="file" /> 
    <input name="new_image[]" type="file" /> 
    <input name="new_image[]" type="file" /> 
    <input name="new_image[]" type="file" /> 
    <input name="new_image[]" type="file" /> 
<button name="submit" type="submit">Upload</button> 
</form> 
<?php 

$img_error = '0'; 
$fill_img_count = '0'; 
if(isset($_POST['submit'])) 
{ 
    $img_count = count($_FILES['new_image']); 
    echo "Total : ".$img_count."<br>"; 
    for ($i=0 ; $i<=$img_count ; $i++) 
    { 
     if (isset($_FILES['new_image']) && !empty($_FILES['new_image']['name'][$i])) 
     { 
      $fill_img_count++; 
     } 
    } 
    echo "Set : ".$fill_img_count."<br>"; 
} 
?> 

Answer 1

我建議你測試對每個UPLOAD_ERR_OK關鍵['error']。

Answer 2

你不需要有name="new_image[]"作爲名字...只是new_image就足夠了。如果您發佈1或多個，在PHP端，你會看到$ _FILES []

你一些有用的鏈接：

• http://php.net/manual/en/reserved.variables.files.php
• http://www.php.net/manual/en/features.file-upload.php

一些代碼：

if (empty($_FILES)) { echo "0 files uploaded"; } 
    else { echo count($_FILES) . " files uploaded"; } 

編輯基於評論：

• How to handle multiple file upload using PHP

從該職位：

echo count($_FILES['file']['tmp_name']); 

Answer 3

<?php 
$count = 0; 
foreach($_FILES['new_image']['error'] as $status){ 
    if($status === UPLOAD_ERR_OK) { 
     $count++; 
    } 
} 
var_dump($count); 
?> 
<form action="test.php" method="post" enctype="multipart/form-data"> 
    <input name="new_image[]" type="file" /> 
    <input name="new_image[]" type="file" /> 
    <input name="new_image[]" type="file" /> 
    <input name="new_image[]" type="file" /> 
    <input name="new_image[]" type="file" /> 
<button name="submit" type="submit">Upload</button> 
</form> 

Answer 4

$count_files = 0; 
foreach ($_FILES['picture']['error'] as $item) { 
    if ($item != 4) { 
     $count_files++; 
    } 
} 
echo $count_files;