Question

我想计算上传文件集。这是我的使用代码。有没有更好的方法来做到这一点。感谢。

<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

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

Answer 2

我建议针对UPLOAD_ERR_OK测试每个['error']密钥。

Answer 3

您不需要name="new_image[]"作为名称......只需要new_image即可。如果您发布1或多个，在PHP端，您将看到$ _FILES []

一些有用的链接：

一些代码：

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

根据评论

修改：

How to handle multiple file upload using PHP

从那篇文章：

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

Answer 4

<?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>

PHP文件上传计数设置文件

4 个答案: