如何检查用户是否在PHP中上传了文件？

I do some form validation to ensure that the file a user uploaded is of the right type. But the upload is optional, so I want to skip the validation if he didn't upload anything and submitted the rest of the form. How can I check whether he uploaded something or not? Will $_FILES['myflie']['size'] <=0 work?

我做了一些表单验证，以确保用户上传的文件类型正确。但上传是可选的，所以如果他没有上传任何内容并提交表单的其余部分，我想跳过验证。我如何检查他是否上传了某些内容？ $ _FILES ['myflie'] ['size'] <= 0会工作吗？

7 个解决方案

#1

110

You can use is_uploaded_file():

您可以使用is_uploaded_file（）：

if(!file_exists($_FILES['myfile']['tmp_name']) || !is_uploaded_file($_FILES['myfile']['tmp_name'])) {
    echo 'No upload';
}

From the docs:

来自文档：

Returns TRUE if the file named by filename was uploaded via HTTP POST. This is useful to help ensure that a malicious user hasn't tried to trick the script into working on files upon which it should not be working--for instance, /etc/passwd.

如果文件名命名的文件是通过HTTP POST上传的，则返回TRUE。这有助于确保恶意用户不会欺骗脚本处理不应该工作的文件 - 例如/ etc / passwd。

This sort of check is especially important if there is any chance that anything done with uploaded files could reveal their contents to the user, or even to other users on the same system.

如果对上传文件所做的任何事情都有可能向用户或甚至同一系统上的其他用户显示其内容，则此类检查尤其重要。

EDIT: I'm using this in my FileUpload class, in case it helps:

编辑：我在我的FileUpload类中使用它，以防它有帮助：

public function fileUploaded()
{
    if(empty($_FILES)) {
        return false;       
    } 
    $this->file = $_FILES[$this->formField];
    if(!file_exists($this->file['tmp_name']) || !is_uploaded_file($this->file['tmp_name'])){
        $this->errors['FileNotExists'] = true;
        return false;
    }   
    return true;
}

#2

@karim79 has the right answer, but I had to rewrite his example to suit my purposes. His example assumes that the name of the submitted field is known and can be hard coded in. I took that a step further and made a function that will tell me if any files were uploaded without having to know the name of the upload field.

@ karim79有正确的答案，但我不得不重写他的例子以适应我的目的。他的例子假设提交的字段的名称是已知的并且可以被硬编码。我更进了一步并且创建了一个函数，它将告诉我是否上传了任何文件而不必知道上传字段的名称。

/**
 * Tests all upload fields to determine whether any files were submitted.
 * 
 * @return boolean
 */
function files_uploaded() {

    // bail if there were no upload forms
   if(empty($_FILES))
        return false;

    // check for uploaded files
    $files = $_FILES['files']['tmp_name'];
    foreach( $files as $field_title => $temp_name ){
        if( !empty($temp_name) && is_uploaded_file( $temp_name )){
            // found one!
            return true;
        }
    }   
    // return false if no files were found
   return false;
}

#3

This code worked for me. I am using multiple file uploads so I needed to check whether there has been any upload.

这段代码对我有用。我正在使用多个文件上传，所以我需要检查是否有任何上传。

HTML part:

HTML部分：

<input name="files[]" type="file" multiple="multiple" />

PHP part:

PHP部分：

if(isset($_FILES['files']) ){  


foreach($_FILES['files']['tmp_name'] as $key => $tmp_name ){

      if(!empty($_FILES['files']['tmp_name'][$key])){

    //  things you want to do
    }
}

#4

I checked your code and think you should try this:

我检查了你的代码并认为你应该试试这个：

if(!file_exists($_FILES['fileupload']['tmp_name']) || !is_uploaded_file($_FILES['fileupload']['tmp_name'])) 
    {
        echo 'No upload';
    }   
    else
        echo 'upload';

#5

<!DOCTYPE html>
<html>
<body>

<form action="#" method="post" enctype="multipart/form-data">
    Select image to upload:
    <input name="my_files[]" type="file" multiple="multiple" />
    <input type="submit" value="Upload Image" name="submit">
</form>


<?php

 if (isset($_FILES['my_files']))
  {
    $myFile = $_FILES['my_files'];
    $fileCount = count($myFile["name"]);


        for ($i = 0; $i <$fileCount; $i++)
         {
           $error = $myFile["error"][$i]; 

            if ($error == '4')  // error 4 is for "no file selected"
             {
               echo "no file selected";
             }
            else
             {

               $name =  $myFile["name"][$i];
               echo $name; 
               echo "<br>"; 
               $temporary_file = $myFile["tmp_name"][$i];
               echo $temporary_file;
               echo "<br>";
               $type = $myFile["type"][$i];
               echo $type;
               echo "<br>";
               $size = $myFile["size"][$i];
               echo $size;
               echo "<br>";



               $target_path = "uploads/$name";   //first make a folder named "uploads" where you will upload files


                 if(move_uploaded_file($temporary_file,$target_path))
                  {
                   echo " uploaded";
                   echo "<br>";
                   echo "<br>";
                  }
                   else
                  {
                   echo "no upload ";
                  }




              }
        }  
}
        ?>


</body>
</html>

refer http://www.techzigzag.com/how-to-check-that-user-has-upload-any-file-or-not-in-php/

参考http://www.techzigzag.com/how-to-check-that-user-has-upload-any-file-or-not-in-php/

But be alert. User can upload any type of file and also can hack your server or system by uploading a malicious or php file. In this script there should be some validations. Thank you.

但要警惕。用户可以上传任何类型的文件，也可以通过上传恶意或php文件来破解您的服务器或系统。在这个脚本中应该有一些验证。谢谢。

#6

You should use $_FILES[$form_name]['error']. It returns UPLOAD_ERR_NO_FILE if no file was uploaded. Full list: PHP: Error Messages Explained

你应该使用$ _FILES [$ form_name] ['error']。如果没有上传文件，则返回UPLOAD_ERR_NO_FILE。完整列表：PHP：错误消息说明

function isUploadOkay($form_name, &$error_message) {
    if (!isset($_FILES[$form_name])) {
        $error_message = "No file upload with name '$form_name' in form.";
        return false;
    }
    $error = $_FILES[$form_name]['error'];

    // List at: http://php.net/manual/en/features.file-upload.errors.php
    if ($error != UPLOAD_ERR_OK) {
        switch ($error) {
            case UPLOAD_ERR_INI_SIZE:
                $error_message = 'The uploaded file exceeds the upload_max_filesize directive in php.ini.';
                break;

            case UPLOAD_ERR_FORM_SIZE:
                $error_message = 'The uploaded file exceeds the MAX_FILE_SIZE directive that was specified in the HTML form.';
                break;

            case UPLOAD_ERR_PARTIAL:
                $error_message = 'The uploaded file was only partially uploaded.';
                break;

            case UPLOAD_ERR_NO_FILE:
                $error_message = 'No file was uploaded.';
                break;

            case UPLOAD_ERR_NO_TMP_DIR:
                $error_message = 'Missing a temporary folder.';
                break;

            case UPLOAD_ERR_CANT_WRITE:
                $error_message = 'Failed to write file to disk.';
                break;

            case UPLOAD_ERR_EXTENSION:
                $error_message = 'A PHP extension interrupted the upload.';
                break;

            default:
                $error_message = 'Unknown error';
            break;
        }
        return false;
    }

    $error_message = null;
    return true;
}

#7

is_uploaded_file() is great to use, specially for checking whether it is an uploaded file or a local file (for security purposes).

is_uploaded_file（）很好用，特别用于检查它是上传文件还是本地文件（出于安全目的）。

However, if you want to check whether the user uploaded a file, use $_FILES['file']['error'] == UPLOAD_ERR_OK.

但是，如果要检查用户是否上传了文件，请使用$ _FILES ['file'] ['error'] == UPLOAD_ERR_OK。

See the PHP manual on file upload error messages. If you just want to check for no file, use UPLOAD_ERR_NO_FILE.

有关文件上载错误消息，请参阅PHP手册。如果您只想检查无文件，请使用UPLOAD_ERR_NO_FILE。

#1

110