Question

我在使用$ _FILES和$ _POST时遇到问题，因为我有一个表单来上传图像和一些数据总线，当我使用其中一个时，它可以工作，但是当我使用另一个id没有＆＃39工作。我的代码是：

    <?php
include 'debugging.php';

//phpinfo();
echo '<br />';

echo '<h1>Image Upload</h1>';

//create a form with a file upload control and a submit button
echo <<<_END
<br />
<form method='post' action='uplaodex.php' enctype='multipart/form-data'>
Select File: <input type='file' name='picName' size='50' />

name: <input type='text' name='usName' size='50' />
username : <input type='text' name='usUsername' size='50' />
pass: <input type='password' name='usPass' size='50' />
email: <input type='text' name='usEmail' size='50' />
<br />
<input type='submit' value='Upload' />
<input type="hidden" name="submit" value="1" />
</form>
<br />
_END;
//above is a special use of the echo function - everything between <<<_END 
//and _END will be treated as one large echo statement

//$_FILES is a PHP global array similar to $_POST and $_GET
if (isset($_FILES['picName'])and isset($_POST['submit'])) {
//we access the $_FILES array using the name of the upload control in the form created above    
//

    //create a file path to save the uploaded file into     
    $name = "images//" . $_FILES['picName']['name']; //unix path uses forward slash
    //'filename' index comes from input type 'file' in the form above
    //
    //move uploaded file from temp to web directory
    if (move_uploaded_file($_FILES['picName']['tmp_name'], $name)) {
       // Create the file DO and populate it.
        include 'Do_profile.php';
        $file = new Do_profile();

        //we are going to store the file type and the size so we get that info
        $type = $_FILES['picName']['type'];
        $size = $_FILES['picName']['size'];
        $usName = trim($_POST['usName']);
        $usUsername = trim($_POST['usUsername']);
        $usPass = trim($_POST['usPass']);
        $usEmail = trim($_POST['usEmail']);

        $file->FileName = $name; //$name is initialised previously using $_FILES and file path
        $file->FileSize = $size;
        $file->Type = $type;
        $file->usName = $usName;
        $file->usUsername = $usUsername;
        $file->usPass = $usPass;
        $file->usEmail = $usEmail;

        if ($file->save()) {
//select the ID of the image just stored so we can create a link                

//display success message                      
                echo "<h1> Thankyou </h1><p>Image stored successfully</p>";
//this above line of code displays the image now stored in the images sub directory                      
                echo "<p>Uploaded image '$name'</p><br /><img src='$name' height='200' width='200'/>";
//create alink to the page we will use to display the stored image                      
                echo '<br><a href="Display.php?id=' . $fileId . '">Display image ' .
                $file->FileName . '</a>';
            } else
                echo '<p class="error">Error retrieving file information</p>';
        }
        else {
            echo '<p class="error"> Oh dear. There was a databse error</p>';
        }
    } else {
        //error handling in case move_uploaded_file() the file fails
        $error_array = error_get_last();
        echo "<p class='error'>Could not move the file</p>";

      //  foreach($error_array as $err)
        //    echo $err;
    }


echo "</body></html>";


?>

我不知道问题是什么，有什么帮助??

Answer 1

if (isset($_FILES['picName'])and isset($_POST['submit']))内的所有内容都无法正常工作，因为超级全局$_FILES可能没有名为picName的密钥。要查看此问题，请尝试var_dump $_FILES，例如var_dump($_FILES);

根据var_dump的输出，您可以了解$_FILES内是否有任何内容。如果已填充，只需查看密钥名称，然后使用该密钥访问该文件。

但是如果数组为空，则PHP或APACHE中可能存在一些错误配置。

一个可能的解决方法是在php的ini文件中设置file_uploads = On。

希望它有所帮助！

Answer 2

如果要进行isset，则必须验证文件的大小。我不知道这是否有效，但更好的方法是首先检查验证的大小是否为isset或是否发送到服务器。

然后，在您的<form method='post' action='uplaodex.php' enctype='multipart/form-data'>中，您必须创建另一个名为uplaodex.php的PHP文件，您将在其中发送数据。所以，你的代码如下面的代码并考虑了第1步。这将是你的uploadex.php

$name_file       = $_FILES['picName']['name'];
$type            = $name_file['type'];
$size            = $name_file['size'];
$tmp_folder      = $name_file['tmp'];
$usName          = trim($_POST['usName']);
$usUsername      = trim($_POST['usUsername']);
$usPass          = trim($_POST['usPass']);
$usEmail         = trim($_POST['usEmail']);

if ( $size > 0 ) {
    //REMOVE another slash images//
    $name = "images/" . $name_file; //unix path uses forward slash
    //'filename' index comes from input type 'file' in the form above
    //
    //move uploaded file from temp to web directory
    if ( move_uploaded_file($tmp_folder, $name) ) {
       // Create the file DO and populate it.
        include 'Do_profile.php';
        $file             = new Do_profile();
        $file->FileName   = $name; //$name is initialised previously using $_FILES and file path
        $file->FileSize   = $size;
        $file->Type       = $type;
        $file->usName     = $usName;
        $file->usUsername = $usUsername;
        $file->usPass     = $usPass;
        $file->usEmail    = $usEmail;
            if ($file->save()) {   
                    //USE PRINTF
                    printf('<h1> Thankyou </h1><p>Image stored successfully</p><br> 
                            <p>Uploaded file: %s</p>. <img src="%s" height="200" width="200" />',
                            $name_file, $name );

                    #WHAT IS $fileId? In which moment was define?
                    //echo '<br><a href="Display.php?id=' . $fileId . '">Display image ' .
                    $file->FileName . '</a>';
            } 

            else
                echo '<p class="error">Error retrieving file information</p>';
            }

    else {
        echo '<p class="error"> Oh dear. There was a databse error</p>';
    } //ENDIF OF if (move_uploaded_file($_FILES['picName']['tmp_name'], $name))

} //ENDIF OF  if ( $size > 0 )

#ELSE OF YOUR if ( $size > 0 )
else {
    //error handling in case move_uploaded_file() the file fails
    $error_array = error_get_last();
    echo "<p class='error'>Could not move the file</p>";

  //  foreach($error_array as $err)
    //    echo $err;
}

Answer 3

我解决了这个问题，你不能同时执行$ _FILES和$ _post，或者在另一个内执行其中一个。从$ _Post开始，然后是$ _FILES，在$ _FILES之外运行你的保存功能

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PHP，同时使用$ _FILES和$ _POST

3 个答案: