Question

我正在探索PHP的文件上传和文本解析功能，但我的第一步是错误的，我无法弄明白。代码的目标是显示一个表单来上传文本，然后显示相应的$_FILES数组的值。但是，它不起作用 - 它运行时没有错误但不显示print_r($_FILES['upload'])。我错过了什么？

这是我的Index_txtParsing文件：

<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
    <head>
        <title>test</title>
    </head>
    <body>
        <form enctype="multipart/form-data" class="Txt_upload" method='POST'  action='index_txtParsing.php'>
            <label for="file">Enter upload here:</label>
            <input type='file' name='upload'/>
            <input type='submit' name='submit' value="upload here"/>
        </form>
    </body>
</html>

<?php
    if(isset($_POST['upload'])){
    $upload=$_POST['upload'];
    print_r($_FILES['upload']);
    }else{
    $upload="unknown";
    }
?>

编辑：在合并以下建议后，此代码有效：

<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
    <head>
        <title>test</title>
    </head>
    <body>
        <form enctype="multipart/form-data" class="Txt_upload" method='POST'  action='index_txtParsing.php'>
            <label for="file">Enter upload here:</label>
            <input type='file' name='upload'/>
            <input type='submit' name='submit' value="upload here"/>
        </form>
    </body>
</html>

<?php
    if(isset($_FILES['upload'])){
    echo "Value of FILES['upload'] ";
    print_r($_FILES['upload']);
    echo "<br/>";
    }else{
    }
?>

Answer 1

要获取上传的文件，请使用$ _FILES而不是$ _POST：

<?php
    if(isset($_FILES['upload'])){
    $upload=$_FILES['upload'];
    print_r($_FILES['upload']);
    }else{
    $upload="unknown";
    }
?>

另请阅读 manual on File Uploads 。

Answer 2

您的if语句应如下所示：

if ($_FILES['upload']['name']!="")

Answer 3

不要在POST数组中查找文件，在FILES数组中查找它：

if(isset($_FILES['upload']['tmp_name']) && $_FILES['upload']['size']>0){ //make sure the file has been uploaded correctly
    $upload=$_FILES['upload'];
    print_r($_FILES['upload']);
}else{
    $upload="unknown";
}

$_FILES['name of field']是一个可以按如下方式访问的数组（来自PHP.net）：

$ _ FILES [ 'userfile的'] [ '名称']

The original name of the file on the client machine.

$ _ FILES [ 'userfile的'] [ '类型']

The mime type of the file, if the browser provided this information. An example would be "image/gif". This mime type is however not checked on the PHP side and therefore don't take its value for granted.

$ _ FILES [ 'userfile的'] [ '尺寸']

The size, in bytes, of the uploaded file.

$ _ FILES [ 'userfile的'] [ 'tmp_name的值']

The temporary filename of the file in which the uploaded file was stored on the server.

$ _ FILES ['userfile'] ['error']

Answer 4

尝试清除action =“”的值并检查_POST中的submit元素，而不是在_POST中上传，以便获得：

<form enctype="multipart/form-data" method="POST" action="">
        <input type="file" name="upload" />
        <input type="submit" name="submit" value="upload here"/>
</form>

<?php if ( isset( $_POST['submit'] ) ) { print_r( $_FILES ); } ?>

这应该有用。

Answer 5

首先，当你要查看$ _POST ['submit']时，如果要查看表单是否已提交，请检查$ _POST ['upload']，或者是$ $ _FILES [upload'] ['name']如果您要检查文件名和
当你试图显示$ _POST ['upload']时，因为它是一个文件而无法使用，所以如果你想要文件的名称，请尝试这样的事情：
```
if(isset($_POST['submit'])){
    print_r($_FILES['upload']);
    echo $_FILES['upload']['name'];
}else{
    $upload="unknown";
}
```

更新

正如我在第一点告诉你的，如果你需要检查一个文件是否已被选中进行上传，请尝试更改你的if语句，这样你首先要检查表单是否已经提交，然后如果文件已被提交选自：

if(isset($_POST['submit'])){
    if($_FILES['upload']['name']!="")
            $upload = $_FILES['upload']['name'];
    else
        $upload = "no file";
}else{
    $upload="form not submitted";
}

echo $upload;

这当然只是一个例子，而不是一个完整的解决方案。在实际情况中，您可能想要检查文件大小，文件类型等...

Answer 6

你需要

<input type="hidden" name="MAX_FILE_SIZE" value="same_or_lower_than_in_your_php.ini">

在您的表单中

Answer 7

我相信这是你想要的方向：

// Check if you uploaded a file previously
if(isset($_POST['upload']))
{
    // Load the new file
    if (!empty($HTTP_POST_FILES['ufile']['tmp_name']))
    {
            // Extension? 
        $temp_name = basename($_FILES['ufile']['name']);
        $extension = explode(".", $temp_name);
        $extension = end($extension);
        $filename .= $extension;

        $path = $uploaddir . $filename;

        if ($ufile != none)
        {
            if (copy($HTTP_POST_FILES['ufile']['tmp_name'], $path))
            {
           // It copied to where ever you want it to go!    
            } 
            else
            //error 
        }
        else
        //error
    }
    else
    //error

}

PHP简单文件上传表单不起作用

7 个答案:

更新