在php中获取文件内容

Question

我想获取通过页面浏览的文件的内容。所以我写了一个php脚本来浏览它。下面

<form action="upload.php" method="post" enctype="multipart/form-data"> 
<input type="file" name="my-file" size="50" maxlength="25"> <br> 
<input type="submit" name="upload" value="Upload"> 

代码，并给出它的PHP代码是

if (is_uploaded_file($_FILES['my-file']['tmp_name']) && $_FILES['my-file']['error']==0) { 
echo "The file name was: " . $_FILES['my-file']['name'] . "<br>"; 
echo "The file type is: " . $_FILES['my-file']['type'] . "<br>"; 
$file=$_FILES['my-file']; 
$out = file_get_contents($file); 

然后我需要打印文件的内容， 我也想打印“这个字电脑的任何一句话”

但它不会打印我浏览的文件的内容，但会打印文件名和文件类型的详细信息。

是否有任何方式使用file_get_contents（）函数

得到.txt文件一个文件的结果来打印文件的值，它是可下载的格式

Answer 1

试试这个

if (is_uploaded_file($_FILES['my-file']['tmp_name']) && $_FILES['my-file']['error']==0) { 
    echo "The file name was: " . $_FILES['my-file']['name'] . "<br>"; 
    echo "The file type is: " . $_FILES['my-file']['type'] . "<br>"; 
    $file=$_FILES['my-file']; 
    $out = file_get_contents($file['tmp_name']); 
    print_r($out); 
    } 

Answer 2

if (is_uploaded_file($_FILES['my-file']['tmp_name']) && $_FILES['my-file']['error']==0) { 
echo "The file name was: " . $_FILES['my-file']['name'] . "<br>"; 
echo "The file type is: " . $_FILES['my-file']['type'] . "<br>"; 
$file=$_FILES['my-file'][''];//use ['tmp_name'] 
$out = file_get_contents($file);