获取使用输入类型“file”选择的图像的文件位置

upload.php

<?php
$host = 'localhost'; 
$user = 'root'; 
$pw = ''; 
$db = 'thepillar'; 
$pic = $_GET['pic'];
$ext = $_GET['ext'];

mysql_connect($host,$user,$pw); 

mysql_select_db($db); 

$handle = fopen($pic, "rb");
$img = fread($handle, filesize($pic));
fclose($handle);


$pic = base64_encode($pic); 

$sql = "insert into infopics values(null,'$pic','$ext');"; 

mysql_query($sql) or die('Bad Query at '.mysql_error()); 

echo "Success! You have inserted your picture!";
?>

In the code above I am fetching $pic from my uploader.php page where I use

<input type="file" name="pic">

The problem here is that fopen() cannot execute its function because by using $_GET['pic'] it only retrieves the filename of the image that was chosen from the other page.

is there a way that I can retrieve the whole file path so that fopen can function?

use

$_FILES["nameOfFileControl"]["tmp_name"] to get path of file

I think you need to review your concept of file and file upload. Please Read From Here. Its basics