Php 获取使用“输入类型”选择的图像的文件位置;文件";
upload.phpPhp 获取使用“输入类型”选择的图像的文件位置;文件";,php,mysql,html,Php,Mysql,Html,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
<?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!";
?>
我认为你需要重新审视你对文件和文件上传的概念。请它的基本功能
$\u FILES[“nameoffelecontrol”][“tmp\u name”]
获取文件路径。我应该在“nameoffelecontrol”和“tmp\u name”中添加什么?这里您必须输入pic,文件上传器html的名称。$\u FILES[“pic”[“tmp\u name”]将给出上传文件的路径。文件将上载到tmp路径,直到您将其移动到其他位置。
<input type="file" name="pic">