class.upload.php获取错误
嗨,我用我的项目,我必须说这是非常有用的。 然而,我被这个错误所困扰 没有正确上载的源文件。不能 进行一个过程 谁能指导我如何解决这个问题。谢谢 这是我认为错误所在的代码。谢谢class.upload.php获取错误,php,class,upload,Php,Class,Upload,嗨,我用我的项目,我必须说这是非常有用的。 然而,我被这个错误所困扰 没有正确上载的源文件。不能 进行一个过程 谁能指导我如何解决这个问题。谢谢 这是我认为错误所在的代码。谢谢 <?php require_once('class.upload.php'); if((isset($_POST['step']))&&($_POST['step']=='process')){ $pictime = strtotime(date('Y-m-d H:i:s')); $p
<?php
require_once('class.upload.php');
if((isset($_POST['step']))&&($_POST['step']=='process')){
$pictime = strtotime(date('Y-m-d H:i:s'));
$pic = "picture";
$id = $_POST['id'];
$category = $_POST['category'];
$username = $_POST['username'];
$path = $_SERVER['DOCUMENT_ROOT'].$_POST['tempfile'];
//connect to the server
$conn = mysql_connect('localhost','root','');
if(!$conn)
{
echo "Could not connect to the server";
}
//connect to the database
$db = mysql_select_db("enzeon_db" , $conn);
if(!$db)
{
echo "Could not connect to the database";
}
//query the database to get the imagepath and the thumbpath
$query = "SELECT * FROM machine_db WHERE id = '$id' and m_category = '$category' and username = '$username'";
$result = mysql_query($query) or die("Some error occured" . mysql_error());
$num = mysql_num_rows($result);
if($num == 1)
{
$row = mysql_fetch_array($result);
$imagepath = $row['m_imagepath'];
$imagethumb = $row['m_thumb_path'];
}
$handle = new Upload($_SERVER['DOCUMENT_ROOT'].$_POST['tempfile']);
if ($handle->uploaded) {
$handle->file_src_name_body = $pic; // hard name
$handle->file_new_name_body = 'enzeon_'.$pictime;
$handle->file_overwrite = false;
$handle->file_auto_rename = false;
$handle->image_resize = true;
$handle->file_src_pathname = true;
$handle->image_x = 200; //size of final picture
$handle->image_y = 200; //size of final picture
$handle->jcrop = true;
$handle->rect_w = $_POST['w'];
$handle->rect_h = $_POST['h'];
$handle->posX = $_POST['x'];
$handle->posY = $_POST['y'];
$handle->jpeg_quality = 100;
$handle->Process($_SERVER['DOCUMENT_ROOT'].'/LoginSystem/upload_pic/');
//thumb-50
$handle->file_src_name_body = $pic; // hard name
$handle->file_new_name_body = 'idrish_'.$pictime;
$handle->file_overwrite = false;
$handle->file_auto_rename = false;
$handle->image_resize = true;
$handle->image_x = 100;
$handle->image_y = 100; //size of picture
$handle->jcrop = true;
$handle->rect_w = $_POST['w'];
$handle->rect_h = $_POST['h'];
$handle->posX = $_POST['x'];
$handle->posY = $_POST['y'];
$handle->jpeg_quality = 100;
$handle->Process($_SERVER['DOCUMENT_ROOT'].'/LoginSystem/upload_pic/');
if($handle->processed)
{
echo "gimme some sunshine";
}
else
{
echo 'error' . $handle->error;
}
$handle->clean();
}
else {
echo "here" . $handle->error;
}
}
echo "yes";
//header("location:".$_SERVER["PHP_SELF"]);
?>
应该是
$handle = new Upload($_SERVER['DOCUMENT_ROOT'].$_FILES['tempfile']);
只要这样做
print\r($\u POST);打印(美元文件)
您将看到为什么只是$handle=newupload($_FILES['tempfile'])代码>@antpaw谢谢你的回复。但是我正在从另一个文件调用tempfile路径,因此我认为$u POST是正确的。另外,print\r($\文件)提供一个空数组。我还尝试将echo$handle->log;在我的代码中,我得到了以下输出,错误为“没有正确的上传源文件。无法继续处理”您可以提出其他建议吗..这让我抓狂…可能您的表单标记缺少enctype=“multipart/form data”
$handle = new Upload($_SERVER['DOCUMENT_ROOT'].$_FILES['tempfile']);