我有一段使用Ajax和PHP上传图像的javascript。
我正在几个 html 页面的 head 部分中使用 javascript,并且想知道是否有一种方法可以将值从这些页面传递到 Upload.php 文件,而无需为每个网页创建单独的 Upload.php 文件。
javascript的部分:
<script type="text/javascript" >
$(function(){
var btnUpload=$('#me');
var mestatus=$('#mestatus');
var files=$('#files');
new AjaxUpload(btnUpload, {
action: '../UploadDoc/Upload.php',
name: 'uploadfile',
我想将值传递给的上传.php文件部分:
<?php
/* Set the location to upload the file to */
$uploaddir = '../documents/userdocs/';
/* Set the name and file extension of the file being saved */
$file = $uploaddir ."HtmlPage1_".basename($_FILES['uploadfile']['name']);
$file_name= "HtmlPage1_".$_FILES['uploadfile']['name'];
其中"HtmlPage1_"将是我需要传入的信息,因为它是唯一更改的部分。
任何帮助非常感谢。
非常感谢
在你的 Javascript 中做
$(function(){
var btnUpload=$('#me');
var mestatus=$('#mestatus');
var files=$('#files');
new AjaxUpload(btnUpload, {
action: '../UploadDoc/Upload.php',
name: 'uploadfile',
params: {
pageKey: 'HtmlPage1_'
}
......ETC-CODE......................
或者你的JavaScript代码
$(function(){
var btnUpload=$('#me');
var mestatus=$('#mestatus');
var files=$('#files');
new AjaxUpload(btnUpload, {
action: '../UploadDoc/Upload.php?pageKey=HtmlPage1_',
name: 'uploadfile',
......ETC-CODE......................
在你的PHP中做:
<?php
/* Set the location to upload the file to */
$uploaddir = '../documents/userdocs/';
/* Set the name and file extension of the file being saved */
$file = $uploaddir .$_REQUEST['pageKey'].basename($_FILES['uploadfile']['name']);
$file_name= $_REQUEST['pageKey'].$_FILES['uploadfile']['name'];
......ETC-CODE......................
然后,您将每个页面中的javascript代码修改为"pageKey"参数。