【发布时间】:2016-05-27 07:11:40
【问题描述】:
为了让我在上传之前轻松浏览正确的文件,这是我想要完成的。该怎么做?
我的输入代码: 表单内部有两个输入。并具有可能相似的文件名。提交按钮将触发“uploadnow”功能。
<td>UPLOAD#1: AGL_001.txt <input type="file" name="upload1" id="upload1"></td>
<td>UPLOAD#2: AGL_0001.txt <input type="file" name="upload2" id="upload2"></td>
function uploadnow(){
$allowed_upload1 = ['AGL_001.txt']; // added
$allowed_upload2 = ['AGL_0001.txt']; //added
if(isset($_FILES['upload1']['name'])){
//$errors= array();
$file_name = $_FILES['upload1']['name'];
$file_size =$_FILES['upload1']['size'];
$file_tmp =$_FILES['upload1']['tmp_name'];
$file_type=$_FILES['upload1']['type'];
$file_ext=strtolower(end(explode('.',$_FILES['upload1']['name'])));
//$img_loc = $file_name.'.'.$file_ext;
if (in_array($file_name, $allowed_upload1)) {
move_uploaded_file($file_tmp,"uploads/".$file_name);
} else {
$message = "Sorry, wrong filename on UPLOAD#1";
echo "<script type='text/javascript'>alert('$message');</script>";
}
}
if(isset($_FILES['upload2']['name'])){
//$errors= array();
$file_name = $_FILES['upload2']['name'];
$file_size =$_FILES['upload2']['size'];
$file_tmp =$_FILES['upload2']['tmp_name'];
$file_type=$_FILES['upload2']['type'];
$file_ext=strtolower(end(explode('.',$_FILES['upload2']['name'])));
//$img_loc = $file_name.'.'.$file_ext;
if (in_array($file_name, $allowed_upload2)) {
move_uploaded_file($file_tmp,"uploads/".$file_name);
} else {
$message = "Sorry, wrong filename on UPLOAD#2";
echo "<script type='text/javascript'>alert('$message');</script>";
}
}
}
【问题讨论】:
-
目前还不清楚您要达到的目标。您是否正在寻找仅显示具有给定文件名的文件的客户端解决方案?或者如果它们与文件名不匹配,您是否试图禁止上传服务器端?
-
@hypeJunction 你说的都是。如果客户端解决方案是不可能的,也许我会寻找服务器端的方法。
标签: php function file file-upload fileopendialog