【发布时间】:2020-08-11 07:33:09
【问题描述】:
如何将此表单中的文件上传到发送消息的表单。当前,正在发送消息,但没有文件 我无法将值从 JS 传递到表单中
我的网站:https://fck-auto.de/ankaufahrzeuge/
表格:https://jsfiddle.net/alexjamesbrown/2nzL9f7g/
PHP 代码:
if(count($_FILES['upload']['name']) > 0){
$rand = rand();
$createFolder = uniqid();
mkdir('uploads/'.$createFolder);
for($i=0; $i<count($_FILES['upload']['name']); $i++) {
$tmpFilePath = $_FILES['upload']['tmp_name'][$i];
if($tmpFilePath != ""){
$shortname = $_FILES['upload']['name'][$i];
$explode = explode(".", $_FILES['upload']['name'][$i]);
$filePath = "uploads/".$createFolder. '/' . rand().'.'.$explode[1];
if(move_uploaded_file($tmpFilePath, $filePath)) {
$files[] = $shortname;
}
}
}
}
【问题讨论】:
标签: javascript php forms