时间:2021-07-01 10:21:17 帮助过:22人阅读
File Upload
function processFile($files, $type) {
$uploadName = null;
foreach ($files as $name => $value) {
$originalName = $value['name'];
$arr = explode(".", $originalName);
$postfix = $arr[count($arr) - 1];
$tmpPath = $value['tmp_name'];
$tmpType = $value['type'];
$tmpSize = $value['size'];
}
$newname = EhlStaticFunction::generateRandomStr(40).".".$postfix;
switch ($type) {
case 1 :
// 处理声音文件
$destination = VIDEOUPLOADDIR.$newname;
break;
case 2 :
// 处理图像文件
$destination = IMAGEUPLOADDIR.$newname;
break;
}
move_uploaded_file($tmpPath, $destination);
}http://www.bkjia.com/PHPjc/664289.htmlwww.bkjia.comtruehttp://www.bkjia.com/PHPjc/664289.htmlTechArticle前言 这星期一直再搞php,涉及到文件上传的部分有些遗忘,这里记录一下 HTML的form表单 用html的表单模拟一个文件上传的post请求,代码如下...