public function uploadImg(){
import('ORG.Util.UploadFile');
$upload = new UploadFile();
$upload->maxSize = '3145728';
$upload->savePath = './Home/Public/Uploads/';
$upload->saveRule = time;
$upload->uploadReplace = true;
$upload->allowExts = array('jpg','jpeg','png','gif');//准许上传文件后缀
$upload->allowType = array('image/png','image/jpg','image/pjpeg','image/gif','image/jpeg');//检测mime类型
$upload->thumb = true; //是否开启图片文件缩略图
$upload->thumbMaxWidth = '67,100';
$upload->thumbMaxHeight = '68,100';
$upload->thumbPrefix = 's_,m_';
$upload->autoSub = true;
$upload->subType = 'date';
$upload->dateFormat = 'y-m-d';
$upload->thumbRemoveOrigin = 1;
if($upload->upload()){
//第一种方式
$info = $upload->getUploadFileInfo();
return $info;
}else {
$this->error($upload->getErrorMsg());
}
}
//上传图片end上传图片的信息, 保存到数据库的名字是["savename"]=> string(23) "13-09-05/1378351148.jpg"
实际保存的图片的名字是
如果不用子目录上传的话.在前台读图片直接可以只用
<div id="img"><img src="__PUBLIC__/Uploads/m_{savename}" /></div>
就直接解析成了
<div id="img"><img src="__PUBLIC__/Uploads/m_1378348035.jpg" /></div>但是现在有子目录.就解析成这样了<img src="__PUBLIC__/Uploads/m_13-09-05/1378348035.jpg">我想要的结果是<img src="__PUBLIC__/Uploads/13-09-05/m_1378348035.jpg">我要怎么做,才能实现? 最佳答案
