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php上传增多域名

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Release: 2016-06-13 12:20:16
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php上传增加域名

function ajaxfile() {<br />    foreach (lib_request::$files as $filename => $value) {<br />        $picname = lib_request::$files[$filename]['name'];<br />        $picsize = lib_request::$files[$filename]['size'];<br />        if ($picname != "") {<br />            if ($picsize > 1024000) {<br />                echo json_encode(array(<br />                    'code' => '-1',<br />                    'title' => '图片上传',<br />                    'msg' => '图片大小不能超过1M'<br />                ));<br />                exit();<br />            }<br />            $type = get_extension($picname);<br />            if ($type != "gif" && $type != "jpg" && $type != "jpeg" && $type != "png") {<br />                echo json_encode(array(<br />                    'code' => '-1',<br />                    'title' => '图片上传',<br />                    'msg' => '图片格式不对!'<br />                ));<br />                exit();<br />            }<br />	$upload_path = '/up';<br />			<br />            $rand = rand(100, 999);<br />            $pics = date("YmdHis") . $rand . '.' . $type;<br />            $dir = date('/YmdH/');<br />            $pic_path = $upload_path  . $dir;<br />            $pic_url = $pic_path . $pics;<br />            if (!file_exists(PATH_ROOT . $pic_path)) {<br />                creatFolder(PATH_ROOT . $pic_path);<br />            }<br />            $PATH_ROOT = PATH_ROOT . $pic_url;<br />            move_uploaded_file(lib_request::$files[$filename]['tmp_name'], $pic_path);<br />            if (request('type') == 'goods') {<br />                makeThumbnail($pic_path, $pic_path . '_290x290.jpg', 290, 190);<br />            }<br />        }<br />        $arr[$filename] = array(<br />            'name' => $picname,<br />            'pic' => $pic_url,<br />            'size' => $picsize<br />        );<br />    }<br />    echo json_encode($arr);<br />    exit();<br />}
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以上上传后得到的路径是/up/xxxxx/xxxxx.jpg
我想修改为:http://www.xxxx.com//up/xxxxx/xxxxx.jpg


------解决思路----------------------
估计是想得到一个改变的url ?
<br /><br /> $pic_url = ‘http://www.xxxx.com/’.$pic_path . $pics;<br /><br />
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是这样么?提问说的不清晰 哈哈。
------解决思路----------------------
你说的是什么意思??不是很明白
如果是数据库里面的路径 你update就可以了
如果是上传到其他服务器的,这应该不行把

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