PHP如何实现网页预览功能

发布网友发布时间：2022-04-23 09:17

共1个回答

热心网友时间：2022-04-06 05:47

html文件
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head><meta http-equiv="Content-Type" content="text/html; charset=utf-8" /><title>无标题文档</title><style type="text/css">#yl{ width:200px; height:300px; background-image:url(images/timg1.jpg); background-size:200px 300px;}#file{ width:200px; height:300px; float:left; opacity:0;}</style></head> <body> <form id="sc" action="ylchuli.php" method="post" enctype="multipart/form-data" target="shangchuan"> <input type="hidden" name="tp" value="" id="tp" /> <div id="yl"> <input type="file" name="file" id="file" onchange="document.getElementById('sc').submit()" /> </div> </form> <iframe style="display:none" name="shangchuan" id="shangchuan"></iframe> </body> <script type="text/javascript"> //回调函数,调用该方法传一个文件路径，改变背景图function showimg(url){ var div = document.getElementById("yl"); div.style.backgroundImage = "url("+url+")"; document.getElementById("tp").value = url;} </script> </html>
php文件<?php if($_FILES["file"]["error"]){ echo $_FILES["file"]["error"];}else{ if(($_FILES["file"]["type"]=="image/jpeg" || $_FILES["file"]["type"]=="image/png")&& $_FILES["file"]["size"]<1024000) { $fname = "./images/".date("YmdHis").$_FILES["file"]["name"]; $filename = iconv("UTF-8","gb2312",$fname); if(file_exists($filename)) { echo "<script>alert('该文件已存在！');</script>"; } else { move_uploaded_file($_FILES["file"]["tmp_name"],$filename); $delurl = iconv("UTF-8","gb2312",$_POST["tp"]); unlink($delurl); //删除文件 echo "<script>parent.showimg('{$fname}');</script>"; } }}