PHP 如何实现图片上传预览？
首先监听input标签的onchange事件；然后使用AJAX将文件上传到服务端；接着在PHP中接收上传的文件，并将文件保存起来；最后将文件访问路径返回，并使用JS渲染即可。
示例代码
<html><head><meta http-equiv="Content-Type" content="text/html; charset=utf-8" /><title>上传头像</title><style type="text/css"> *{ font-family:"微软雅黑";} #zong{ /*border:1px solid black;*/ position:relative; width:52%; height:500x; left:24%} .nr{ float:left; margin-right:30px;} #yl{width:240px; height:240px; background-size:240px 240px;} #file{width:240px; height:240px; float:left; opacity:0;}</style></head><body><div id="zong"><form id="sc" action="2.php" method="post" enctype="multipart/form-data" target="shangchuan"> <input type="hidden" name="tp" value="" id="tp" /> <div id="yl" style="background-image:url(./image/1.jpg)" class="nr">//头像显示的位置 <input type="file" name="file" id="file" onchange="document.getElementById('sc').submit()" /> </div> <div class="nr"> </div> </form><iframe style="display:none" name="shangchuan" id="shangchuan"></iframe></div></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><?phpsession_start();$uid = $_SESSION["uid"];if($_FILES["file"]["error"]){ echo $_FILES["file"]["error"];}else{ if(($_FILES["file"]["type"]=="image/jpeg" || $_FILES["file"]["type"]=="image/png")&& $_FILES["file"]["size"]<1024000) { $fname = "./a/image/".date("YmdHis").$_FILES["file"]["name"]; //头像存储的路径 $filename = iconv("UTF-8","gb2312",$fname); if(file_exists($filename)) { echo "<script>alert('该文件已存在！');</script>"; } else { move_uploaded_file($_FILES["file"]["tmp_name"],$filename); unlink($_POST["tp"]); echo "<script>parent.showimg('{$fname}');</script>"; } }}推荐奖教程：《PHP》
