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><?php
session_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》
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