php实现头像上传预览功能

5年以前 | 阅读数：463 次 | 编程语言：PHP

本文实例为大家分享了php实现头像上传预览功能的具体代码，供大家参考，具体内容如下

主页面1.php


    <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>

处理页面2.php


    <?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>";
        }

      }
    }

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