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php实现文件预览功能

时间:2021-07-01 10:21:17 帮助过:31人阅读

上一篇博客是上传功能,本篇是上传后图片预览和更改:

代码如下:

1.yulan.php

<!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>

2.ylchuli.php

<?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>";
    }
    
  }
}

以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持脚本之家。

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