Home >Backend Development >PHP Problem >How to implement image upload preview in PHP?

How to implement image upload preview in PHP?

Guanhui
GuanhuiOriginal
2020-07-21 15:12:553884browse

How to implement image upload preview in PHP?

PHP How to implement image upload preview?

First listen to the onchange event of the input tag; then use AJAX to upload the file to the server; then receive the uploaded file in PHP and save the file; finally return the file access path, and Just use JS to render.

Sample code

<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(&#39;sc&#39;).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($filename))
    {
      echo "<script>alert(&#39;该文件已存在!&#39;);</script>";
    }
    else
    {
      move_uploaded_file($_FILES["file"]["tmp_name"],$filename);
      
      unlink($_POST["tp"]);
      echo "<script>parent.showimg(&#39;{$fname}&#39;);</script>";
    }
    
  }
}

Recommended Award Tutorial: "PHP"

The above is the detailed content of How to implement image upload preview in PHP?. For more information, please follow other related articles on the PHP Chinese website!

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn