php ajax upload image example code without refresh_php skills

This article shares the example code of php combined with ajax to upload images without refreshing. I share it with everyone. I hope everyone can learn and make progress together with the editor.

1. Import files

<!--图片上传begin-->
<script type="text/javascript" src="/js/jquery.form.js"></script>
<script type="text/javascript" src="/js/uploadImg.js"></script>
<link href="/css/uploadImg.css" rel="stylesheet" type="text/css" />
<!--图片上传end-->

2.html part

<div class="upimg">
       <input name="icon" type="text" class="imgsrc" value="<!--{$contents.icon}-->" />
       <div class="showimg">
        <!--{if $contents.icon}-->
        <img  src="/static/imghwm/default1.png"  data-src="<!--{$contents.icon}-- alt="php ajax upload image example code without refresh_php skills" >"  class="lazy" height="120px">
        <!--{/if}-->
       </div>          
       <div class="btn" style="height:20px;">
          <span>添加图片</span>
          <input class="fileupload" type="file" name="pic[]">
       </div>
       </div> 

3. Add a form to fileupload

/*图片上传*/
  $(".fileupload").wrap("<form action='/bookstore/book/uploadpic' method='post' enctype='multipart/form-data'></form>"); //函数处理

4.ajax file upload

jQuery(function ($) { 
  $(".fileupload").change(function(){ //选择文件 
    if ('' === $(this).val()) return;
    var upimg = $(this).parent().parent().parent();
    var showimg = upimg.find('.showimg');
    var btn = upimg.find('.btn span');
    var imgsrc = upimg.find('.imgsrc');
    $(this).parent().ajaxSubmit({ 
      //dataType: 'json', //数据格式为json 
      beforeSend: function() { //开始上传 
        showimg.empty(); //清空显示的图片 
        imgsrc.val("");
        btn.html("上传中..."); //上传按钮显示上传中 
      }, 
      uploadProgress: function(event, position, total, percentComplete) { 
      }, 
      success: function(data) { //成功 
        //获得后台返回的json数据，显示文件名，大小，以及删除按钮 
        var img = data;
        //显示上传后的图片 
        imgsrc.val("");
        imgsrc.val(img);
        showimg.html("<img  src="/static/imghwm/default1.png"  data-src="http://files.jb51.net/file_images/article/201511/20151117145516729.png&#63;20151017145526?x-oss-process=image/resize,p_40"  class="lazy"     style="max-width:90%"  style="max-width:90%"+img+"' alt="php ajax upload image example code without refresh_php skills" >"); 
        btn.html("上传成功"); //上传按钮还原 
      }, 
      error:function(xhr){ //上传失败 
        btn.html("上传失败"); 
      } 
    }); 
  }); 
}); 

5. Processing in the background

public function uploadpicAction(){ //图片上传和显示
    $data = "";
    $src = $this->uploadFiles2($imgpath = "/upload/book" ,$filesname = "pic");      
    isset($src[0]['src']) && $src[0]['src'] ? $data = $this->concaturl($src[0]['src']) : null;
    echo $data; 
  }

6. Hand the returned data to the front end for some processing.

Then submit it to the backend database.

I hope this article will be helpful to everyone learning PHP programming.