如何使用jquery压缩图像

时间:2018-04-01 07:31:18

标签: php jquery ajax image-uploading form-data

我无法访问php.ini文件,因此我需要在上传的图像到达服务器端之前对其进行压缩。我还需要同时处理许多文本输入(注释,组名,事件名)。

我的HTML如下:

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Uploads</title>
<link type="text/css" href="style.css">
</head>
<body>
<div class="container">
     <form onClick="return submit_form_up()" enctype="multipart/form-data" id="up-form">

           <input type="file" name="img" id="img" accept="image/*"><br>
           <textarea name="text" cols="40" rows="4" placeholder="Enter text here" id="comment"></textarea><br>
           <select name = "group_name" id="gr_name">
                <optgroup label="Group_name">

                  <?php

                  include ('conn.php');

                  $sql = "SELECT * FROM groups ORDER BY id ASC";
                  $result = mysqli_query($conn , $sql);
                  while($row = mysqli_fetch_array($result , MYSQLI_ASSOC))
                  {
                      echo '<option value="' . $row['group_name'] . '">' . $row['group_name'] . '</option>';
                  }
                  ?>
                </optgroup>
            </select>

            <select name="event_name" id="ev_name">
                <optgroup label="Event_name">

                  <?php

                  include ('conn.php');

                  $sql = "SELECT * FROM event_name ORDER BY id ASC";
                  $result = mysqli_query($conn , $sql);
                  while($row = mysqli_fetch_array($result , MYSQLI_ASSOC))
                  {
                      echo '<option value="' . $row['events'] . '">' . $row['events'] . '</option>';
                  }
                  ?>
                </optgroup>
            </select>
            <br>      
           <button type="submit" name="sub-button" id="sub-button">Submit</button>

     </form>
 </div>    
<script type="text/javascript" src="updates_js.js"></script>     
</body>
</html>

当前的jquery文件看起来像这样,请你帮我解释一下如何不仅要压缩图像一定程度,还要正确传输必要的数据

$(document).ready(function (){
    "use strict";
    var filedata = $('#img').prop('files')[0];
    var form_data = new FormData();                  
    form_data.append('file', file_data);
    alert(form_data);                             
    $.ajax({
        url: 'upload.php', // point to server-side PHP script 
        dataType: 'text',  // what to expect back from the PHP script, if anything
        cache: false,
        contentType: false,
        processData: false,
        data: form_data,                         
        type: 'post',
        success: function(php_script_response){
            alert(php_script_response); // display response from the PHP script, if any
        }
     });
});

1 个答案:

答案 0 :(得分:1)