刷新浏览器后,如何让数组和会话变量回到空白状态?

时间:2012-04-27 17:16:27

标签: php javascript json

我有一个问题,每当我刷新浏览器时,我希望包含$ _SESSION变量的数组回到空白状态。暂时可以说我上传了2个文件,然后刷新了浏览器,当我上传另一个文件时,它显示了以前上传的文件的名称。如果刷新浏览器,如何让数组和会话变量返回空白?

以下是代码:

  function stopImageUpload(success){

          var imageNameArray = new Array();
// WHEN PAGE IS REFRESH, ARRAY SHOULD GO BACK TO BEING BLANK

 imageNameArray = <?php echo json_encode(isset($_FILES ['fileImage']['name']) ? $_FILES ['fileImage']['name'] : null); ?>;
//RETRIEVES THE SESSION VARIABLE FROM THE PHP SCRIPT OF THE FILE NAMES WHICH HAVE BEEN UPLOADED

          var result = '';



          if (success == 1){
             result = '<span class="msg">The file was uploaded successfully!</span><br/><br/>';

                for(var i=0;i<imageNameArray.length;i++)  //LOOP THROUGH ALL UPLOADED FILE NAMES
        {
             $('.listImage').append(imageNameArray[i]+ '<br/>');//APPEND FILE NAME

         }

          }
          else {
             result = '<span class="emsg">There was an error during file upload!</span><br/><br/>';
          }

    return true;

    }

下面是php脚本,它从上面的javascript函数上传另一个页面上的文件:

<?php

    session_start();

    $result = 0;
    $errors = array ();
    $dirImage = "ImageFiles/";


if (isset ( $_FILES ['fileImage'] ) && $_FILES ["fileImage"] ["error"] == UPLOAD_ERR_OK) {

$fileName = $_FILES ['fileImage'] ['name'];

$fileExt = pathinfo ( $fileName, PATHINFO_EXTENSION );
$fileExt = strtolower ( $fileExt );


$fileDst = $dirImage . DIRECTORY_SEPARATOR . $fileName;

        if (count ( $errors ) == 0) {
            if (move_uploaded_file ( $fileTemp, $fileDst )) {
                $result = 1;


            }
        }

    }

    ?>

<script language="javascript" type="text/javascript">window.top.stopImageUpload(<?php echo $result;?>);</script>

3 个答案:

答案 0 :(得分:3)

提交表单并且您的PHP已完成其工作后,您需要执行303重定向到您希望显示的页面。然后,当用户刷新页面时,您将不会再次提交表单。

例如:

header('Location: ' . $_SERVER['PHP_SELF'], true, 303);

答案 1 :(得分:0)

虽然我同意@Reza Sanaie的评论,但您可以通过在成功上传后取消设置变量来解决您的特定问题:

     if (success == 1)
     {
         result = '<span class="msg">The file was uploaded successfully!</span><br/><br/>';

         for(var i=0;i<imageNameArray.length;i++)  //LOOP THROUGH ALL UPLOADED FILE NAMES
         {
             $('.listImage').append(imageNameArray[i]+ '<br/>');//APPEND FILE NAME

         }

         $_SESSION ['fileImage'] = NULL; // or use unset()
     }

答案 2 :(得分:0)

只需在PHP中进行错误检查,并将数组传递到HTML页面,以显示页面中上传文件的列表。你不需要javascript。

查看How to upload files in PHP