Question

我想上传文件并发送到服务器而不刷新页面。

我的HTML文件中有以下行

<form  id="FileUploader" enctype="multipart/form-data" >
    <input type="file" name="mfile" id="mfile" style='width:100%;' onchange="uploaded()">
</form>

function uploaded()
    {
        alert($('form#FileUploader')[0]);
        var formData=new FormData($('form#FileUploader')[0]);
        //alert(formData);
          $.ajax({
            url: "<?php echo $_SESSION['webpage']."/upload" ?>",
            type: "POST",
            async: true,
            dataType: "JSONP",
            data : formData
            })
            .success (function(response){
                alert(response);
            })
            .error   (function()     { alert("Error")   ; }) ;


    }

upload.php文件

 if ($_FILES["mfile"]["error"] >0 )
    {
        echo "Error: " ;
    }
    else
    {
        if (file_exists("upload_email_files/" . $_FILES["mfile"]["name"]))
          {
          echo $_POST["file"]. " already exists. ";
          }
        else
          {
          $otp= move_uploaded_file('$_FILES["mfile"]"name"]','/../upload_templates/');
          }
    }

它不起作用。任何人都可以帮我这个吗？它不会进入upload.php并给我错误的非法调用。

谢谢，希里什

Answer 1

如果是“没有刷新页面”，则可以使用隐藏的iframe。

供参考，

http://viralpatel.net/blogs/ajax-style-file-uploading-using-hidden-iframe/，

http://joekuan.wordpress.com/2009/06/12/ajax-a-simplified-version-of-file-upload-form-using-iframe/

Answer 2

检查一下我没有测试过它希望这会有效http://blog.new-bamboo.co.uk/2012/01/10/ridiculously-simple-ajax-uploads-with-formdata

<form  id="FileUploader" enctype="multipart/form-data" >
    <input type="file" name="mfile" id="mfile" style='width:100%;' onchange="uploaded()">
</form>

function uploaded()
    {
        alert($('form#FileUploader')[0]);
        var formData=new FormData($('form#FileUploader'));//remove [0]
        //alert(formData);
          $.ajax({
            url: "<?php echo $_SESSION['webpage']."/upload" ?>",
            type: "POST",
            //async: true,//Remove this line
            //dataType: "JSONP",//Remove this line
            data : formData
            })
            .success (function(response){
                alert(response);
            })
            .error   (function()     { alert("Error")   ; }) ;


    }

Answer 3

试试这个：）

<form  id="FileUploader" enctype="multipart/form-data" >
   <input type="file" name="mfile" id="mfile" style='width:100%;' onchange="uploaded()">
</form>
$(document).ready(function() {
$("#form-geninfo").submit(function(e)
{
    e.preventDefault();
    alert($('form#FileUploader')[0]);
    var formData=new FormData($('form#FileUploader')[0]);
    //alert(formData);
      $.ajax({
        url: "<?php echo $_SESSION['webpage']."/upload" ?>",
        type: "POST",
        async: true,
        dataType: "JSONP",
        data : formData
        })
        .success (function(response){
            alert(response);
        })
        .error   (function()     { alert("Error")   ; }) ;


}
});

使用ajax，jquery，php上传文件

3 个答案: