6

我有一个小表格，我想将文件上传到我的 CF 服务器。我过去可以通过传统的操作页面提交我的 CFFORM 来完成这项工作。但是，我想改为使用 AJAX 上传文件。

我在处理页面上收到如下错误：cffile action="upload" 要求表单使用 enctype="multipart/form-data" 即使我已经这样定义了我的表单。

从 google'ng 周围，我认为这可能是因为 Cffile 需要 filefield 属性，但是因为没有传递给coldfusion的表单对象。可能的类似问题。我不太喜欢发布的解决方案。

无论如何我可以解决这个错误吗？

这是我的 AJAX：

<!---Script to upload file link --->     
<cfoutput>
<script>
$(function(){
//Add a new note to a link
$("##upload-file").submit(function(event){
   // prevent native form submission here
   event.preventDefault();
        $.ajax({
            type: "POST",
            data: $('##upload-file').serialize(),
            url: "actionpages/file_upload_action.cfm",
            beforeSend: function(){
                $('.loader').show();
            },
            complete: function(){
                 $('.loader').hide(3000);
            },
            success: function() {
                PopulateFileUploadDiv();
                $("##upload").val('');
                $("##response").append( "File successfully Uploaded." );
              }    
        });
        return false;           
    });
});
</script>
</cfoutput>

我的表格：

<form method="post" name="upload-file" id="upload-file" enctype="multipart/form-data">

      <input tabindex="0" size="50" type="file" id="upload" name="upload" accept="image/bmp, image/jpg, image/jpeg, image/tiff, image/gif, image/png, text/richtext, text/plain, text/css, text/html, application/rtf, application/msword, application/x-excel, application/zip, multipart/x-zip, audio/wav" value="Upload an additional file" />

      <br />
      <input name="submitForm" id="submitForm" type="submit" value="Attach File To Ticket">

      </form>

处理页面：

<!---File Upload Logic --->

        <!---CFFile Tag located in template file --->
        <!---CFFile code --->
        <cffile action="upload" filefield="upload" destination="c:\inetpub\wwwroot\ticket-uploads\" accept="image/bmp, image/jpeg, image/jpg, image/tiff, image/gif, image/png, text/richtext, text/plain, text/css, text/html, text/xml, application/x-zip-compressed, application/xml, application/mspowerpoint, application/powerpoint, application/vnd.openxmlformats-officedocument.presentationml.presentation , application/vnd.openxmlformats-officedocument.spreadsheetml.sheet, application/vnd.ms-powerpoint, application/x-mspowerpoint, application/octet-stream, application/pdf, application/rtf, application/msword, application/x-excel, application/zip, multipart/x-zip, audio/wav" nameconflict="makeunique">

       <!---Retrieve Uploaded filename --->
        <cfoutput>
        <cfset Uploaded_File_Name = #cffile.ServerFile#>
        </cfoutput>

        <!--- Add file details to file_uploads table --->
        <cfquery name="uploads" datasource="#datasource#">
        insert into file_uploads (ticket_id, file_path)
        values(#form.ticket_id#, '#Uploaded_File_Name#')
        </cfquery>
4

1 回答 1

5

正如评论中提到的@cfqueryparam，关键是javascript代码。特别是contentType和processData设置。操作页面可以用任何服务器端语言编写。

只是为了演示，这个线程中的示例工作正常。至少在较新的浏览器中。除了将状态转储到 a 之外div，我唯一更改的是输入名称。那是因为 FILE 是 CF 中的关键字。

上传

<!DOCTYPE html>
<html>
<head>
    <title>Image Upload Form</title>
    <script src="//code.jquery.com/jquery-1.9.1.js"></script>
    <script type="text/javascript">
        function submitForm() {
            console.log("submit event");
            var fd = new FormData(document.getElementById("fileinfo"));
            $.ajax({
              url: "action.cfm",
              type: "POST",
              data: fd,
              enctype: 'multipart/form-data',
              processData: false,  // tell jQuery not to process the data
              contentType: false   // tell jQuery not to set contentType
            }).done(function( response ) {
                // display response in DIV
                $("#output").html( response.toString());
            })
           .fail(function(jqXHR, textStatus, errorMessage) {
                // display error in DIV
                $("#output").html(errorMessage);
            })            
            return false;
        }
    </script>
</head>

<body>
    <form method="post" id="fileinfo" name="fileinfo" onsubmit="return submitForm();">
        <label>Select a file:</label><br>
        <input type="file" name="upload" required />
        <input type="submit" value="Upload" />
    </form>
    <div id="output"></div>
</body>
</html>

动作.cfm

<cffile action="upload" filefield="upload" 
    destination="c:/temp/" 
    nameconflict="makeunique"
    result="uploadResult">
<!--- Quick and dirty response dump for DEV/Debugging only --->
<cfoutput>#serializeJSON(uploadResult)#</cfoutput>
于 2014-11-06T21:46:15.380 回答