从 UploadiFive 向 Codeigniter 控制器发送数据

Sending data to Codeigniter controller from UploadiFive

我正在尝试使用 UploadiFive 上传一些文件,并在上传文件时将有关它们的信息添加到数据库中。用户在表单中输入一些详细信息,然后单击上传,此时文件被上传,表单中的信息以相应的文件名添加到数据库中。

我可以上传文件,但每次文件上传完成时我都需要 post 表格。它正在 posting 表单,但我正在努力从上传的文件中获取文件名。代码如下:

HTML 页面:

<?php echo form_open_multipart('upload/do_upload', 'id="upload_form" name="upload_form"');?>
    <div id="queue"></div>
    <input id="file_upload" name="file_upload" type="file" multiple="true">
    <div id="target"></div>
</form>

<script type="text/javascript">
    <?php $timestamp = time();?>
    $(function() {
        $('#file_upload').uploadifive({
            'auto'             : true,
            'checkScript'      : '<? echo base_url();?>uploadify/check-exists.php',
            'formData'         : {
                                   'timestamp' : '<?php echo $timestamp;?>',
                                   'token'     : '<?php echo md5('unique_salt' . $timestamp);?>'
                                 },
            'queueID'          : 'queue',
            'onError'          : function(errorType) {
                                    alert('The error was: ' + errorType);
                                },

            'uploadScript'     : '<? echo base_url();?>uploadify/uploadifive.php',
            'onUploadComplete' : function (event, queueID, fileObj, response, data, file) {

                //Post response back to controller
                $.post('<?php echo site_url('upload/do_upload');?>', {
                    field1: $("#field1").val(),
                    field2: $("#field2").val(),
                    field3: $("#field3").val(),
                    field4: $("#field4").val(),
                    checkbox1: $("#checkbox1:checked").val(),
                    field5: $("#field5").val(),
                    filearray : response},
                    function(info){
                $("#target").append(info);  //Add response returned by controller
                });
            }
        });
    });
</script>

然后我的控制器:

//Decode JSON returned
$file = $this->input->post('filearray');
$json_decoded = json_decode($file);

// Get the image filename & full filename with path
$image_file = $json_decoded->{'file_name'};
$path = "assets/photos/highres/".$image_file;

echo "IMAGE FILE NAME: " . $image_file; die;

出于调试目的,我只是回显了 $image_file

它似乎正在提交除了来自 uploadifive.php 脚本的响应之外的所有内容。当我使用 Firebug 时,我可以看到我确实得到了一个响应,它看起来是正确的,但是响应 (filearray) 没有被 post 编辑到要解码的表单。

关于为什么我无法从响应中得到 filename 的任何想法?

TL;DR

onUploadComplete:function(event,data){}

中使用event.name

如果你真的需要服务器的响应,那么你可能想使用 data(不幸的是,我无法测试它,但文档不会说谎,对吗?)。

详情

onUploadComplete 的文档告诉我们以下内容:

onUploadComplete

Input Type
function

Overridable
N/A

Triggered once for each file upload that completes. Arguments

  • file
    The file object that was uploaded
  • data
    The data returned from the server-side upload script (echoed in uploadifive.php)

Demo

$(function() {
    $('#file_upload').uploadifive({
        'uploadScript'     : '/uploadifive.php'
        'onUploadComplete' : function(file, data) {
            alert('The file ' + file.name + ' uploaded successfully.');
        }
    });
});

这与您的代码中的完全不同:

'onUploadComplete' : function (event, queueID, fileObj, response, data, file) {...}

我无法测试 UploadiFive,但用 Uploadify 进行了快速检查:

'onUploadComplete' : function(event) {
    console.log(JSON.stringify(event,null,4));
}

返回此输出:

{
    "size": 34405,
    "post": {},
    "modificationdate": "2015-09-21T02:24:51.597Z",
    "name": "Tire-wheel-advisor1.jpg",
    "creationdate": "2015-09-21T02:24:51.539Z",
    "id": "SWFUpload_0_0",
    "type": ".jpg",
    "filestatus": -4,
    "index": 0
}