Ajax表单提交无法获取FormData

时间:2017-04-04 10:54:44

标签: php jquery codeigniter

当我使用Ajax提交时我无法获取我的FormData ..Form提交没有FormData ..我想我错过了一些东西..但是不能得到它..请帮助我。 我的代码............................................... < / p>

<form id="regfrm" name="regfrm" class="form-horizontal" method="POST" action="<?php echo site_url('settings/finance'); ?>">
                    <div class="form-group row">
                        <label for="txt_sqdcp" class="col-sm-2 col-form-label">Select File to Upload</label>
                        <div class="col-sm-10">
                            <div class="input-group">
                                <span class="btn btn-file btn-info" title="">
                                    <span class="fileupload-new">Click here to choose file</span>
                                    <span class="fileupload-exists">Change</span>
                                    <input type="file" name="userfile" id="userfile" />
                                </span>
                                <div style="padding-top: 6px;">
                                    <span id="vop_fileName"></span>
                                </div>
                            </div>
                        </div>
                    </div>
                    <div class="form-group row">
                        <label for="txt_sqdcp" class="col-sm-2 col-form-label">Excel Sheet Name</label>
                        <div class="col-sm-10">
                            <div class="input-group">
                               <input class="form-control" type="text" id="txt_excel" name="txt_excel" value="" required="required"> &nbsp;&nbsp;Eg : VOP Feb 17
                               <!-- <div class="loadingGIF" id="vop_loader"><img src="<?php // echo base_url(); ?>assets/logo/loading.gif" /></div> -->
                               <span id="excelcheck" class="text-yellow"></span>
                           </div>
                       </div>
                   </div>
                   <div class="form-group row pull-right">
                        <input type="hidden" name="txt_id" id="txt_id" value="<?php if(!empty($edit_sqdcp)) echo @$cat; ?>" />
                        <input type="submit" id="sub" name="sub" class="btn <?php if(empty($edit_sqdcp)) echo "btn-success"; else echo "btn-warning";?>" value="<?php if(empty($edit_sqdcp)) echo "submit"; else echo"update";?>"/>
                    </div>
                <?php // echo form_close(); ?>
                </form>

脚本

$("#regfrm").submit(function(e){
            var formObj = $(this);
            var formURL = formObj.attr("action");
            var formData = new FormData(this);
            console.log(formData);
            $.ajax({
            url: formURL,
            type: 'POST',
            data:  formData,
            mimeType:"multipart/form-data",
            contentType: false,
            cache: false,
            processData:false,
            success: function(data, textStatus, jqXHR)
            {

            },
            error: function(jqXHR, textStatus, errorThrown) 
            {
            }          
            });
        });

0 个答案:

没有答案