如何在提交另一份表格的ajax回复后提交表格？

Question

我有一个文件输入表单：

<form method="post" id='careerform' action="<?php echo Yii::app()->createUrl('site/career'); ?>" enctype="multipart/form-data">
   <input type="file" name="cv" id="cv">
   <input type="button" id="submitcareer" class="btn btn-default pull-right" data-toggle="modal" data-target=".bs-example-modal-sm" value="Submit">
</form>

当用户点击提交时，将检查表单以进行验证：

<script type="text/javascript">

    $(document).ready(function() {

$('#careerform').bootstrapValidator({

        feedbackIcons: {

            valid: 'glyphicon glyphicon-ok',

            invalid: 'glyphicon glyphicon-remove',

            validating: 'glyphicon glyphicon-refresh'

        }, 

        fields: {

    cv: {

        validators: {

            notEmpty: {

                message: 'CV is required.'

            },

            file: {

                extension: 'doc,docx,pdf,zip,rtf',

                type: 'application/pdf,application/msword,application/vnd.openxmlformats-officedocument.wordprocessingml.document,application/rtf,application/zip',

                maxSize: 5 * 1024 * 1024,

                message: 'The selected file is not valid, it should be (doc,docx,pdf,zip,rtf) and 5 MB at maximum.'

            },

        }

        }
    }
    }); 
$('#submitcareer').click(function(){      

                $('#careerform').data('bootstrapValidator').validate();     

                if($("#careerform").data('bootstrapValidator').isValid()){
            $('#modalbtn').click();         

                }        

                }); 

});
    </script>

然后会出现一个模态，模态包含一个包含google recaptcha输入的表单：

                <div class="modal-body">
<form method="post" id="modalform" enctype="multipart/form-data">
                        <?php                       

                            require_once(Yii::app()->basePath.'/extensions/recaptchalib.php');

                            $publickey = "...";

                            echo recaptcha_get_html($publickey);

                        ?>                      

                        <button type="submit" class="btn btn-primary" id="modalsubmit" >Submit</button>                 

                    </form>                   

                </div>

                <div class="modal-footer">

                    <script type="text/javascript">         

                        $(document).ready(function() {                                  

                                $("#modalsubmit").click(function(event){ 
                                    event.preventDefault();
                                    $.ajax({
                                        type:"post",
                                        url:"<?php echo Yii::app()->createUrl('site/verify'); ?>",
                                        data:$("#modalform").serialize(),
                                        dataType: 'json',
                                        async: false,
                                        success:function(response){

                                            if(response == 2){

                                                    $('#careerform').data('bootstrapValidator').validate();
                                                        if($("#careerform").data('bootstrapValidator').isValid()){      
                                                            $('#careerform').submit();
                                                        }

                                            }

                                        }
                                    });
                                });

                        });

                    </script>

                </div> 

如果用户正确输入验证码，则ajax响应为2，则应提交第一个表单。

当我得到响应值= 2时，我的问题是第一个表单没有提交，那么这里的问题是什么？还有另一种方法可以做我想要的吗？

Answer 1

确定响应的类型。如果dataType设置为JSON，jQuery正试图将响应转换为JSON对象。