如何在单个Ajax调用中发送表单字符串数据和图像数据

如何合并两个Ajax调用，以便它们使用formData API传递表单输入数据以及表单图像数据？

查看Ajax调用：

$("#submitbmsymptom").click(function(h){
    h.preventDefault();
    var radioValue;
    $('#radio1, #radio2, #radio3, #radio4').each(function(){
        if($(this).is(':checked')){
            radioValue = $(this).val();
        }
    });

    console.log(radioValue +
    $("#bmdate").val() +
    $("#bmtime").val() +
    $("#bmscale").val() +
    $("#bmpainlevel").val() +
    $("#bmobs").val()
    );

    $.ajax({
        url:'urllocation',
        method: 'POST',
            data:{
                bmdate_data : $("#bmdate").val(),
                bmtime_data : $("#bmtime").val(),   
                bmscale_data : $("#bmscale").val(),
                bmcontents_data : radioValue,
                bmpainlevel_data : $("#bmpainlevel").val(),
                bmobs_data : $("#bmobs").val(),
            },
        }
        ).done(function(regresult){

    });
})

$("#submitbmsymptom").click(function(h){
    h.preventDefault();

    var pooPicture = $('input[name=poopic]');
    var fileUpload = pooPicture[0].files[0];

    var formData = new FormData();
    formData.append("file", fileUpload);

    $.ajax({
        url:'urllocation',
        method: 'POST',
        data: formData,
        processData: false,
        contentType: false,
        success: function() {
            alert("File uploaded");
        }
    });
})

Controller方法接收数据，将文件保存到服务器，然后使用关联数组将字符串数据和文件名插入数据库：

function bowelmovements(){
    $bmuserid = $this->session->userdata('id');
    $bmsymptomdate = $this->input->post('bmdate_data');
    $bmsymptomtime = $this->input->post('bmtime_data');
    $bmsymptomtype = $this->input->post('bmscale_data');
    $bmsymptomlocation = $this->input->post('bmcontents_data');
    $bmsymptompainlevel = $this->input->post('bmpainlevel_data');
    $bmsymptomobs = $this->input->post('bmobs_data');

    $config['upload_path'] = APPPATH.'/assets/uploads/';
    $config['allowed_types'] = "gif|jpg|png|mp4";

    $this->load->library('upload', $config);
    if($this->upload->do_upload("file")){
        $imageData = $this->upload->data();
        $fileName = $imageData[file_name];
        $bmdata = array(
            'userid' => $bmuserid,
            'bmdate' => $bmsymptomdate ,
            'bmtime' => $bmsymptomtime,
            'bmscale' => $bmsymptomtype,
            'bmcontents' => $bmsymptomlocation ,
            'bmpainlevel' => $bmsymptompainlevel,
            'bmobs' => $bmsymptomobs,
            'bmimage' => $fileName
        );

    $insertBM = $this->poomonitormodel->insertBM($bmdata);
    }
    else{
        echo "File not uploaded";
    }

}

目前，这仅提交第二个Ajax调用，并仅将文件名插入数据库，但是我需要将这两组数据提交到表中 .

谢谢 .