Codeigniter jQuery在AJAX文件上传内部不起作用

Question

我在codeigniter中有一个AJAX文件上传代码。 问题是我将简单表单​​提交更改为文件提交。 但是之后，JQUERY已停止工作。 响应即将成功，但同时调用了ajax错误函数。 我不知道我的代码有什么问题。

这是我的控制器。

public function ajax_add() {
    $this->_validate();

    $config = [
    'upload_path' => './assets/game_images/',
    'allowed_types' => 'gif|png|jpg|jpeg'
    ];
    $this->load->library('upload', $config);
    if ($this->upload->do_upload('image')) {
        $file = $this->upload->data();
        $file_name = $file['file_name'];

        if ($file_name == '') {
            $data['error_string'][] = 'Please upload an image.';
            $data['status'] = FALSE;
            echo json_encode($data);
            exit();
        }
    } else {
        $data['inputerror'][] = 'image';
        $data['error_string'][] = $this->upload->display_errors();
        $data['status'] = FALSE;
        echo json_encode($data);
        exit();
    }

    $data = array(
        'title' => $this->input->post('title'),
        'iframe' => $this->input->post('iframe'),
        'status' => $this->input->post('status'),
        'category_id' => $this->input->post('category_id'),
        //'image' => $file_name
        );
    $insert = $this->game->save($data);
    echo json_encode(array("status" => TRUE));
}

private function _validate() {
    $data = array();
    $data['error_string'] = array();
    $data['inputerror'] = array();
    $data['status'] = TRUE;

    if ($this->input->post('title') == '') {
        $data['inputerror'][] = 'title';
        $data['error_string'][] = 'Game Title is required';
        $data['status'] = FALSE;
    }

    if ($this->input->post('iframe') == '') {
        $data['inputerror'][] = 'iframe';
        $data['error_string'][] = 'Game Iframe is required';
        $data['status'] = FALSE;
    }

    if ($this->input->post('status') == '') {
        $data['inputerror'][] = 'status';
        $data['error_string'][] = 'Status is required';
        $data['status'] = FALSE;
    }

    if ($this->input->post('category_id') == '') {
        $data['inputerror'][] = 'category_id';
        $data['error_string'][] = 'Please select category';
        $data['status'] = FALSE;
    }

    if ($data['status'] === FALSE) {
        echo json_encode($data);
        exit();
    }
}

这是我的HTML

if (save_method == 'add') {
            url = "<?php echo site_url('game/ajax_add') ?>";
        } else {
            url = "<?php echo site_url('game/ajax_update') ?>";
        }
var formData = new FormData($('#form')[0]);

        $.ajax({
            url: url,
            type: 'JSON',
            data: formData,
            async: false,
            success: function (data)
            {
                if (data.status) //if success close modal and reload ajax table
                {
                    $('#modal_form').modal('hide');
                    reload_table();
                } else
                {
                    for (var i = 0; i < data.inputerror.length; i++)
                    {
                        $('[name="' + data.inputerror[i] + '"]').parent().parent().addClass('has-error'); //select parent twice to select div form-group class and add has-error class
                        $('[name="' + data.inputerror[i] + '"]').next().text(data.error_string[i]); //select span help-block class set text error string
                    }
                }
                $('#btnSave').text('save'); //change button text
                $('#btnSave').attr('disabled', false); //set button enable 
            },
            error: function (jqXHR, textStatus, errorThrown)
            {
                alert('Error adding / update data');
                $('#btnSave').text('save'); //change button text
                $('#btnSave').attr('disabled', false); //set button enable 

            },
            cache: false,
            contentType: false,
            processData: false
        });

Answer 1

$.ajax({
   type: 'POST',
   url: url,
   dataType: 'JSON',
   contentType: 'application/json; charset=utf-8'
})