Question

我一直在通过返回json编码的数组作为响应来处理成功/错误消息，但我突然想到这可能不是处理通知的正确方法。

例如，我的控制器将如下所示：

public function controller_name() {
    //validate form input
    $this->form_validation->set_rules('id', 'id', 'required|is_natural_no_zero');

    // if validation was successful with no errors
    if ($this->form_validation->run() && $this->model_name->method()) {

        $this->data['status'] = 'success';
        $this->data['message'] = 'This is the success message';
        echo json_encode($this->data);

    } else {
        $this->data['status'] = 'error';
        $this->data['message'] = validation_errors();
        echo json_encode($this->data);
    }
}

然后是jQuery：

$.ajax({
    url: url,
    type: 'POST',
    data: data,
    success: function (r) {

        json = $.parseJSON(r);

        if (json.status == 'success') {
            if (json.message == 'added') {
                $this.addClass('success');
            } else {
                $this.removeClass('success');
            }
        } else {
            console.log('There was an error')
        }

这样做的最佳做法是什么？我可以抛出异常来使用ajax错误吗？

Answer 1

发送错误的http状态代码应该触发jQuery ajax错误处理程序：

public function controller_name() {
    //validate form input
    $this->form_validation->set_rules('id', 'id', 'required|is_natural_no_zero');

    // if validation was successful with no errors
    if ($this->form_validation->run() && $this->model_name->method()) {
        $this->data['message'] = 'This is the success message';
    } else {
        $this->output->set_status_header('400'); //Triggers the jQuery error callback
        $this->data['message'] = validation_errors();

    }
    echo json_encode($this->data);
}

JS：

$.ajax({
    url: url,
    type: 'POST',
    data: data,
    success: function (r) {
        var json = $.parseJSON(r);

    },

    error: function( jqXhr ) {
        if( jqXhr.status == 400 ) { //Validation error or other reason for Bad Request 400
            var json = $.parseJSON( jqXhr.responseText );
        }
    }
});

使用Codeigniter处理ajax响应

1 个答案: