如果视图中的记录为空，我该如何说“是”或“否” - How do I say if record empty yes else no in a view?

How do I say if record empty yes else no in a view?

本文关键字：视图记录如果 | 更新日期: 2023-09-27

在我的代码点火器PHP模型中，我有

if  ($this->input->post('questions') != "")
        {
            if($this->input->post('questions') == "Yes")
            {
                $this->db->where('webinar_event.questions !=',"");
                $this->db->where('webinar_event.questions IS NOT ', null, false);
            }
            else
            {
                $this->db->where('webinar_event.questions',"");
                $this->db->where('webinar_event.questions IS', null, true);
            }

然而，当我运行echo $this->db->last_query();我得到这个错误

'Error Number: 1064
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 5
SELECT * FROM (`health_professional`) JOIN `webinar_event` ON `webinar_event`.`hpid` = `health_professional`.`hpid` WHERE `webinar_event`.`questions` = '' AND `webinar_event`.`questions` IS
Filename: D:'Development'PfizerWebinar'web'system'database'DB_driver.php
Line Number: 330'

基本上，我想做的是，如果我搜索"确实问过问题"，我会得到任何不为空的东西，如果我想搜索他们是否问过问题，则向我显示所有不为空。

    Try this 

      if  ($this->input->post('questions') != "")
      {
        if($this->input->post('questions') == "Yes")
        {
            $this->db->where('webinar_event.questions !=',"");
            $this->db->where('webinar_event.questions IS NOT NULL', null, false);
        }
        else
        {
            $this->db->where('webinar_event.questions',"");
            $this->db->where('webinar_event.questions IS NULL', null, true);
        }
      }

if(!empty($var)){
  //do stuff
}

在这里，我们检查变量是否不为空，因此！或者你可以使用：

if(isset($var)){
 //do stuff
}

这将检查变量是否已设置。或：

if(is_null($var)){
}else{
  //do stuff
}

无论你喜欢什么，或者isset和！空通常是我想要的。