PHP和Codeigniter - 如何检查模型是否存在和/或不抛出错误？ [英] PHP and Codeigniter - How do you check if a model exists and/or not throw an error?

问题描述

示例1

bschaeffer的回答这个问题 - 在他的最后一个例子：

  $ this-> load-> model（'table'）; 
 $ data = $ this-> table-> some_func（）; 
 $ this-> load-> view（'view'，$ data）; 
  

'table'不存在？

---

示例＃2

  try {
 $ this-> load-> model（'serve_'。$ model_name，'my_model'）; 
 $ this-> my_model-> my_fcn（$ prams）; 
 
 //模型存在
 
} catch（Exception $ e）{
 //模型不存在
} 
  > 
 
 
 
 
 遇到错误 
 
 
 
无法找到模型您已指定：serve_forms 
 
 
 
 
 
 
 
 
我通过以下方式调用此函数：
 
 
 
 1）获取一些JSON：
 
 
  model_1：{function_name：{pram_1：1，pram_2：1}} 
  
 
 
 2）将它转换为函数调用：
 
 
 $ this-> load->模型（'serve_'。model_1，'my_model'）; 
 
 
 3）我在哪里调用：
 
 
  $ this-> my_model-> function_name（pram_1 = 1，pram_2 = 1）; 
  
 
 
 
 
 
 
 
 
解决方案
 
 
 
问题在于CodeIgniter的 show_error（...）函数显示错误，然后 exit;  ...不冷却...所以我overrode： model（...）  - >  my_model（..）（如果你只是覆盖它会收到错误），并删除 show_error（...）你不能覆盖它 - 奇怪的Codeigniter）。然后在 my_model（...）中引发异常
 
 
 我的个人意见：调用函数 return 
 show_error（message）; 其中show_error返回 FALSE  ---或
你可以取出退出;   - 并使 show_error（...） / code> 
 overridable 
 
 
 
 
解决方案
 
  $ model ='my_model'; 
 if（file_exists（APPPATH。models / $ model.php））{
 $ this-> load-> model（$ model）; 
 $ this-> my_model-> my_fcn（$ prams）; 
} 
 else {
 //模型不存在
} 
  
 
Example #1


bschaeffer'sanswer to this question - in his last example:
$this->load->model('table');
$data = $this->table->some_func();
$this->load->view('view', $data);
How do you handle this when 'table' doesn't exist?



Example #2

    try {
        $this->load->model('serve_' . $model_name, 'my_model');
        $this->my_model->my_fcn($prams);

        // Model Exists

    } catch (Exception $e) {
        // Model does NOT Exist
    }
But still after running this (obvously the model doesn't exist - but sometimes will) it fails with the following error:

  
An Error Was Encountered
  
  
Unable to locate the model you have specified: serve_forms




I am getting this function call by:

1) Getting some JSON:

"model_1:{"function_name:{"pram_1":"1", "pram_2":"1"}}

2) And turning it into the function call:

  
$this->load->model('serve_' . "model_1", 'my_model');
3) Where I call:

$this->my_model->function_name(pram_1=1, pram_2=1);





SOLUTION

The problem lies in the fact that CodeIgniter's show_error(...) function displays the error then exit; ... Not cool ... So I overrode: model(...) -> my_model(..) (you'll get errors if you just override it) and removed the show_error(...) because for some reason you can't override it - weird for Codeigniter). Then in my_model(...) made it throw an Exception

  
My personal opinion: the calling function should return
  show_error("message"); where show_error returns FALSE --- that or
  you could take out the exit; - and make show_error(...)
  overridable

解决方案
You can see if the file exists in the models folder.
$model = 'my_model';
if(file_exists(APPPATH."models/$model.php")){
   $this->load->model($model);
   $this->my_model->my_fcn($prams);
}
else{
  // model doesn't exist
}


                        
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