如果PHP / YII中不为空，则显示字段

问题描述：

I used to code with Groovy ... I found that this 'feature' (no idea what they called it) so fun n nice (I heard they implemented this on C# too).

for example ... I want to display person neighbour name

I just type

println person?.neigbour?.name ;

it means if the neighbour is empty / blank .. it didn't display anything .

how to do this in php 5/yii?

example:

instead of typing long codes like '/>

Would it be better to type like '/>

我曾经用Groovy编码...我发现这个'功能'（不知道他们叫什么）非常好玩（我听说他们也在C＃上实现了这一点）。 p>

例如...我想显示人物邻居名称 p>

我只需输入 p>

  println person？.neigbour？.name;  
  code>  pre> 
 
 这意味着如果邻居为空/空白..它没有显示任何内容。 p> 
 
 
如何执行此操作 在PHP 5 / yii？   p> 
 
 
示例： p> 
 
 
而不是输入长代码，例如
 
'/＆gt;  p> 
 
 
会不会 最好输入类似
'/＆gt;  p> 
  div>

答

In the first place I think this is no Yii issue, but simply PHP. Assuming, that you use Yii with nice models, there it would go sth like this:

if($person && $person->neighbour && !empty($person->neighbour->name))  {
   echo $person->neighbour->name;
}

a shortcut for this may be (not so nice):

echo $person ? ($person->neighbour ? ($person->neighbour->name ? $person->neighbour->name : "" ) : ""  ) : "";

答

use empty() check:

if(!empty($variable))
{
//show fields here
}

如果PHP / YII中不为空，则显示字段

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