为什么当我从函数返回假值时，父函数在 php 中将其设为真？

Question

我是 PHP 新手，也许这是一个愚蠢的问题，但是当我使用函数检查数据库中是否存在记录时，使用以下命令：

function isGameExisted($con,$name) 
 {        
     $query = "SELECT * from Games WHERE NAME ='$name' LIMIT 1";
     $check = mysqli_query($con,$query);              
//       var_dump($check);
     if(mysqli_num_rows($check)){

       //echo " EXISTS ";
      return true;
     }else
     {      
     //echo " OFFER DOES NOT EXIST  ";   
     return false;  
     }
 }

到目前为止，它在名称不存在时返回 false

但在父函数中

$isExisting = isGameExisted($con,$name);
if($isExisting)
    {

       $json['Result'] = "Fail";
       $json['message'] = "Error in registering. Probably the name already exists";

    }else
    {

    }

我说实话了！消息注册时出错...我在这里缺少什么

谢谢

Answer 1

问题是isset($check) 总是正确的。如果查询找到匹配行，$check 将被设置为一个数组。如果查询没有找到任何东西，$check 将被设置为NULL。

你应该这样做：

return $check != false;

整个函数应该是：

function isGameExisted($con,$name) 
{        
     $query = "SELECT 1 from Games WHERE NAME ='$name' LIMIT 1";
     $check = mysqli_fetch_array(mysqli_query($con,$query));        
     return $check != null; 
}     

Answer 2

Please try with this

function isGameExisted($con,$name) 
 {        
     $query = "SELECT * from Games WHERE NAME ='$name' LIMIT 1";
     $excQuery = mysqli_query($con,$query);              
     if(mysqli_num_rows($excQuery)){
        //echo " EXISTS ";
          return true;
     }else
     {      
     //echo " GAME DOES NOT EXIST  ";    
     return false;  
     }       
 }