PHP：如何运行这个IF语句

Question

我运行了这个脚本，但我一直收到我放入的“没有工作”的信息。我知道，在分贝值是正确的，它必须是东西在我的PHP ...

  $getadmin = "SELECT role FROM user WHERE user_id=$uid";  
      $showadmin = @mysqli_query ($dbc, $getadmin); // Run the query. 

      $admin = mysqli_fetch_assoc($showadmin); 
      if($admin == 'admin'){ 

      echo 'You are an admin!'; 

      } else { 

      echo 'Did not work'; 

      } 

Answer 1

试试这个：

 $getadmin = "SELECT role FROM user WHERE user_id=$uid";  
     $showadmin = @mysqli_query ($dbc, $getadmin); // Run the query. 

     $role = mysqli_fetch_assoc($showadmin); 
     if($role['role'] == 'admin'){ 

     echo 'You are an admin!'; 

     } else { 

     echo 'Did not work'; 

     } 

Answer 2

未测试

$row = mysqli_fetch_row($showadmin); 
if($row[0] == 'admin'){ .. 

Answer 3

 $getadmin = "SELECT role FROM user WHERE user_id=$uid";  
     $showadmin = @mysqli_query ($dbc, $getadmin); // Run the query. 

     $admin = mysqli_fetch_assoc($showadmin); 
     if($admin && $admin[0]['role'] == 'admin'){ 

      echo 'You are an admin!'; 

     } else { 

      echo 'Did not work'; 

     } 

Answer 4

fetch_assoc返回一个数组，而不是单个字符串

您需要检查$ admin ['role']而不是$ admin作为值admin。

Answer 5

$getadmin = "SELECT role FROM user WHERE user_id={$uid}"; //use {} around variables for interpolation 
$showadmin = @mysqli_query ($dbc, $getadmin); // Run the query. 

if(!$showadmin) 
{ 
    die('<p>ERROR: DB result is null</p>'); 
} 

while($ROW=mysqli_fetch_assoc($showadmin))//returns $ROW associative array 
{ 
    if($ROW['role'] == 'admin') 
    { 
     echo 'You are an admin!'; 
    } 
    else //$ROW['role']!='admin' 
    { 
     echo 'Did not work because it is '.$ROW['role']; 
    } 
}