我想從一個函數在PHP返回兩個結果

Question

這是我在DB_Functions.php功能，我想從單一功能只有在這裏兩個不同的表，可能得到兩個不同的值是代碼是我迄今爲止嘗試過，但值將變爲空值。

public function getUserMetvalue($exname,$fname) { 

    $result = mysql_query("SELECT metvalue FROM fitnessactivitylist WHERE activityname='$exname'") or die(mysql_error()); 

    $result1 = mysql_query("SELECT weight FROM users WHERE name='$fname'") or die(mysql_error()); 

     // check for result 

     $no_of_rows = mysql_num_rows($result); 
     $no_of_rowss = mysql_num_rows($result1); 


     if ($no_of_rows > 0) { 
      $result = mysql_fetch_array($result); 
      if ($no_of_rowss > 0) { 
       $result1 = mysql_fetch_array($result1); 
       return $result1; 
      }       
      return $result; 
     } else { 
     //exercise name not found 
     return false; 
     } 
    } 

，這裏是我的index.php

 //TAG METVALUE 
     if ($tag == 'metvalue') { 

     $exname = $_POST['exname']; 

     $fname = $_POST['fname']; 

     $usermetvalue = $db->getUserMetvalue($exname,$fname); 

     if ($usermetvalue != false) { 

     $response["success"] = 1; 

    $response["usermetvalue"]["exname"] = $usermetvalue["exname"]; 

     $response["usermetvalue"]["fname"] = $usermetvalue["fname"]; 

     echo json_encode($response); 

     } 

       else { 

$response["error"] = 1; 
$response["error_msg"] = "No exercise found!"; 
echo json_encode($response); 

      } 

    } 

Answer 1

現貨的差異：

$result = mysql_query("SELECT metvalue etc... 
           ^^^^^^^^ 
$result1 = mysql_query("SELECT weight etc... 
           ^^^^^^ 


$response["usermetvalue"]["exname"] = $usermetvalue["exname"]; 
                ^^^^^^ 
$response["usermetvalue"]["fname"] = $usermetvalue["fname"]; 
                ^^^^^ 

你獲取了這些不晚使用的字段，然後嘗試訪問哪些不是場取出擺在首位...

Answer 2

function a_function() { 
    $a = 'Learn'; 
    $b = 'Programming.'; 

    return array($a, $b); 
} 

list($one, $two) = a_function(); 

echo $one . ' ' . $two;