PHP函數不是結果工作同統計

Question

你好，我需要幫助大約一個功能我試圖解釋：

這是代碼：

// here user-a START // 
function get_car_speed_a($carid) 
{ 
    $sqle = mysql_query("SELECT speed FROM `user_cars` WHERE userid='$urow[id]' AND carid='$carid'"); 
    $row1 = mysql_fetch_array($sqle); 
    return $row1['speed']; 
} 

function get_car_stability_a($carid) 
{ 
    $sqle = mysql_query("SELECT stability FROM `user_cars` WHERE userid='$urow[id]' AND carid='$carid'"); 
    $row1 = mysql_fetch_array($sqle); 
    return $row1['stability']; 
} 

function get_car_acceleration_a($carid) 
{ 
    $sqle = mysql_query("SELECT acceleration FROM `user_cars` WHERE userid='$urow[id]' AND carid='$carid'"); 
    $row1 = mysql_fetch_array($sqle); 
    return $row1['acceleration']; 
} 
// here user-a END // 



// here user-b START // 
function get_car_speed_b($carid) 
{ 
    $sqle = mysql_query("SELECT speed FROM `user_cars` WHERE userid='$row2[id]' AND carid='$carid2'"); 
    $row1 = mysql_fetch_array($sqle); 
    return $row1['speed']; 
} 

function get_car_stability_b($carid) 
{ 
    $sqle = mysql_query("SELECT stability FROM `user_cars` WHERE userid='$row2[id]' AND carid='$carid2'"); 
    $row1 = mysql_fetch_array($sqle); 
    return $row1['stability']; 
} 

function get_car_acceleration_b($carid) 
{ 
    $sqle = mysql_query("SELECT acceleration FROM `user_cars` WHERE userid='$row2[id]' AND carid='$carid2'"); 
    $row1 = mysql_fetch_array($sqle); 
    return $row1['acceleration']; 
} 
// here user-b END // 

而這正是召回函數的代碼：

// battle function start// 

    $apoints = get_car_speed_a($useracar) + get_car_stability_a($useracar) + get_car_acceleration_a($useracar); 
    $bpoints = get_car_speed_b($userbcar) + get_car_stability_b($userbcar) + get_car_acceleration_b($userbcar); 


    // battle function end// 

的問題很容易在網站時，我就與用戶的爭奪戰VS用戶b它不採取正確的速度，穩定性和accelleration所以pratically用戶b（敵人）會失去...

有人可以幫助我嗎？非常感謝你 問候 艾米

Answer 1

在 「B」 的功能，你宣佈$ carid，並在SQL中使用$ carid2，改變這種：

// here user-b START // 
function get_car_speed_b($carid) 
{ 
    $sqle = mysql_query("SELECT speed FROM `user_cars` WHERE userid='$row2[id]' AND carid='$carid'"); 
    $row1 = mysql_fetch_array($sqle); 
    return $row1['speed']; 
} 

function get_car_stability_b($carid) 
{ 
    $sqle = mysql_query("SELECT stability FROM `user_cars` WHERE userid='$row2[id]' AND carid='$carid'"); 
    $row1 = mysql_fetch_array($sqle); 
    return $row1['stability']; 
} 

function get_car_acceleration_b($carid) 
{ 
    $sqle = mysql_query("SELECT acceleration FROM `user_cars` WHERE userid='$row2[id]' AND carid='$carid'"); 
    $row1 = mysql_fetch_array($sqle); 
    return $row1['acceleration']; 
} 
// here user-b END // 

並檢查變量$ urow [ '身份證' ]和$ 2行[「身份證」]作爲用戶名和$ carid，如果他們有正確的價值

UPDATE

你可以放在一起功能於一體

function get_car_info($userid, $carid) 
{ 
    $res = mysql_query("SELECT speed, stability, acceleration FROM `user_cars` WHERE userid='$userid' AND carid='$carid'"); 
    return mysql_fetch_array($res); 
} 

而得到這樣

$carinfoA = get_car_info($useridA, $caridA); 
$carinfoB = get_car_info($useridB, $caridB); 

$apoints = $carinfoA['speed'] + $carinfoA['stability'] + $carinfoA['acceleration']; 
$bpoints = $carinfoB['speed'] + $carinfoB['stability'] + $carinfoB['acceleration']; 

我創造了你發送的測試表中的結果，這裏的結果是正確的。

如果錯誤仍然存​​在，錯誤應該在代碼的另一部分。