2014-03-01 64 views
-1

解析错误:语法错误,意想不到的“如果”(T_IF)在StatusIG.php在线25PHP - 解析错误:语法错误,意想不到的T_IF

我真的不看在我的代码的任何问题,这是为什么发生,请帮助。在PHP IM初学者

我的第25行是>>如果($状态== 0)

 

    $accountdb_ip = "ip"; //Accountserver-IP 
    $accountdb_login = "acc"; //Accountserver-Loginname 
    $accountdb_pw = "psw"; //Accountserver-Passwort 
    $db_ip = "ip"; //DB-Server-IP 
    $db_login = "acc"; //DB-Server-Loginname 
    $db_pw = "psw"; //DB-Server-Passwort 

    $con = mysql_connect($db_ip, $db_login, $db_pw); 
    $con_account = mysql_connect($accountdb_ip, $accountdb_login, $accountdb_pw); 
    if (!$con) 
     { 
     die('Could not connect: ' . mysql_error()); 
     } 

    $name = mysql_real_escape_string($_GET["name"], $con); 

    $result_id = mysql_query("SELECT * FROM player.player WHERE name='".$name."' LIMIT 1", $con); 
    $player_acc_id = mysql_result($result_id, 0, "account_id"); 

    $result = mysql_query("SELECT * FROM account.account WHERE id='".$player_acc_id."'", $con); 
    $status = mysql_result($result, 0, "raiguard") 

    if($status == 0) 
     {echo "0";} 
    elseif($status == 1) 
     {echo "1";} 
    elseif($status == 2) 
     {echo "2";} 

    mysql_close($con); 
    mysql_close($con_account); 

+5

这个问题似乎是脱离主题,因为它是关于基本的语法错误。 – lifetimes

回答

3

你错过了结束分号:

$status = mysql_result($result, 0, "raiguard") 

将其替换为:

$status = mysql_result($result, 0, "raiguard"); 
+0

如果您在此处发布错误,那对我们来说会容易得多。 – Timo

+0

感谢,但现在不同的错误 说明:未定义指数:姓名RaiGuardStatusIG.php线路17 警告:mysql_result():无法跳到上的MySQL结果索引4 RaiGuardStatusIG.php线行0 20 警告:mysql_result ():无法跳到第23行的RaiGuardStatusIG.php中的MySQL结果索引5上的行0 第17行:$ name = mysql_real_escape_string($ _GET [“name”],$ con); 第20行:$ player_acc_id = mysql_result($ result_id,0,“account_id”); 第23行:$ status = mysql_result($ result,0,“raiguard”); – user3368877

+0

@ user3368877如何生成您的链接以获取GET名称? –

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