解析错误:语法错误,第25行上StatusIG.php中的意外'if'(T_IF)
我真的没有看到我的代码中有任何问题,为什么会发生这种情况,请帮忙。我是PHP的初学者
我的LINE 25是>> if($ status == 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);
答案 0 :(得分:3)
你错过了分号:
$status = mysql_result($result, 0, "raiguard")
将其替换为:
$status = mysql_result($result, 0, "raiguard");