Question

嘿，由于某些原因，代码不起作用，它不能回应成功，这很奇怪，它插入了mysql信息，但只是没有回应成功，如果你的家伙知道问题，请发布修复或感谢很多人的东西！< / p>

<?php

//Initiliaze Database connection
require("config.php");    


//IMPORTANT STUFF
$username = mysql_real_escape_string(stripslashes($_POST["strUsername"]));
$password = mysql_real_escape_string(stripslashes($_POST["strPassword"]));
$pass1 = gen_token($password, $username);
$age = mysql_real_escape_string(stripslashes($_POST["intAge"]));
$dob = mysql_real_escape_string(stripslashes($_POST["strDOB"]));
$email = mysql_real_escape_string(stripslashes($_POST["strEmail"]));
$gender = mysql_real_escape_string(stripslashes($_POST["strGender"]));
$classid = mysql_real_escape_string(stripslashes($_POST["ClassID"]));
$eyecolor = mysql_real_escape_string(stripslashes($_POST["intColorEye"]));
$skincolor = mysql_real_escape_string(stripslashes($_POST["intColorSkin"]));
$haircolor = mysql_real_escape_string(stripslashes($_POST["intColorHair"]));
$hairid = $_POST['HairID'];


//Checks if Email has Already been used
$emailcheck = mysql_query("SELECT id FROM users WHERE Email='$email'") or die("status=Error&strReason=" . mysql_error());
if (mysql_num_rows($emailcheck) != 0) {
        die("status=Taken&strReason=The email is already in used by another user.");
}

//Checks If Username has been Taken
$sql = mysql_query("SELECT * FROM users WHERE Username = '$username'") or die("status=Error&strReason=" . mysql_error());
if (mysql_num_rows($sql) !=0) {
        die("status=Taken&strReason=The username is already in use by another character.");
} else {

//Sets Hairname & hairfile
switch ($hairid) {
        //MALE HAIR
        case 52:
                $hairname = 'Default';
                $hairfile = 'hair/M/Default.swf';
                break;
        case 55:
                $hairname = 'Goku1';
                $hairfile = 'hair/M/Goku1.swf';
                break;
        case 58:
                $hairname = 'Goku2';
                $hairfile = 'hair/M/Goku2.swf';
                break;
        case 64:
                $hairname = 'Normal2';
                $hairfile = 'hair/M/Normal2.swf';
                break;
        case 92:
                $hairname = 'Ponytail8';
                $hairfile = 'hair/M/Ponytail8.swf';
                break;

        //FEMALE HAIR
        case 14:
                $hairname = 'Pig1Bangs1';
                $hairfile = 'hair/F/Pig1Bangs1.swf';
                break;
        case 18:
                $hairname = 'Pig2Bangs2';
                $hairfile = 'hair/F/Pig2Bangs2.swf';
                break;
        case 26:
                $hairname = 'Pony2Bangs2';
                $hairfile = 'hair/F/Pony2Bangs2.swf';
                break;
        case 83:
                $hairname = 'Bangs2Long';
                $hairfile = 'hair/F/Bangs2Long.swf';
                break;
        case 84:
                $hairname = 'Bangs3Long';
                $hairfile = 'hair/F/Bangs3Long.swf';
                break;
}
$time = date("Y-m-d");
//Inserts Character Info into DB
$sql2 = mysql_query("INSERT INTO `users` (`Username`, `Password`, `Access`, `ActivationFlag`, `Age`, `Gender`, `Email`, `Level`, `Gold`, `Coins`, `Exp`, `ColorHair`, `ColorSkin`, `ColorEye`, `ColorBase`, `ColorTrim`, `ColorAccessory`, `DateCreated`, `UpgradeExpire`, `UpgradeDays`, `BankSlots`, `HouseSlots`, `BagSlots`, `HairID`, `HairFile`, `HairName`, `Permamute`, `Quests`, `Settings`, `Achievement`, `Country`, `AchievementID`, `CurrentServer`) VALUES ('$username', '$pass1', '0', '5', '15', '$gender', '$email', '1', '0', '0', '0', '$haircolor', '$skincolor', '$eyecolor', '0', '0', '0', '$time', '$time', '-1', '0', '20', '150', '$hairid', '$hairfile', '$hairname', '0', '00000000000000000000000000000000000000000000000000', '0', '0', 'US', '', 'Offline');")  or die("status=Error&strReason=" . mysql_error());

//Selects New User ID
$sql3 = mysql_query("SELECT * FROM users WHERE Username='$username'") or die("status=Error&strReason=" . mysql_error());
$user = mysql_fetch_assoc($sql3) or die("status=Error&strReason=" . mysql_error());
$userId = $user['id'];

//Add's Starting Armor
switch ($classid) {
        case 2:
                $addarmour = mysql_query("INSERT INTO users_items (itemid, userid, equipped, equipment, level) VALUES ('2', '$userId', '1', 'ar', '1')");
                break;
        case 4:
                $addarmour = mysql_query("INSERT INTO users_items (itemid, userid, equipped, equipment, level) VALUES ('4', '$userId', '1', 'ar', '1')");
                break;
        case 3:
                $addarmour = mysql_query("INSERT INTO users_items (itemid, userid, equipped, equipment, level) VALUES ('3', '$userId', '1', 'ar', '1')");
                break;
        case 5:
                $addarmour = mysql_query("INSERT INTO users_items (itemid, userid, equipped, equipment, level) VALUES ('5', '$userId', '1', 'ar', '1')");
                break;
                }

// ADDS DEFAULT WEAPON
$addweapon = mysql_query("INSERT INTO users_items (itemid, userid, equipped, equipment, level) VALUES ('1', '$userId', '1', 'Weapon', '1')" );


// ADDS USERS FRIEND LIST
$addfriends = mysql_query("INSERT INTO users_friends (userid, friends) VALUES ($userId, '')" );


//SUCCESS      
echo "status=Success";
}

function gen_token($pass, $salt) {
        $salt = strtolower($salt);
        $str = hash("sha512", $pass.$salt);
        $len = strlen($salt);
        return strtoupper(substr($str, $len, 17));
}
?>

Answer 1

我会计算总是会发生的4个INSERT命令（假设成功）。这是正确的，还是只有部分工作？

没有什么是不正确的。您将mysql_query结果存储在变量中，但不对它们执行任何操作。也许尝试测试这些值以查看MySQL返回的内容？

您也没有测试是否有任何POST的值为空，空或无效。始终验证正在POST的内容 - 为了快速调试，你可以做一个print_r（$ _ POST）;.

PHP MYSQL回声不起作用

1 个答案: