使用php unity3d从数据库中检索多个值

时间:2015-03-11 11:40:27

标签: c# php database unity3d

我正在制作RPG游戏,用户可以登录并保存和加载数据。我可以登录并保存。但是,当我想加载数据时,我很难加载它。

我将仅发布必要的代码(PHP代码(Load.php)):

    <?PHP

    $Username = $_POST['Username'];
    $Location = $_POST['Location'];
    $Gold = $_POST['Gold'];
    $Level = $_POST['Level'];
    $Attack = $_POST['Attack'];
    $Defense = $_POST['Defense'];
    $MagicDefense = $_POST['MagicDefense'];
    $Evasion = $_POST['Evasion'];
    $Health = $_POST['Health'];
    $MaxHealth = $_POST['MaxHealth'];
    $Mana = $_POST['Mana'];
    $MaxMana = $_POST['MaxMana'];
    $Exp = $_POST['Exp'];
    $NextExp = $_POST['NextExp'];
    $AcceptedChiefQuest = $_POST['AcceptedChiefQuest'];
    $AddedChiefQuest = $_POST['AddedChiefQuest'];

    $con = mysql_connect("SERVER_NAME","DATABASE_NAME","PASSWORD") or ("Cannot connect!"  . mysql_error());

    if (!$con)
        die('Could not connect: ' . mysql_error());

    mysql_select_db("DATABASE_NAME" , $con) or die ("Could not load the database" . mysql_error());

    $check = mysql_query("SELECT * FROM Information WHERE `Username` = '".$Username."'");

    $numrows = mysql_num_rows($check);

    if ($numrows > 0)
    {
        while($row = mysql_fetch_assoc($check))
        {
            $Username = $row['Username'];
            $Location = $row['Location'];
            $Gold = $row['Gold'];
            $Level = $row['Level'];
            $Attack = $row['Attack'];
            $Defense = $row['Defense'];
            $MagicDefense = $row['MagicDefense'];
            $Evasion = $row['Evasion'];
            $Health = $row['Health'];
            $MaxHealth = $row['MaxHealth'];
            $Mana = $row['Mana'];
            $MaxMana = $row['MaxMana'];
            $Exp = $row['Exp'];
            $NextExp = $row['NextExp'];
            $AcceptedChiefQuest = $row['AcceptedChiefQuest'];
            $AddedChiefQuest = $row['AddedChiefQuest'];

            echo $Username;
            echo $Location;
        }

        die();
    }

    else
    {
        die ("Data does not exist!");
    }


    ?>

在这里我通过Unity中的脚本访问它(问题所在的必要代码):

    IEnumerator LoadCoroutine(WWW _www)
        {
            yield return _www;

            if (_www.error == null)
            {
                message = _www.text;

                if (_www.text == "Data does not exist!")
                {
                    message = "Data does not exist";

                    clickedLoadGame = false;
                }

                else if (_www.text == "Saved data does not match!")
                {
                    message = "Saved data does not match";

                    clickedLoadGame = false;
                }

                else
                {
                    if (_www.text == "Village")
                    {
                        GameManager.CurrentLocation = "Village";

                        GameManager.Loader = 1;
                    }

                    else if (_www.text == "Yein Plain")
                    {
                        GameManager.CurrentLocation = "Yein Plain";

                        GameManager.Loader = 4;
                    }

                    message = "Successfully Loaded.";

                    mainMenu = false;

                    yield return new WaitForSeconds(3);

                    Reset();

                    GameManager.LoadLevel("Loading Scene");
                }
            }

            else
            {
                message = "Error while trying to connect to the server.\nMessage: " + _www.error;

                clickedLoadGame = false;
            }
        }

脚本只返回基于PHP代码的单个值 $ Location ,当我想要检索 $ Username 时, _www。文字将是 VillageYunnan 这是我不想要的,我只希望单个值会打印在 _www.text 上,就像村庄云南而不是 VillageYunnan ,这只是2个值,如果我想要检索PHP中的所有 $ ,它将是喜欢 VillageYunnan15101010100 等等,这将非常难以维护。

我的问题是我只希望 _www.text 只打印像单个值,而不是像我之前提到的那样组合在一起。

任何帮助?

真的很感激你的回答。

谢谢

已编辑:(必填代码)

GameManager.cs

public static void WebsiteConnect(string link)
{
    wwwForm = new WWWForm();

    wwwForm.AddField("Username", LoggedInUsername);
    wwwForm.AddField("Gold", UserInformation.CurrentGold);
    wwwForm.AddField("Level", Status.level);
    wwwForm.AddField("Attack", Status.attack);
    wwwForm.AddField("Defense", Status.defense);
    wwwForm.AddField("MagicDefense", Status.magicDefense);
    wwwForm.AddField("Evasion", Status.evasion);
    wwwForm.AddField("Health", Status.hp);
    wwwForm.AddField("MaxHealth", Status.maxHp);
    wwwForm.AddField("Mana", Status.mp);
    wwwForm.AddField("MaxMana", Status.maxMp);
    wwwForm.AddField("Exp", Status.CurrentExperience);
    wwwForm.AddField("NextExp", Status.NextLevelExperience);
    wwwForm.AddField("AcceptedChiefQuest", QuestManager._acceptedChiefQuest);
    wwwForm.AddField("AddedChiefQuest", QuestManager._addedChiefQuest);
    }

    www = new WWW(link, wwwForm);
}

我这样称呼它:

private void OnGUI()
{
if (GUILayout.Button("Continue", customStartButton.customStyles[0]))
            {
                clickedLoadGame = true;

                audio.PlayOneShot(audioSEClip);

                Load();
            }
}

private void Load()
    {
        GameManager.WebsiteConnect("WEBSITE_ADDRESS/Load.php");
        StartCoroutine(LoadCoroutine(GameManager.www));
    }

IEnumerator LoadCoroutine(WWW _www)
    {
        yield return _www;

        if (_www.error == null)
        {
            message = _www.text;

            var jsonObject = JSON.Parse(_www.text);

            //string jsonString = jsonObject.ToString();

            Debug.Log(jsonObject);

            if (_www.text == "Data does not exist!")
            {
                message = "Saved data does not match!";

                mainMenu = false;

                fadeAudio = false;

                yield return new WaitForSeconds(3);

                clickedLoadGame = false;

                Reset();
            }

            else
            {
                if (_www.text == "Village")
                {
                    message = "Successfully Loaded.";

                    GameManager.CurrentLocation = "Village";

                    GameManager.Loader = 1;

                    mainMenu = false;

                    fadeAudio = true;

                    yield return new WaitForSeconds(3);

                    Reset();

                    GameManager.LoadLevel("Loading Scene");
                }

                else if (_www.text == "Yein Plain")
                {
                    message = "Successfully Loaded.";

                    GameManager.CurrentLocation = "Yein Plain";

                    GameManager.Loader = 4;

                    mainMenu = false;

                    fadeAudio = true;

                    yield return new WaitForSeconds(3);

                    Reset();

                    GameManager.LoadLevel("Loading Scene");
                }

                else
                {
                    message = "Data does not exist!";

                    mainMenu = false;

                    fadeAudio = false;

                    yield return new WaitForSeconds(3);

                    clickedLoadGame = false;

                    Reset();
                }
            }
        }

        else
        {
            message = "Error while trying to connect to the server.\nMessage: " + _www.error;

            clickedLoadGame = false;
        }
    }

这是PHP代码(Load.php):

<?PHP

$Username = $_POST['Username'];
$Location = $_POST['Location'];
$Gold = $_POST['Gold'];
$Level = $_POST['Level'];
$Attack = $_POST['Attack'];
$Defense = $_POST['Defense'];
$MagicDefense = $_POST['MagicDefense'];
$Evasion = $_POST['Evasion'];
$Health = $_POST['Health'];
$MaxHealth = $_POST['MaxHealth'];
$Mana = $_POST['Mana'];
$MaxMana = $_POST['MaxMana'];
$Exp = $_POST['Exp'];
$NextExp = $_POST['NextExp'];
$AcceptedChiefQuest = $_POST['AcceptedChiefQuest'];
$AddedChiefQuest = $_POST['AddedChiefQuest'];

$con = mysql_connect("WEBSITE_ADDRESS","DATABASE_NAME","PASSWORD") or ("Cannot connect!"  . mysql_error());

if (!$con)
    die('Could not connect: ' . mysql_error());

mysql_select_db("DATABASE_NAME" , $con) or die ("Could not load the database" . mysql_error());

$check = mysql_query("SELECT * FROM Information WHERE `Username` = '".$Username."'");

$numrows = mysql_num_rows($check);

if ($numrows > 0)
{
    while($row = mysql_fetch_assoc($check))
    {
        if ($Username == $row['Username'])
        {
            $Username = $row['Username'];
            $Location = $row['Location'];
            $Gold = $row['Gold'];
            $Level = $row['Level'];
            $Attack = $row['Attack'];
            $Defense = $row['Defense'];
            $MagicDefense = $row['MagicDefense'];
            $Evasion = $row['Evasion'];
            $Health = $row['Health'];
            $MaxHealth = $row['MaxHealth'];
            $Mana = $row['Mana'];
            $MaxMana = $row['MaxMana'];
            $Exp = $row['Exp'];
            $NextExp = $row['NextExp'];
            $AcceptedChiefQuest = $row['AcceptedChiefQuest'];
            $AddedChiefQuest = $row['AddedChiefQuest'];

            json_encode($row);
        }

        else
        {
            die ("Saved data does not match!");
        }
    }

    die();
}

else
{
    die ("Data does not exist!");
}


?>

当我删除所有$Username等等并仅用json_encode($row)替换它时,如果PHP代码保持如上,结果仍然相同,则会出现{{1}的错误} string jsonString = jsonObject.ToString();为空。

2 个答案:

答案 0 :(得分:1)

您可以使用一些通用格式(如xml或json)在不同的应用程序之间共享结构化数据。

在PHP代码中替换:

$Username = $row['Username'];
$Location = $row['Location'];
$Gold = $row['Gold'];
$Level = $row['Level'];
$Attack = $row['Attack'];
$Defense = $row['Defense'];
$MagicDefense = $row['MagicDefense'];
$Evasion = $row['Evasion'];
$Health = $row['Health'];
$MaxHealth = $row['MaxHealth'];
$Mana = $row['Mana'];
$MaxMana = $row['MaxMana'];
$Exp = $row['Exp'];
$NextExp = $row['NextExp'];
$AcceptedChiefQuest = $row['AcceptedChiefQuest'];
$AddedChiefQuest = $row['AddedChiefQuest'];

echo $Username;
echo $Location;

使用:

echo json_encode($row);

在C#代码中,使用一些JSON库(如SimpleJSON)来解析JSON字符串并检索编码数据。

答案 1 :(得分:0)

我不是100%确定我得到了问题,但我认为这是导致问题的原因:

        echo $Username;
        echo $Location;

这会输出YunnanVillage,这很糟糕。替换为:

        echo '[' . $Username . ',' . $Location . ']';

这将输出[Yunnan, Village]。您现在可以更轻松地解析此字符串(它是JSON格式)。请注意,这是一个非常简单的示例,有更好的方法可以在PHP中输出数组,例如:

json_encode(array('Username' => $Username, 'Location' => $Location));

我在这里是因为c#标签但知道PHP足以看到代码中的错误。希望这会有所帮助。