使用php unity3d从数据库中检索多个值 [英] Retrieving multiple values from the database using php unity3d

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问题描述

我正在制作类似于RPG游戏的游戏，用户可以登录并保存和加载数据。我可以登录并保存。但是，当我要加载数据时，我很难加载它。

I am making like RPG game where user can login and save and load the data. I can login and save. But, when I want to load the data, I am having a trouble to load that.

我只会发布必要的代码（PHP代码（Load.php）） ：

I will post the necessary code only (PHP code (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，我在Unity中通过脚本访问它（问题所在的必需代码）：

And here where I access it by script in Unity (necessary code where the problem is):

    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;
            }
        }

脚本仅返回单个值，即基于PHP代码的$ Location ，例如，当我想检索 $ Username 时， _www.text 将是 VillageYunnan 这不是我想要的，我只希望将单个值打印在 _www.text 上，例如 Village 或 Yunnan ，而不是 > VillageYunnan ，并且仅用于2个值，如果我想在PHP中检索所有 $ ，则类似于 VillageYunnan15101010100 ，依此类推，

The script only return single value which is $Location based on PHP code, and when I want to retrieve the $Username for example, the _www.text will be VillageYunnan which is I dont want, I want only the single value will be printed on _www.text, like Village or Yunnan and not VillageYunnan and that is only for 2 values, if I want to retrieve all of the $ in PHP, it will be like VillageYunnan15101010100 and so on, which will be very hard to maintain.

我的问题是我只希望 _www.text 仅打印单个值，而不希望将其合并在一起我以前提到过。

My question is I only want the _www.text only print like a single value and not combined together like I mention before.

有帮助吗？

真的很感谢您的回答。

谢谢

编辑后：（必需的鳕鱼e）

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） ：

And here is the PHP code (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 并等等，仅将其替换为 json_encode（$ row），如果PHP代码仍然如上，结果仍然相同，则指向 string jsonString = jsonObject.ToString（）; 和 var jsonObject = JSON.Parse（_www.text）; 为空。

When I delete all the $Username and so on and replace it with json_encode($row) only and if the PHP code stays like above, the result still same, there is an error pointed at string jsonString = jsonObject.ToString(); and the var jsonObject = JSON.Parse(_www.text); is null.

推荐答案

您可以使用某些通用格式（例如xml或json）在不同应用程序之间共享结构化数据。

You can use some common format (like xml or json) to share structured data between different applications.

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;

具有：

echo json_encode($row);

在C＃代码中使用一些JSON库（例如 SimpleJSON ）来解析JSON字符串并检索编码后的数据。

In C# code use some JSON library (like SimpleJSON) to parse JSON string and retreive encoded data.

这篇关于使用php unity3d从数据库中检索多个值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！

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