目前我的问题简化了:
下面的代码在统一编辑器中工作,我已经多次检查过php,现在发送的url工作,并且它返回正确的值,但是将内容上传到要在facebook app中使用的空间中断由于某种原因,img提供了显示停止的地方。
其他一些细节:
环顾所有迹象表明它是v3.0的错误(版本I'我正在使用v4.3.3f1),但是虽然我的代码在编辑器中工作没有问题,但它赢得了#39 ;当我将其构建上传到我提供的html空间时,t继续通过yield return www请求。
在下面的块之前登录工作的当前方式是初始化FB游戏对象,登录到Facebook,检查没有错误,然后传回FB用户ID,然后传递给下面的代码 - 其按预期工作
快速编辑 - 数据库连接器包含主URL,下面的代码添加到php get的必要字符串上。
我不知道从哪里开始尝试修复它,除了如图所示更改正在显示的文本,这有助于将其分解为我的问题。
粘贴的登录屏幕图像:http://i.imgur.com/ti2YLrW.png?1
代码:
using UnityEngine;
using System.Collections;
using System;
public class ConnectToDataBase : MonoBehaviour {
public bool bDatabaseConnected = false;
public bool bConnectionFailed = false;
public string sFacebookID;
public WWWForm wwwForm;
public void vStartConnection()
{
var text = GameObject.FindGameObjectWithTag ("Other");
text.guiText.text = "Login : connectToDatabase: prepping query";
//call databasequeries and get sDBConnect
string sConnectPhp = GameObject.FindGameObjectWithTag("DBConnector").GetComponent<DatabaseQueries>().sDBConnect;
//now ready the url with the necessary code for php's get, and the FacebookID
string url = sConnectPhp + "?UserID=" + sFacebookID;
text.guiText.text = "Login : connectToDatabase: url being sent:\n" + url;
WWW wwwGet = new WWW(url);
text.guiText.text = "Login : connectToDatabase: wwwGet created";
text.guiText.text = "Login : connectToDatabase: starting coroutine";
StartCoroutine(Connect(wwwGet));
}
IEnumerator Connect(WWW www)
{
var text = GameObject.FindGameObjectWithTag("Other");
text.guiText.text = "Login : connectToDatabase: coroutine started - sending www request";
yield return www;
text.guiText.text = "Login : connectToDatabase: wwwGet yield return";
string sTemp = www.text;
if(www.error == null)
{
text.guiText.text = "Login : connectToDatabase: wwwGet has not errored";
string newString = sTemp.ToString();
int newInt = Convert.ToInt32(newString);
//print (newString);
text.guiText.text = "Login : connectToDatabase: checking wwwGet return as int";
if(newInt == 0)
{
//if successfully connected set to true
print ("connectToDatabase: olduser successful");
text.guiText.text = "Login : connectToDatabase: olduser successful";
bDatabaseConnected = true;
yield break;
}
else if(newInt == 1)
{
//if successfully connected set to true
print ("connectToDatabase: newuser successful");
text.guiText.text = "Login : connectToDatabase: newuser successful";
bDatabaseConnected = true;
yield break;
}
else if(newInt == 2)
{
//game connection has failed
print ("connectToDatabase: failed");
text.guiText.text = "Login : connectToDatabase: failed";
bDatabaseConnected = false;
bConnectionFailed = true;
yield break;
}
else
{
text.guiText.text = "Login : connectToDatabase: php did not return a 0/1/2 value";
}
}
else
{
//game connection has failed
print ("connectToDatabase: failed");
text.guiText.text = "Login : connectToDatabase: wwwGet has errored:\n" + www.error;
bDatabaseConnected = false;
bConnectionFailed = true;
yield break;
}
text.guiText.text = "Login : connectToDatabase: wwwGet if statement skipped entirely";
}
}
答案 0 :(得分:1)
在浏览器中运行时有不同的规则。如上所述,here应用程序处于沙盒模式,因此您需要添加crossdomain策略(我知道该链接适用于Flash Player,但适用相同的原则)。
最后,您需要保存在Web服务器根目录中 crossdomain.xml 文件中的此类内容。
<?xml version="1.0"?>
<cross-domain-policy>
<allow-access-from domain="*" secure="false"/>
</cross-domain-policy>