無法團結從PHP接收數據

Question

我有這個劇本與我的數據庫是MySQL的

string dataSubmitURL = "http://localhost/HighScore/AddScore.php?"; 
string dataGetURL = "http://localhost/HighScore/GetScore.php?"; 

     // Use this for initialization 
     void Start() { 
      StartCoroutine(GetScores()); 
     } 

     // Update is called once per frame 
     void Update() { 

     } 

     IEnumerator PostScores(string playerName, int score) { 
      string submitURL = dataSubmitURL + "name="+ playerName + "&score=" + score; 
      print(submitURL); 
      WWW submitData = new WWW(submitURL); 
      yield return submitData; 
      if (submitData.error != null) 
      { 
       Debug.LogError("Error occur when Submitting data : " + submitData.error); 
       Debug.LogError("Error occur when Submitting data : " + submitData.text); 
       Debug.LogError("Error occur when Submitting data : " + submitData.responseHeaders); 
       //submitData.text 
      } 
      else { 
       print(" Submitted"); 
      } 

     IEnumerator GetScores() { 
      WWW getData = new WWW(dataGetURL); 
      yield return getData; 
      if (getData.error != null) 
      { 
       Debug.LogError("There was an error getting the high score: " + getData.error); 
      } 
      else { 
       print(getData.text); 
      } 
     } 
     } 

但問題是我得到

時出錯獲取互動高分：從服務器空回覆

雖然這兩個URL

string dataSubmitURL = "http://localhost/HighScore/AddScore.php?"; 
string dataGetURL = "http://localhost/HighScore/GetScore.php?"; 

在瀏覽器中正常工作時，我直接把它。我還在根文件夾中添加了這個crossDomain.xml，以便所有人都可以訪問它。我做錯了什麼。

<?xml version="1.0"?> 
<cross-domain-policy> 
<allow-access-from domain="*"/> 
</cross-domain-policy> 

Answer 1

我不知道是什麼問題所有的代碼是正確的，它的工作原理是，只要更換IP地址。

string dataSubmitURL = "http://YouripAddress/HighScore/AddScore.php?"; 
string dataGetURL = "http://YouripAddress/HighScore/GetScore.php?";