通过PHP+从MySql获取状态；JSON

我正在构建一个android游戏，在这个游戏中，我试图将用户的个人资料从MySql数据库下载到本地SQLite数据库中。我可以将配置文件发送到服务器，但无法检索它

这里是相关的android方法，它们应该没有问题，它们可以正常工作，尽管逻辑本身可能是关闭的

    //LOAD PROFILE CLASS
    userFunctions.getServerGameState(email, monster, qty, exp);
    db.saveLocalGameSate(monster, qty, exp);
    //FUNCTIONS CLASS
    public JSONObject getServerGameState(String email, String monster, int qty, int exp) {
    // Building Parameters
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("tag", get_game_state));
    params.add(new BasicNameValuePair("email", email));
    params.add(new BasicNameValuePair("monster", monster));
    params.add(new BasicNameValuePair("qty", Integer.toString(qty)));
    params.add(new BasicNameValuePair("exp", Integer.toString(exp)));

    // getting JSON Object
    JSONObject json = jsonParser.getJSONFromUrl(monsterURL, params);
    // return json
    return json;
}
    //DB HANDLER CLASS
    public void saveLocalGameSate(String monster, int qty, int exp){

    SQLiteDatabase db = this.getWritableDatabase();

    ContentValues values = new ContentValues();
    values.put(KEY_MONSTER, monster); // Do they own it
    values.put(KEY_QTY, qty); // Number of things they own
    values.put(KEY_EXP, exp); // monster's level

    // Inserting Row
    db.insert(USERPROFILES_TABLE, null, values);
    db.close(); // Closing database connection

}

functions.php

public function getGameState($email, $monster, $qty, $exp) {
    $result = mysql_query("SELECT monster AND qty AND exp WHERE email = '$email'");
    $row = mysql_fetch_assoc($result);

        if (mysql_num_rows($result) > 0) {
        $gamestate = array();
        $gamestate["monster"] = $result["monster"];
        $gamestate["qty"] = $result["qty"];
        $gamestate["exp"] = $result["exp"];

        // success
        $response["success"] = 1;

        // user node
        $response["GameState"] = array();

        array_push($response["GameState"], $gamestate);

        // echoing JSON response
        echo ($response);
        } else {
            return false;
        }

下面是我收到的一些错误消息

06-12 14:15:52.046: E/JSON(10577): <b>Warning</b>:  mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in 

    <b>/home/content/40/8529140/html/webapps/monster/include/DB_Functions.php</b> on line <b>145</b><br />
06-12 14:15:52.046: E/JSON(10577): <b>Warning</b>:  mysql_num_rows() expects parameter 1 to be resource, boolean given in <b>/home/content/40/8529140/html/webapps/monster/include/DB_Functions.php</b> on line <b>147</b><br />
06-12 14:15:52.046: E/JSON(10577): {"tag":"get_game_state","success":0,"error":1,"error_msg":"Error getting game state from server"}
06-12 14:15:52.046: E/JSON Parser(10577): Error parsing data org.json.JSONException: Value <br of type java.lang.String cannot be converted to JSONObject

$result=mysql_查询（“选择monster和数量以及exp，其中email='$email'）

---

你能分享一下服务器端（PHP）的“请求”结构吗？你的mysql\u query（）失败了。使用mysql_error（）找出原因。谢谢这就是SQL语句失败的原因。愚蠢的错误。请查看底部的我的更新以及相关错误消息。您是如何解析json响应的？为了不发布整个类，我使用的是同一个类，就像您控制+F“json PARSER class”一样。我使用的是确切的类。错误在于解析不正确。所以请检查解析响应。JSONObject json_user=json.getJSONObject（“用户”）；user是您的json数组名。

06-12 14:15:52.046: E/JSON(10577): <b>Warning</b>:  mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in 

    <b>/home/content/40/8529140/html/webapps/monster/include/DB_Functions.php</b> on line <b>145</b><br />
06-12 14:15:52.046: E/JSON(10577): <b>Warning</b>:  mysql_num_rows() expects parameter 1 to be resource, boolean given in <b>/home/content/40/8529140/html/webapps/monster/include/DB_Functions.php</b> on line <b>147</b><br />
06-12 14:15:52.046: E/JSON(10577): {"tag":"get_game_state","success":0,"error":1,"error_msg":"Error getting game state from server"}
06-12 14:15:52.046: E/JSON Parser(10577): Error parsing data org.json.JSONException: Value <br of type java.lang.String cannot be converted to JSONObject

06-12 15:03:48.216: E/JSON Parser(11427): Error parsing data org.json.JSONException: Value Array of type java.lang.String cannot be converted to JSONObject

You missed table name . It should be as follows :

$result = mysql_query("SELECT monster,qty,exp from table_name WHERE email = '$email'")