我正在构建一个 android 游戏,我试图将用户的配置文件从 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。
这是相关的android方法,它们应该没有问题,尽管逻辑本身可能已关闭,但它们可以正常运行。
//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
}
这是PHP,我相信我的错误在于这两个文件之一。
索引.php
else if ($tag == 'get_game_state') {
$email = $_POST['email'];
$monster = $_GET['monster'];
$qty = $_GET['qty'];
$exp = $_GET['exp'];
$result = $db->getGameState($email, $monster, $qty, $exp);
if (!empty($result)) {
echo json_encode($response);
} else {
$response["error"] = 1;
$response["error_msg"] = "Error getting game state from server";
echo json_encode($response);
}
函数.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;
}
这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
最后,请不要告诉我我的 php 不安全,我知道这个PHP 不是我的强项,我正在寻找一个测试版本并运行。PDO在我的开发路线图上并将实施。
谢谢。
更新:
修复 SQL 语句以从正确的表中选择后
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