我正在尝试从包含Facebook ID的字符串中获取“长”值,这是 10位数字。这是我的功能:
@Override
public void onItemClick(AdapterView<?> arg0, View v, int position, long arg3) {
try {
final long opponentId;
opponentId = jsonArray.getJSONObject(position).getLong("uid");
String opponentName = jsonArray.getJSONObject(position).getString("name");
Toast.makeText(this, "UID: " + opponentId + "\n" + "Name: " + opponentName, Toast.LENGTH_SHORT).show();
SharedPreferences myPrefs = getSharedPreferences("user", MODE_WORLD_READABLE);
String playerUidString = myPrefs.getString("uid", null);
if (playerUidString == null) {
Log.d(GlobalVars.TAG, "Empty UID");
return;
}
Log.d(GlobalVars.TAG, "ID: " + playerUidString);
Log.d(GlobalVars.TAG, "ID Long: " + String.valueOf(Long.getLong(playerUidString)));
// long player_id = Long.getLong(playerUidString);
// new GameRequest(getBaseContext(), player_id, opponentId);
} catch (JSONException e) {
Log.d(GlobalVars.TAG, e.getMessage());
}
}
通常,我会设置long player_id等于playerUidString到long的转换,但是Long.getLong(playerUidString)会导致null,而playUidString则等于10位ID ID。出现这种情况的原因是什么?
编辑:我的Log.d输出:
06-29 11:06:37.823: D/I See It(16053): ID: 1089490706
06-29 11:06:37.823: D/I See It(16053): ID Long: null
答案 0 :(得分:2)
您错过了Long.getLong(String str)
不应将String解析为long的事实,而是返回由该字符串表示的系统属性的长值。
正如其他人所建议的那样,您实际需要的是Long.parseLong(String str)
答案 1 :(得分:1)
你必须使用Long.parseLong(yourstring)
我希望我能帮助你。