Question

我这里有一段Java代码，它应该从数据库查询中检索结果，并且ResultSet应该迭代这些值，以便为ResultSet的每个条目检索某些API数据。但是，问题是我只能检索ResultSet的第一个条目的API数据。

此代码完全按预期工作，并返回所有数据库条目。

try {
    ResultSet rs;
    rs = stat.executeQuery("select * from schedule");
    while (rs.next()) {
        model.addRow(new Object[]{rs.getString("SHOW"), rs.getString("SEASON")});
    }
} catch (Exception e) {
    console.append(e.getMessage() + '\n');
}

但是，此代码仅返回第一个条目。

try {
    ResultSet rs = stat.executeQuery("select * from schedule");
    while (rs.next()) {
        String show = rs.getString("SHOW");
        String season = rs.getString("SEASON");
        String api_url = "<API_URL>/" + show + "/" + season;

     URL url = new URL(api_url);
        HttpURLConnection con = (HttpURLConnection) url.openConnection();
        con.setRequestMethod("GET");
        con.setRequestProperty("User-Agent", USER_AGENT);
        int responseCode = con.getResponseCode();
        if (responseCode == 200) {
            conn_stat.setText("Connection Status : OK");
        } else {
            conn_stat.setText("Connection Status : ERR");
        }
        BufferedReader in = new BufferedReader(
                new InputStreamReader(con.getInputStream()));
        String inputLine;
        StringBuilder response = new StringBuilder();
        while ((inputLine = in.readLine()) != null) {
            response.append(inputLine);
        }
        in.close();
        String s = response.toString();
        JsonArray json = JsonArray.readFrom(s);
        for (int i = 0; i < json.size(); i++) {
            JsonObject show_json = json.get(i).asObject();
            int episode = show_json.get("episode").asInt();
            String date = show_json.get("first_aired_iso").asString();
            String title = show_json.get("title").asString();
            String date_formatted = date.substring(0, date.indexOf("T"));
            SimpleDateFormat original = new SimpleDateFormat("yyyy-MM-dd");
            SimpleDateFormat target = new SimpleDateFormat("dd-MMM-yyyy");
            Date unformatteddate = original.parse(date_formatted);
            String dateStart = target.format(unformatteddate);
            Date curr_date = new Date();
            String dateStop = target.format(curr_date);
            Date d1 = null;
            Date d2 = null;
            d1 = target.parse(dateStart);
            d2 = target.parse(dateStop);
            long diff = d2.getTime() - d1.getTime();
            long diffDays = diff / (24 * 60 * 60 * 1000);
            if (diffDays < 0) {
               alert_model.addRow(new Object[]{show + " - " + episode, title, dateStart});
            }
        }
    }
} catch (Exception e) {
    console.append(e.getMessage() + '\n');
}

Answer 1

尝试分解你的逻辑。这是您从数据库创建所有节目/季节的列表的方法。

    try {
        ResultSet rs = stat.executeQuery("select * from schedule");
        List<String[]> list = new ArrayList<>();
        while (rs.next()) {
            String show = rs.getString("SHOW");
            String season = rs.getString("SEASON");
            list.add(new String[]{show, season});
        }
    } 
    catch (Exception e) {
        e.printStackTrace();
    }

在结果集（可能还有连接）关闭之后，您应该使用此列表构造ULR并查询外部服务。

循环访问数据库ResultSet

1 个答案: