php - Android应用程序解码来自PHP MySql数据库的JSON响应，没有特殊字符

问题描述

我正在尝试从 mydatabase中检索一些数据android app，但某些特殊字符发送不正确。数据实际上首先从我的应用程序发送，存储，然后检索。当我发送它时，数据以正确的方式注册，但在检索时似乎存在编码/解码问题。为了正确存储数据，我使用mysqli_set_charset($conn, 'utf8');了命令，因此数据Situație 4在发送时存储。发送数据的 PHP 代码：

$stmt->bind_result($id, $latitudine, $longitudine, $tip_problema, $data_ora, $ora_catalogare, $validare, $grad_incredere);
                while($stmt->fetch()) {
                    $locatie[] = array (
                            "id_alerta"=>$id,
                            "latitudine"=>$latitudine, 
                            "longitudine"=>$longitudine, 
                            "tip_problema"=>$tip_problema,
                            "data_ora"=>$data_ora,
                            "ora_catalogare"=>$ora_catalogare,
                            "stare_problema"=>$validare,
                            "grad_incredere"=>$grad_incredere
                            );
            $response['error'] = false; 
            $response['message'] = 'Alerte reimprospatate';
            $response['locatie']= $locatie;
                }

我需要的信息存储在 $locatie 中，所以我可以在我的应用程序上获得 JSON 响应。

输入 JSON 响应的应用代码：

@Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
            //hiding the progressbar after completion

            try {
                //converting response to json object
                JSONObject obj = new JSONObject(s);

                //if no error in response
                if (!obj.getBoolean("error")) {
                    Toast.makeText(getApplicationContext(), obj.getString("message"), Toast.LENGTH_SHORT).show();

                    JSONArray locatieArray = obj.getJSONArray("locatie");
                    for (int i = 0; i < locatieArray.length(); i++) {
                        JSONObject locatie = locatieArray.getJSONObject(i);
                        // check latitudine and longitudine is not null and if not null then cast these values and call the addMarker() method.
                        if (!locatie.isNull("latitudine") && !locatie.isNull("longitudine")) {
                            latitudine_sql = Double.valueOf(locatie.getString("latitudine"));
                            longitudine_sql = Double.valueOf(locatie.getString("longitudine"));
                            tip_problema_sql = locatie.getString("tip_problema");
                            id = locatie.getString("id_alerta");
                            data_ora_raportare = locatie.getString("data_ora");
                            stare_alarma = locatie.getString("stare_problema");
                            data_ora_ack = locatie.getString("ora_catalogare");
                            grad_incredere = locatie.getString("grad_incredere");
                            addMarker(latitudine_sql, longitudine_sql);

                        }
                    }

                } else {
                    Toast.makeText(getApplicationContext(), "Some error occurred", Toast.LENGTH_SHORT).show();
                }
            } catch (JSONException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    }

棘手的部分来了。json 响应如下所示：

{
"error": false,
"message": "Alerte reimprospatate",
"locatie": [{
    "id_alerta": 43,
    "latitudine": "37.41593630609526",
    "longitudine": "-122.09065474569798",
    "tip_problema": "Situa\u021bie 4",
    "data_ora": "2020-01-14 15:44:37",
    "ora_catalogare": null,
    "stare_problema": "Nevalidat",
    "grad_incredere": 65.75
}]

}

尽管该tip_problema列已Situație 4存储在数据库中，但当它被发送到应用程序时，它会转换为Situa\u021bie 4.

要建立连接，我使用RequestHandler：

public class RequestHandler {
public String sendPostRequest(String requestURL, HashMap<String, String> postDataParams) {
    URL url;

    StringBuilder sb = new StringBuilder();
    try {
        url = new URL(requestURL);
        HttpURLConnection conn = (HttpURLConnection) url.openConnection();
        conn.setReadTimeout(15000);
        conn.setConnectTimeout(15000);
        conn.setRequestMethod("POST");
        conn.setDoInput(true);
        conn.setDoOutput(true);

        OutputStream os = conn.getOutputStream();

        BufferedWriter writer = new BufferedWriter(
                new OutputStreamWriter(os, "UTF-8"));
        writer.write(getPostDataString(postDataParams));

        writer.flush();
        writer.close();
        os.close();
        int responseCode = conn.getResponseCode();

        if (responseCode == HttpsURLConnection.HTTP_OK) {

            BufferedReader br = new BufferedReader(new InputStreamReader(conn.getInputStream()));
            sb = new StringBuilder();
            String response;

            while ((response = br.readLine()) != null) {
                sb.append(response);
            }
        }

    } catch (Exception e) {
        e.printStackTrace();
    }
    return sb.toString();
}

private String getPostDataString(HashMap<String, String> params) throws UnsupportedEncodingException {
    StringBuilder result = new StringBuilder();
    boolean first = true;
    for (Map.Entry<String, String> entry : params.entrySet()) {
        if (first)
            first = false;
        else
            result.append("&");
        result.append(URLEncoder.encode(entry.getKey(), "UTF-8"));
        result.append("=");
        result.append(URLEncoder.encode(entry.getValue(), "UTF-8"));
    }
    return result.toString();
}

}

我在哪里都设置了UTF-8 collation. 任何建议将不胜感激！更新：我已添加UTF-8到 BufferReader 上，InputStream但响应是相同的。 BufferedReader br = new BufferedReader(new InputStreamReader(conn.getInputStream(),"UTF-8"));

更新：重新安装应用程序后，即使响应相同Situa\u021bie 4，应用程序似乎也能理解消息所指的信息，Situație 4因此它现在可以正常工作。

标签： phpandroidjson

php - Android应用程序解码来自PHP MySql数据库的JSON响应，没有特殊字符

问题描述

解决方案

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