Question

我正在尝试将一个字符串（例如＆＃34; name＆＃34;）传递给我的PHP脚本，该脚本需要获取＆＃34; name＆＃34;来自Android应用程序的$ _POST，然后在SELECT查询中使用它并将其传回Android应用程序。

我已经完成了所有工作，除了我只能将字符串传递给PHP脚本，但我无法在SELECT查询中使用它。这是我的代码：

PHP代码：

<?php
include "../dbc.php";
//include "get_id.php";

$value = $_POST["value"];
$response = array();

$q=$dbh->prepare("SELECT * FROM users WHERE user_name = :value");
$q->bindParam(':value', $value);
$q->execute();
while ($row = $q->fetch(PDO::FETCH_ASSOC))
{
    $first_name=$row['first_name'];
    $last_name=$row['last_name'];
    $company_name=$row['company_name'];
    $loc=$row['loc'];
    $tel=$row['tel'];
    $website=$row['website'];
    $aboutme=$row['aboutme'];
    array_push($response, array("first_name"=>$first_name,"last_name"=>$last_name,"company_name"=>$company_name,"loc"=>$loc,"tel"=>$tel,"website"=>$website,"aboutme"=>$aboutme));
}

echo json_encode(array("server_response"=>$response));
?>

Android代码：

class GetUserSettings extends AsyncTask<Void,Void,String> {
    String json_get_settings_url;
    @Override
    protected void onPreExecute() {
        json_get_settings_url = "http://url.com/json.php";
    }

    @Override
    protected void onProgressUpdate(Void... values) {
        super.onProgressUpdate(values);
    }

    @Override
    protected String doInBackground(Void... Voids) {
        URL url = null;
        try {
            url = new URL(json_get_settings_url);
        } catch (MalformedURLException e) {
            e.printStackTrace();
        }
        try {
            HttpURLConnection httpURLConnection = (HttpURLConnection)url.openConnection();
            InputStream inputStream = httpURLConnection.getInputStream();
            BufferedReader bufferedReader = new BufferedReader (new InputStreamReader(inputStream));
            StringBuilder stringBuilder = new StringBuilder();
            while((JSON_STRING = bufferedReader.readLine()) != null) {
                stringBuilder.append(JSON_STRING+"\n");
            }
            bufferedReader.close();
            inputStream.close();
            httpURLConnection.disconnect();
            return stringBuilder.toString().trim();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }
    @Override
    protected void onPostExecute(String JSON_STRING) {
        SharedPreferences settings = getSharedPreferences(PREFS_NAME, MODE_PRIVATE);
        String value = settings.getString("user_name", "");
        //this is where it starts parsing the JSON from the URL
        try {
            JSONObject jsonResponse = new JSONObject(JSON_STRING);
            JSONArray jsonMainNode = jsonResponse.optJSONArray("server_response");

            for (int i = 0; i < jsonMainNode.length(); i++) {
                JSONObject jsonChildNode = jsonMainNode.getJSONObject(i);
                user_name = jsonChildNode.optString(value);
                et_user_name.setText(value);
                first_name = jsonChildNode.optString("first_name");
                et_first_name.setText(first_name);
                last_name = jsonChildNode.optString("last_name");
                et_last_name.setText(last_name);
                company_name = jsonChildNode.optString("company_name");
                et_company_name.setText(company_name);
                loc = jsonChildNode.optString("loc");
                et_loc.setText(loc);
                tel = jsonChildNode.optString("tel");
                et_tel.setText(tel);
                website = jsonChildNode.optString("website");
                et_website.setText(website);
                aboutme = jsonChildNode.optString("aboutme");
                et_aboutme.setText(aboutme);
            }
        } catch (JSONException e) {
            e.printStackTrace();
            Toast.makeText(getApplicationContext(),"Something is wrong. Please restart the app or contact us.",Toast.LENGTH_LONG).show();
        }
    }
}

所以从代码中可以看出，我正在尝试传递String＆＃34;值＆＃34;到我的PHP代码，然后分配＆＃34;值＆＃34; $ value然后在SELECT查询中使用$ value但我怎么能传递String＆＃34;值＆＃34;然后执行查询，以便我可以使用JSON来解析数据？

Answer 1

您可以使用ContentValues（）将字符串传递给php amd返回select查询。

ContentValues values = new Content Values（）; values.put（＆＃34;串＆＃34;，变量）;

你可以传递网址

http://example.com/dir/page.php?var=variable

在php端

$ var = $ _GET [＆＃39;变量＆＃39;]：现在，您可以在select查询中传递此$ var。

希望这有帮助。

将字符串传递给PHP文件以查询SELECT语句

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