java - 将参数发布到 PHP 文件

Question

我有一个代码 fragment 可以帮助我从我的 mySQL 数据库中获取数据。它也可以正常工作!但是现在我想为我的 sql 查询添加参数，因为到目前为止这段代码只适用于此:

mysql_query("SELECT * FROM tablename ORDER BY 5 DESC");

所以一切都在输出。例如我想使用这样的查询:

mysql_query("SELECT * FROM tablename WHERE name LIKE "a%" ORDER BY 5 DESC");

这是我在 Android 中的代码 fragment :

    public class get_data extends AsyncTask<Void, Void, Boolean>
{

    @Override
    protected Boolean doInBackground(Void... params) {

        String result = null;
        InputStream is = null;
        StringBuilder sb = null;

        str_name_list = new ArrayList<String>();
        str_nname_list = new ArrayList<String>();

        try {
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("Here is my Link");
            HttpResponse response = httpclient.execute(httppost);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();
        } catch (Exception e) {
            Log.e("log_tag", "Error in http connection" + e.toString());
        }
        // convert response to string
        try {
            BufferedReader reader = new BufferedReader(new InputStreamReader(
                    is, "iso-8859-1"), 8);
            sb = new StringBuilder();
            sb.append(reader.readLine() + "\n");
            String line = "0";
            while ((line = reader.readLine()) != null) {
                sb.append(line + "\n");
            }
            is.close();
            result = sb.toString();
        } catch (Exception e) {
            Log.e("log_tag", "Error converting result " + e.toString());
        }
        // paring data
        JSONArray jArray;
        try {
            jArray = new JSONArray(result);
            JSONObject json_data = null;

            int anz = jArray.length();
            if (anz > 15)
                anz = 15;


            for (int i = 0; i < anz ; i++) {
                json_data = jArray.getJSONObject(i);
                str_name_list.add(json_data.getString("NAME"));
                str_nname_list.add(json_data.getString("AGE"));
            }
        } catch (JSONException e1) {

        } catch (ParseException e1) {
            e1.printStackTrace();
        }


        return null;
    }
}

我发现我必须在 HttpEntity 中添加它，但真的不知道如何才能不破坏我的代码。提前致谢!

Answer 1

在你的HTTPResponse之前，尝试添加；

 List<NameValuePair> params = new ArrayList<NameValuePair>();
  params.add(new BasicNameValuePair("your value", value));

然后;

 httpPost.setEntity(new UrlEncodedFormEntity(params));

希望这能奏效!