Question

我在我的Android应用程序中有这两个活动，其中第一个是用户输入询问信息的地方（到edittext），另一个是它将发送数据的地方（我用过{ {1}}将数据从第一个活动传输到第二个活动到MySQL数据库，稍后将在putExtra显示结果。问题是，当第二个活动开始时（考虑到我已经在第一个活动上输入了某些内容）并且在进度对话框显示之后，没有显示任何内容，或者结果没有显示。但是当我尝试启动第二个活动（第一个活动中的ListView为空）时，它会显示结果。我不确定导致问题的原因是应用程序还是用于获取数据的PHP文件？

以下是代码：

MainActivity.java

edittext

SearchResultsAcivity.java

//first activity
public class SearchFragment extends Fragment implements View.OnClickListener {

Button butt;
EditText destination;
String d;

public SearchFragment() {
    // Required empty public constructor
}

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
                         Bundle savedInstanceState) {

    // Inflate the layout for this fragment
    View view =  inflater.inflate(R.layout.fragment_search, container, false);

    butt = (Button) view.findViewById(R.id.searchBUTTon);
    butt.setOnClickListener(this);

    destination = (EditText) view.findViewById(R.id.destinationTO);

    return view;
}

@Override
public void onClick(View v) {
    d = destination.getText().toString();
    Intent a = new Intent(getActivity(), SearchResultsActivity.class);
    a.putExtra("to", d);
    startActivity(a);

}
}

RequestHandler.java

//second activity
public class SearchResultsActivity extends AppCompatActivity implements ListView.OnItemClickListener {

private ListView listView;
private String JSON_STRING;

String destination;
TextView d;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_search_results);

    Intent a = getIntent();
    destination = a.getStringExtra("to");

    d = (TextView) findViewById(R.id.textView3);
    d.setText(destination);

    listView = (ListView) findViewById(R.id.listView);
    listView.setOnItemClickListener(this);

    getJSON();

}

private void showBusList() {

    JSONObject jsonObject = null;
    ArrayList<HashMap<String, String>> list = new ArrayList<HashMap<String, String>>();

    try {
        jsonObject = new JSONObject(JSON_STRING);
        JSONArray result = jsonObject.getJSONArray(config.TAG_JSON_ARRAY);

        for (int i = 0; i < result.length(); i++) {
            JSONObject jo = result.getJSONObject(i);

            //get strings
            String id = jo.getString(config.TAG_ID);
            String busName = jo.getString(config.TAG_BUSNAME);
            String terminal = jo.getString(config.TAG_TERMINAL);

            HashMap<String, String> busDetails = new HashMap<>();
            busDetails.put(config.TAG_ID, id);
            busDetails.put(config.TAG_BUSNAME, busName);
            busDetails.put(config.TAG_TERMINAL, terminal);
            list.add(busDetails);
        }
    } catch (JSONException e) {
        e.printStackTrace();
    }

    ListAdapter adapter = new SimpleAdapter(
            SearchResultsActivity.this, list, R.layout.result_list_item, new String[] {
            config.TAG_ID, config.TAG_BUSNAME, config.TAG_TERMINAL}, new int[] {R.id.id, R.id.busName,
            R.id.terminal});
    listView.setAdapter(adapter);


}

private void getJSON() {
    class GetJSON extends AsyncTask<Void, Void, String> {
        ProgressDialog loading;

        @Override
        protected void onPreExecute() {
            super.onPreExecute();

            loading = ProgressDialog.show(SearchResultsActivity.this, "Message", "Fetching data... Please wait.", false, false);

        }

        @Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
            loading.dismiss();
            JSON_STRING = s;
            showBusList();
        }

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

            RequestHandler rh = new RequestHandler();
            String s = rh.sendGetRequestParam(config.URL_SEARCH, destination);

            return s;

        }
    }

    GetJSON gj = new GetJSON();
    gj.execute();
}




@Override
public void onItemClick(AdapterView<?> parent, View view, int position, long id) {

}

}

PHP文件

//handles requests

public String sendGetRequestParam(String requestURL, String id){
    StringBuilder sb =new StringBuilder();
    try {
        URL url = new URL(requestURL+id);
        HttpURLConnection con = (HttpURLConnection) url.openConnection();
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(con.getInputStream()));

        String s;
        while((s=bufferedReader.readLine())!=null){
            sb.append(s+"\n");
        }
    }catch(Exception e){
    }
    return sb.toString();
}

来自logcat的错误

<?php
    $connection = mysqli_connect("mysql.hostinger.ph", "u679871488_bus", "Damnyoufudge20", "u679871488_bus") or die("Error " . mysqli_error($connection));

    $des = $_GET['destination'];

$sql = "SELECT * from appDB WHERE route LIKE  '%".$des."%'";
$result = mysqli_query($connection, $sql) or die ("Error in Selecting " . mysqli_error($connection));

$thisArray = array();
while($row = mysqli_fetch_assoc($result)) {
    $thisArray[] = $row;
}

echo json_encode(array('busDetails' => $thisArray));

Answer 1

在PHP代码中使用php中的array_push方法

while($row = mysqli_fetch_assoc($result)) {
    //$thisArray[] = $row;
    array_push($thisArray, $row);
}

我认为它会起作用。尝试一次并检查是否在android中获取此数据。

将用户输入直接附加到php中的sql查询是不好的做法，就像使用'%".$des."%'一样。它会导致SQL Injection Attacks。

始终倾向于mysqli_prepare($sql)以避免SQL Injection Attacks。

更新1

在SearchResultsActivity.java中，尝试替换

destination = a.getStringExtra("to");

这一行

destination = a.getExtras().getString("to");
Log.e("tag", " DESTINATION :: " + destination);

并检查您是否从第一项活动中获取了该文字。

更新2

永远不要忘记你投掷的catch例外情况。

您忘记在sendGetRequestParam方法中捕获异常。抓住并打印出来。因此，您将知道连接到服务器是否有任何错误。

从MySQL数据库解析不在Activity中显示

1 个答案: