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我可以從db中選擇並顯示所有圖像。現在我打算通過id過濾它們。我如何通過ID從數據庫獲取圖像？Android + json通過編號獲取圖像
有沒有辦法從數據庫中選擇了「imovel_id」 JSON響應比較的AsyncTask裏面的字符串「imovel_id」？
或者是更好地得到通過請求字符串「imovel_id」從Android應用程式傳送，然後選擇通過「imovel_id」的所有圖像？
我想，在下面的代碼從發送的AsyncTask的字符串，得到它在PHP請求，然後使用具有WHERE clausule，但我沒有成功的查詢。我得到一個空值。
public class AsyncHttpTask extends AsyncTask<String, Void, Integer> {
@Override
protected Integer doInBackground(String... params) {
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("imovel_id", i_id));
try {
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://meuwebsite.com/panel/json_images.php");
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
inputStream = entity.getContent();
Log.e("pass 1", "connection success ");
} catch (Exception e) {
Log.e("Fail 1", e.toString());
Toast.makeText(getApplicationContext(), "Invalid IP Address",
Toast.LENGTH_LONG).show();
}
try {
BufferedReader reader = new BufferedReader
(new InputStreamReader(inputStream, "UTF-8"));
StringBuilder sb = new StringBuilder();
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
inputStream.close();
result = sb.toString();
Log.e("pass 2", "connection success ");
} catch (Exception e) {
Log.e("Fail 2", e.toString());
}
try {
JSONObject json_data = new JSONObject(result);
i_id = (json_data.getString("imovel_id"));
Log.e("pass 1", "id do imovel = " + i_id);
} catch (Exception e) {
Log.e("Fail 3", e.toString());
}
Integer result = 0;
try {
// Create Apache HttpClient
HttpClient httpclient = new DefaultHttpClient();
HttpResponse httpResponse = httpclient.execute(new HttpGet(params[0]));
int statusCode = httpResponse.getStatusLine().getStatusCode();
// 200 represents HTTP OK
if (statusCode == 200) {
String response = streamToString(httpResponse.getEntity().getContent());
parseResult(response);
result = 1; // Successful
} else {
result = 0; //"Failed
}
} catch (Exception e) {
Log.d(TAG, e.getLocalizedMessage());
}
return result;
}
@Override
protected void onPostExecute(Integer result) {
// Download complete. Lets update UI
if (result == 1) {
mGridAdapter.setGridData(mGridData);
} else {
Toast.makeText(GridViewActivity.this, "Failed to fetch data!", Toast.LENGTH_SHORT).show();
}
//Hide progressbar
mProgressBar.setVisibility(View.GONE);
}
}
String streamToString(InputStream stream) throws IOException {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(stream));
String line;
String result = "";
while ((line = bufferedReader.readLine()) != null) {
result += line;
}
// Close stream
if (null != stream) {
stream.close();
}
return result;
}
/**
* Parsing the feed results and get the list
*
* @param result
*/
private void parseResult(String result) {
try {
JSONObject response = new JSONObject(result);
JSONArray posts = response.optJSONArray("posts");
GridItem item;
for (int i = 0; i < posts.length(); i++) {
JSONObject post = posts.optJSONObject(i);
item = new GridItem();
item.setImage(post.getString("images"));
mGridData.add(item);
}
} catch (JSONException e) {
e.printStackTrace();
}
}
PHP腳本
include_once 'db_connect.php';
$i_id = $_REQUEST["imovel_id"];
$sql = "SELECT * FROM iMoveis WHERE imovel_id='$i_id'";
$result = mysqli_query($mysqli, $sql);
$response = array();
$images = array();
while($row = mysqli_fetch_assoc($result)){
$images[] = array('images' => $row['img1']);
$images[] = array('images' => $row['img2']);
$images[] = array('images' => $row['img3']);
$images[] = array('images' => $row['img4']);
$images[] = array('images' => $row['img5']);
}
$response['posts'] = $images;
echo json_encode($response, JSON_UNESCAPED_SLASHES);
添加JSON數組內的ID，然後將其從所述的TextView比較parseResult方法與字符串「imovel_id」內可能是一個解決方案？
我想，我已經做了，並在查詢過濾器中使用WHERE它會很容易通過namevaluepairs中發送的字符串，但它似乎沒有that's工作那麼容易。
在此先感謝。