【发布时间】：2016-02-04 08:56:50
【问题描述】：
我可以选择并显示数据库中的所有图像。现在我正在尝试按 id 过滤它们。如何通过 id 从 db 中获取图片？
有没有办法将 AsyncTask 中的字符串“imovel_id”与从数据库中选择的“imovel_id”json 响应进行比较？
或者通过android应用程序发送的请求获取字符串“imovel_id”，然后通过“imovel_id”选择所有图像更好？
我在下面的代码中尝试了从 AsyncTask 发送字符串，在 php 请求中获取它，然后使用带有 WHERE clausule 的查询，但我没有成功。我得到一个空值。
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 脚本
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);
在 json 数组中添加 id，然后在 parseResult 方法中将其与 textview 中的字符串“imovel_id”进行比较可能是一种解决方案吗？
我认为通过 nameValuePairs 发送字符串会很容易，就像我已经完成的那样，并在查询中使用 WHERE 进行过滤，但似乎并不那么容易。
提前致谢。
【问题讨论】：
标签： php android json android-asynctask