【发布时间】：2014-05-10 12:02:49
【问题描述】：
作为序言，我想说我对 Android 应用程序开发领域还比较陌生。
我一直在开发一个涉及登录/注册系统的 Android 应用程序，该系统使用 JSON 到 PHP 脚本和 MySQL 从 Android 应用程序连接。登录/注册功能正常，但我正在尝试进行一项活动，该活动能够搜索注册用户并在同一 xml 布局上的 ListView 中显示结果。我的搜索功能 PHP 脚本正常运行。
我几乎在网上到处找了相关结果。如果有人可以帮助我或指出正确的方向，将不胜感激。
提前致谢。
编辑： 这是我正在使用的 JSON Parser 脚本 --
公共类 JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
// constructor
public JSONParser() {
}
public JSONObject getJSONFromUrl(final String url) {
// Making HTTP request
try {
// Construct the client and the HTTP request.
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
// Execute the POST request and store the response locally.
HttpResponse httpResponse = httpClient.execute(httpPost);
// Extract data from the response.
HttpEntity httpEntity = httpResponse.getEntity();
// Open an inputStream with the data content.
is = httpEntity.getContent();
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
// Create a BufferedReader to parse through the inputStream.
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
// Declare a string builder to help with the parsing.
StringBuilder sb = new StringBuilder();
// Declare a string to store the JSON object data in string form.
String line = null;
// Build the string until null.
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
// Close the input stream.
is.close();
// Convert the string builder data to an actual string.
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// Try to parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// Return the JSON Object.
return jObj;
}
// function get json from url
// by making HTTP POST or GET mehtod
public JSONObject makeHttpRequest(String url, String method,
List<NameValuePair> params) {
// Making HTTP request
try {
// check for request method
if(method == "POST"){
// request method is POST
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
httpPost.setEntity(new UrlEncodedFormEntity(params));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}else if(method == "GET"){
// request method is GET
DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
}
【问题讨论】：
-
也许this post 可以提供帮助。
-
感谢您的快速回复。我去看看帖子。
-
我忘了说搜索是通过 AsyncTask 完成的，搜索结果显示在同一个活动的 ListView 中。
标签： php android mysql listview search