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作爲前言，我想說我對Android應用程序開發場景比較陌生。Android PHP MySQL搜索結果到ListView
我一直在研究一個Android應用程序，它涉及一個登錄/註冊系統，它使用JSON連接到使用JSON的PHP應用程序和MySQL腳本。登錄/註冊功能，但我試圖有一個活動，能夠搜索註冊用戶，並顯示在同一個xml佈局的ListView結果。我的PHP腳本用於搜索功能。
我幾乎看到網上無處不在，無濟於事的相關結果。如果有人能夠幫助我或指引我正確的方向，將不勝感激。
在此先感謝。
編輯： 這是我使用JSON解析器腳本 -
公共類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;
}
}
也許[此帖]（http://stackoverflow.com/questions/5530672/connecting-android-apps-to-mysql-database）可以提供幫助。 – Joel
感謝您的快速回復。我會看看這篇文章。 – cmk0803
我忘了提及搜索是通過AsyncTask完成的，搜索結果顯示在同一活動的ListView中。 – cmk0803