[英]Android, connecting to MySQL using PHP
我正在嘗試使用php腳本連接到MySQL數據庫。 但是我沒有任何輸出,只有異常代碼。 我不知道問題出在哪里。 我使用了教程代碼。
private EditText outputStream;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
String result = null;
InputStream input = null;
StringBuilder sbuilder = null;
outputStream = (EditText)findViewById(R.id.output);
ArrayList <NameValuePair> nameValuePairs = new ArrayList <NameValuePair>();
try{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://ik.su.lt/~jbarzelis/Bandymas/index.php");
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
input = entity.getContent();
}
catch(Exception e){
Log.e("log_tag","Error in internet connection"+e.toString());
}
try{
BufferedReader reader = new BufferedReader(new InputStreamReader(input,"iso-8859-1"),8);
sbuilder = new StringBuilder();
String line = null;
while((line = reader.readLine()) != null){
sbuilder.append(line + "\n");
System.out.println(line);
}
input.close();
result = sbuilder.toString();
}
catch(Exception e){
Log.e("log_tag", "Error converting result "+e.toString());
}
int fd_id;
String fd_name;
try{
JSONArray jArray = new JSONArray(result);
JSONObject json_data = null;
for(int i=0;i<jArray.length();i++){
json_data = jArray.getJSONObject(i);
fd_id = json_data.getInt("FOOD_ID");
fd_name = json_data.getString("FOOD_NAME");
outputStream.append(fd_id +" " + fd_name + "\n");
}
}
catch(JSONException e1){
Toast.makeText(getBaseContext(), "No food found", Toast.LENGTH_LONG).show();
}
catch(ParseException e1){
e1.printStackTrace();
}
}
PHP腳本:
<?php
mysql_connect("localhost","**********","******");
mysql_select_db("test");
$sql = mysql_query("select FOOD_NAME as 'Maistas' from FOOD where FOOD_NAME like 'A%'");
while($row = mysql_fetch_assoc($sql)) $output[]=$row;
print(json_encode($output));
mysql_close;
?>
任何想法如何解決?
首先,不要使用Exception.toString(),請使用Exception.printStackTrace():
catch (Exception e) {
e.printStackTrace();
}
其次,在您的PHP代碼中,您不檢查任何錯誤。 如果發生任何錯誤,建議您在Android代碼中發出不同的HTTP狀態代碼(例如400):
if (response.getStatusLine().getStatusCode() != 200) {
Log.d("MyApp", "Server encountered an error.);
}
這樣,您將知道服務器上是否發生了某些事情。
希望這可以幫助
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.