【发布时间】:2012-03-02 00:27:46
【问题描述】:
我在实现一个检查 MySQL 数据库的值然后根据值的存在返回“True”或“False”的程序时遇到了一个小问题。我可以让 PHP 使用以下代码返回值(在此示例中以“A”开头)(它还在 Java 日志中显示匹配的行):
<?php
mysql_connect("host","username","password");
mysql_select_db("Deal");
$sql=mysql_query("select * from CITY where CITY_NAME like 'A%'");
while($row=mysql_fetch_assoc($sql))
$output[]=$row;
print(json_encode($output));
mysql_close();
?>
我只是想检查是否有一条记录满足查询(在本例中是一个以“A”开头的城市),然后返回“True”或“False”并将其打印在 ddms 日志中.
我一直在尝试实现以下内容,但我没有返回任何内容。
<?php
mysql_connect("host","username","password");
mysql_select_db("Deal");
$sql = mysql_query("select * from CITY where CITY_NAME like 'A%'") or die(mysql_error());
if ($sql) {
if(mysql_num_rows($sql) == 0) {
$row = "False";
print(json_encode($row);
mysql_close();
}
else {
$row = "True";
//while($row = mysql_fetch_assoc($sql))
//$output[]=$row;
print(json_encode($row);
//print(json_encode($output));
mysql_close();
}
}
?>
这是我的 Android 代码:
public class CityActivity extends ListActivity {
JSONArray jArray;
String result = null;
InputStream is = null;
StringBuilder sb=null;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
//http post
try{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://www.example.com/example.php");
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
is = entity.getContent();
}catch(Exception e){
Log.e("log_tag", "Error in http connection"+e.toString());
}
//convert response to string
try{
BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
sb = new StringBuilder();
sb.append(reader.readLine() + "\n");
String line="0";
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
result=sb.toString();
}catch(Exception e){
Log.e("log_tag", "Error converting result "+e.toString());
}
//paring datag
try{
jArray = new JSONArray(result);
JSONObject json_data=null;
for(int i=0;i<jArray.length();i++){
String myString ="";
json_data = jArray.getJSONObject(i);
int ct_id = json_data.getInt("CITY_ID");
String ct_name = json_data.getString("CITY_NAME");
myString = Integer.toString(ct_id);
Log.i(ct_name, myString);
}
}
catch(JSONException e1){
Toast.makeText(getBaseContext(), "No City Found" ,Toast.LENGTH_LONG).show();
} catch (ParseException e1) {
e1.printStackTrace();
}
}
}
如果您有任何建议,请告诉我。我真的很感激。谢谢!
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标签: java php android mysql httpresponse