【发布时间】:2012-01-02 09:29:47
【问题描述】:
我想在数据库中有一些数据,我的应用程序从那里检索数据。我是初学者,我不知道如何创建我的数据库,尽管我已经编写了 php 脚本和 java 代码来获得这些结果。所以我需要你的帮助来创建那个数据库。我正在使用 WAMP,并且创建了一个数据库。我的问题是这些调用的论点是什么:
mysql_connect("127.0.0.1","root","xxpasswordxx");
还有这个:
public static final String KEY_121 = "http://xx.xx.xxx.xxx/hellomysql/mysqlcon.php";
我在某处读到,我必须将 10.0.0.2 和我的目录放在存储我的 php 文件的 C:\wamp\www 下。
编辑:添加代码
这是我的安卓代码:
@SuppressWarnings("null")
private void DoMain() {
setContentView(R.layout.main);
// Set the text and call the connect function.
//call the method to run the data retreival
getServerData(KEY_121);
}
public static final String KEY_121 ="http://10.0.2.2:8888/example.php"; //what in here??
private String getServerData(String returnString) {
InputStream is = null;
String result = "";
//the year data to send
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("year","1970"));
//http post
try{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(KEY_121);
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);
StringBuilder sb = new StringBuilder();
String line = null;
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());
}
//parse json data
try{
JSONArray jArray = new JSONArray(result);
for(int i=0;i<jArray.length();i++){
JSONObject json_data = jArray.getJSONObject(i);
Log.i("log_tag","id: "+json_data.getInt("id")+
", name: "+json_data.getString("name")+
", sex: "+json_data.getInt("sex")+
", birthyear: "+json_data.getInt("birthyear")
);
//Get an output to the screen
returnString += "\n\t" + jArray.getJSONObject(i);
}
}catch(JSONException e){
Log.e("log_tag", "Error parsing data "+e.toString());
}
return returnString;
}
这是我的 php 文件
<?php
mysql_connect("127.0.0.1","root","xxpasswordxx"); //what in here??
mysql_select_db("peopledata");
$q=mysql_query("SELECT * FROM people WHERE birthyear>'".$_REQUEST['year']."'");
while($e=mysql_fetch_assoc($q))
$output[]=$e;
print(json_encode($output));
mysql_close();
?>
【问题讨论】:
-
您是否创建了 Web 服务来检索或发送数据?
-
如果你的意思是,用 Java 和 php 文件编写代码,是的,我做到了。我的问题是我不知道参数的值。