我有一个应用程序,在listview中显示数据,但我希望第一行通过不同的布局膨胀,我这样做,但因为有大量的listitems,我想优化listview,并有问题.当我在那个方式填充listview时,我无法优化listview,那么我怎样才能把内容应该放在listview标题内的第一行中,通过一些布局来膨胀,这里是Adapter的代码
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
requestWindowFeature(Window.FEATURE_NO_TITLE);
getWindow().setFlags(WindowManager.LayoutParams.FLAG_FULLSCREEN, WindowManager.LayoutParams.FLAG_FULLSCREEN);
setContentView(R.layout.main);
LinearLayout content = (LinearLayout) findViewById(R.id.content);
LinearLayout refLayout = (LinearLayout) findViewById(R.id.refLayout);
refLayout.setVisibility(View.GONE);
mBtnNaslovnica = (Button) findViewById(R.id.mBtnNaslovnica);
mBtnNaslovnica.setSelected(true);
TextView txtView=(TextView) findViewById(R.id.scroller);
txtView.setSelected(true);
loadPage();
ImageButton mBtnRefresh = (ImageButton) findViewById(R.id.btnRefresh);
mBtnRefresh.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
// TODO Auto-generated method stub
new LoadingTask().execute(URL);
}
});
}
public void loadPage(){
ArrayList<HashMap<String, String>> homeList = new ArrayList<HashMap<String, String>>();
JSONObject jsonobj;
try {
jsonobj = new JSONObject(getIntent().getStringExtra("json"));
JSONObject datajson = jsonobj.getJSONObject("data");
JSONArray news …嗨iam begginer在c ++我有静态方法的类,我无法访问它,它会引发我一个错误
1>------ Build started: Project: CPractice, Configuration: Debug Win32 ------
1> Source.cpp
1>Source.obj : error LNK2001: unresolved external symbol "private: static class std::basic_string<char,struct std::char_traits<char>,class std::allocator<char> > CPractice::name" (?name@CPractice@@0V?$basic_string@DU?$char_traits@D@std@@V?$allocator@D@2@@std@@A)
1>c:\users\innersoft\documents\visual studio 2012\Projects\CPractice\Debug\CPractice.exe : fatal error LNK1120: 1 unresolved externals
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========
这是我的代码
#include <iostream>
#include <stdio.h>
#include <cstdlib>
#include <string>
using namespace std;
class CPractice
{
public:
static void setName(string s)
{
name = s;
}
static string getName()
{ …我有一个应用程序,需要使用AsyncTask从URL下载JSON,并在onPostExecute()上使用putExtra方法将该JSON对象传递给下一个Activity,但我不断收到错误,其中显示"putExtra只能用于传递字符串和不是对象"？如何解决这个问题？
我有PHP生成的多维数组和数据库中的数据,但我有像"čćšđž"这样的字符,当我尝试在json中输出它时他只返回null,我做了一些关于它的阅读,它说JSON仅适用于UTF-8.那么如何在UTF-8中转换这些数组呢？但是我仍然需要数组和？
这是我的脚本代码
$sql_main = mysql_connect(DB_HOST, DB_UNM, DB_PSW);
($sql_main)? mysql_select_db(DB_NM) : mysql_error();
$APP_URL_ACCESS = $_GET['app_access_key'];
$sql_app = mysql_query("SELECT * FROM app_sys WHERE APP_OW_C='$APP_URL_ACCESS'") or die(mysql_error());
if(mysql_num_rows($sql_app)==1){
while($row = mysql_fetch_array($sql_app)){
$APP_UA_ID = $row['APP_UA_ID'];
$APP_NM = $row['APP_NM'];
$APP_H_DMN = $row['APP_H_DMN'];
$APP_H = $row['APP_H'];
$APP_H_DB_UNM = $row['APP_H_DB_UNM'];
$APP_DB_NM = $row['APP_DB_NM'];
$APP_H_DB_PSW = $row['APP_H_DB_PSW'];
$APP_H_DB_SRV = $row['APP_H_DB_SRV'];
$APP_ACTIVE = $row['APP_ACTIVE'];
$APP_OW_C = $row['APP_OW_C'];
}
$ROW_APP[] = array(
'APP_UA_ID' => $APP_UA_ID,
'APP_PERMISSION' => $APP_ACTIVE,
'APP_KEY' => $APP_OW_C);
$APP_ARRAY[''] = $ROW_APP;
($APP_ACTIVE == '1')? $sql_connect_app = mysql_connect($APP_H_DB_SRV, …我有网页女巫包含JSON代码与数据库中的数据,代码是由PHP脚本生成的,但JSON代替显示那些字母"čšćđž"显示类似"\ u0161","\ u0111"等,但我不能图它出错了,编码表,数据库,文档,标题是UTF-8
这是http://dmb.site50.net/application.php?app_access_key=c4ca4238a0b923820dcc509a6f75849b页面的链接
这是PHP脚本的来源
$APP_URL_ACCESS = $_GET['app_access_key'];
$sql_app = mysql_query("SELECT * FROM app_sys WHERE APP_OW_C='$APP_URL_ACCESS'") or die(mysql_error());
if(mysql_num_rows($sql_app)==1){
while($row = mysql_fetch_array($sql_app)){
$APP_UA_ID = $row['APP_UA_ID'];
$APP_NM = $row['APP_NM'];
$APP_H_DMN = $row['APP_H_DMN'];
$APP_H = $row['APP_H'];
$APP_H_DB_UNM = $row['APP_H_DB_UNM'];
$APP_DB_NM = $row['APP_DB_NM'];
$APP_H_DB_PSW = $row['APP_H_DB_PSW'];
$APP_H_DB_SRV = $row['APP_H_DB_SRV'];
$APP_ACTIVE = $row['APP_ACTIVE'];
$APP_OW_C = $row['APP_OW_C'];
}
$ROW_APP[] = array(
'APP_UA_ID' => $APP_UA_ID,
'APP_PERMISSION' => $APP_ACTIVE,
'APP_KEY' => $APP_OW_C);
$APP_ARRAY[] = $ROW_APP;
($APP_ACTIVE == '1')? $sql_connect_app = mysql_connect($APP_H_DB_SRV, $APP_H_DB_UNM, $APP_H_DB_PSW) && mysql_select_db($APP_DB_NM): …