我是处理QMainWindow的"Core"对象.
Core.h代码
class Core : public QObject
{
Q_OBJECT
public:
explicit Core(QObject *parent = 0);
~Core();
void appInit();
int getAuth();
public slots:
void appExit();
private slots:
void appMenuTriggered(QAction *action);
private:
void preInit();
MainWindow *mwnd;
};
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Core.cpp代码
Core::Core(QObject *parent) : QObject(parent)
{
qDebug() << "Core::Constructor called";
preInit();
}
Core::~Core()
{
delete mwnd;
qDebug() << "Core::Destructor called";
}
int Core::getAuth()
{
LoginDialog *login = new LoginDialog();
int r = login->exec();
delete login;
return r;
}
void Core::appExit() // connected to qapplication aboutToQuit …Run Code Online (Sandbox Code Playgroud) 我编写了下面的代码 - 我希望DataSource是一个单例,并使用枚举习惯用于单例.Data source rejected establishment of connection, message from server: "Too many connections"经过一段时间后我得到了很多- 我的Singleton模式的实现是错误的还是原因在于其他地方?
import java.sql.Connection;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Statement;
import javax.naming.Context;
import javax.naming.InitialContext;
import javax.sql.DataSource;
class DBConnectionPool {
private DataSource ds = null;
private DBConnectionPool() {
try {
Context context = new InitialContext();
Context envctx = (Context) context.lookup("java:comp/env");
ds = (DataSource) envctx.lookup("jdbc/TestDB");
} catch (Exception e) {
e.printStackTrace();
}
}
private static enum PoolSingleton {
POOL_INSTANCE;
private static final DBConnectionPool singleton = new DBConnectionPool(); …Run Code Online (Sandbox Code Playgroud) 这是我用来获取当前位置的代码:
mgr = (LocationManager) getSystemService(LOCATION_SERVICE);
Criteria criteria = new Criteria();
best = mgr.getBestProvider(criteria, true);
if (best == null) {
//ask user to enable atleast one of the Location Providers
} else {
Location location = mgr.getLastKnownLocation(best);
//But sometimes it returns null
}
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几乎每次都有 best = network
但它有时并没有提供这个位置.
mgr.getLastKnownLocation(best) returns null
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也:
onResume() {
mgr.requestLocationUpdates(best, 15000, 10, this);
}
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和
onPause() {
mgr.removeUpdates(this);
}
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这种情况有替代方案吗?
一种选择可能是
List<String> providers = mgr.getAllProviders();
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存储所有提供商并逐步减少1.但从未在任何地方看到过此推荐.此外,要求最好的提供者是文档建议的.
我是openshift世界的新手.我有wordpress安装了openshift应用程序.我已经通过windows中的putty ssh配置并连接到应用程序.现在我不知道如何访问我的应用程序内的文件.帮我.
Welcome to OpenShift shell
This shell will assist you in managing OpenShift applications.
!!! IMPORTANT !!! IMPORTANT !!! IMPORTANT !!!
Shell access is quite powerful and it is possible for you to
accidentally damage your application. Proceed with care!
If worse comes to worst, destroy your application with 'rhc app delete'
and recreate it
!!! IMPORTANT !!! IMPORTANT !!! IMPORTANT !!!
Type "help" for more info.
Note: Your application supports version 2 cartridges.
[xxxx-cloudrun.rhcloud.com xxxxxxxxxxxxxx]\>
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在这里,我想知道如何访问xxxx/wp-content/...
在MainWindow's title 中显示实际时间(Hh:mm:ss)的简单方法是什么?使用插槽和信号技术。
所以我有一些非分阶段的变化和一些上演的变化.我发了
Welcome to Git (version 1.8.3-preview20130601)
$ git stash save --keep-index
Saved working directory and index state WIP on master: ab0d18d Setup of alarms f
or network service + Logging exceptions + long arithmetic
HEAD is now at ab0d18d Setup of alarms for network service + Logging exceptions
+ long arithmetic
$ git stash
Saved working directory and index state WIP on master: ab0d18d Setup of alarms f
or network service + Logging exceptions + long arithmetic
HEAD is now …Run Code Online (Sandbox Code Playgroud) 我在设置setOnItemClickListener时遇到问题.以下是我的代码.我已经测试了setAdapter的工作情况,并且在UI上显示了列表和项目.在设置setOnItemClickListener时,它不起作用.
cool_simpleAdapter = new SimpleAdapter(this, items,
R.layout.mylistitem, new String[] { "title", "link" }, new int[] {
R.id.textView_title, R.id.textView_link });
cool_listView.setAdapter(cool_simpleAdapter);
Log.d("tag_1", "before setOnItemClickListener");
cool_listView.setOnItemClickListener(new OnItemClickListener() {
public void onItemClick(AdapterView<?> parent, View view,
int position, long id) {
Log.d("tag_setonItemClick", "in onItemClick");
Uri uri = Uri.parse("http://www.google.com");
Intent intent = new Intent(Intent.ACTION_VIEW, uri);
startActivity(intent);
}
});
Log.d("tag_2", "after setOnItemClickListener");
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我把日志跟踪发生了什么:
Log.d("tag_1","before setOnItemClickListener");
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和
Log.d("tag_2","after setOnItemClickListener");
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显示但是
Log.d("tag_setonItemClick","in onItemClick");
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没有显示.我无法点击该项目,也无法打开URL.我不知道该如何解决这个问题.
编辑:添加mylistitem.xml布局
<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:orientation="vertical" >
<LinearLayout
android:layout_width="match_parent"
android:layout_height="wrap_content" > …Run Code Online (Sandbox Code Playgroud) android android-intent android-ui android-layout android-listview
运行这个:
import os
if __name__ == '__main__':
exclude = os.path.join(
r"C:\Dropbox\eclipse_workspaces\python\sync\.git", "info", "exclude")
with open(exclude, 'w+') as excl: # 'w' will truncate
# print excl.read() # empty
# excl.readall() # AttributeError: 'file' object has no attribute
# 'readall' -- this also I do not understand
excl.write('This will be written as expected if I comment the
line below')
print "Garbage\n\n", excl.read()
# if I do not comment the line however, the file contains all the garbage
# excl.read() just printed (edit: …Run Code Online (Sandbox Code Playgroud) class _GhostLink(object):
toGhost = lambda filename: False
class _Mod_AllowGhosting_All(_GhostLink):
def _loop(self):
# ...
if self.__class__.toGhost(fileName) != oldGhost:...
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产生:
class _GhostLink(object):
toGhost = lambda filename: False
class _Mod_AllowGhosting_All(_GhostLink):
def _loop(self):
# ...
if self.__class__.toGhost(fileName) != oldGhost:...
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同时传递一个实例,if self.toGhost(fileName) != ...结果如下:
Traceback (most recent call last):
File "bash\basher\mod_links.py", line 592, in Execute
changed = self._loop()
File "bash\basher\mod_links.py", line 587, in _loop
if self.__class__.toGhost(fileName) != oldGhost:
TypeError: unbound method <lambda>() must be called with _Mod_AllowGhosting_All instance as first argument (got Path …Run Code Online (Sandbox Code Playgroud) 我目前正在使用Windows 10,如果该平台是Windows 10,则需要输入一些代码.因此,我检查了python文档并阅读了有关平台模块的内容.这是文档说的:
platform.win32_ver(release='', version='', csd='', ptype=''):从Windows注册表获取其他版本信息,并返回一个元组(版本,版本,csd,ptype),参考操作系统版本,版本号,CSD级别(服务包)和操作系统类型(多/单处理器)
当我在Windows 10机器上尝试相同的功能时,我得到了以下内容:
>>> platform.win32_ver(?
('8', '6.2.9200', '', u'Multiprocessor Free')
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但是,我期待发布的版本是10而不是8.
那么,任何想法,如果我在这里遗漏了什么?
另外,有人可以告诉我是否有任何其他方法来检测Windows平台是否是Windows 10?