顺序如下.假设我要合并a.db和b.db. 在命令行中,我执行以下操作.
它运作良好,但在引用的网站中,提问者询问加速,答案是使用'begin'和'commit'命令.
然后,我想出了以下python代码来完成同样的事情.我用SQLiteDB抽象SQLite函数调用,其中一个方法是runCommand().即使我删除了self.connector.commit(),我也得到了同样的错误.
# run command
def runCommand(self, command):
self.cursor.execute(command)
self.connector.commit() # same error even though I delete this line
db = SQLiteDB('a.db')
cmd = "attach \"%s\" as toMerge" % "b.db"
print cmd
db.runCommand(cmd)
cmd = "begin"
db.runCommand(cmd)
cmd = "insert into benchmark select * from toMerge.benchmark"
db.runCommand(cmd)
cmd = "commit"
db.runCommand(cmd)
cmd = "detach database toMerge"
db.runCommand(cmd)
但是,我收到以下错误.
OperationalError: cannot commit - …我需要从回调函数中获取JSON对象数据,以便我可以在页面中稍后处理它,而不是像我现在那样在回调中处理它.对其他人来说一定是显而易见的,因为我看不到任何关于它的文章.谁能告诉我怎么做？
这是我的代码:
<script type="text/javascript" src="/site_media/js/jstree/_lib/jquery.js"></script>
<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false"></script>
<script type="text/javascript">
function initialize() {
var myLatlng = new google.maps.LatLng(-25.363882,131.044922);
var myOptions = {
zoom: 4,
center: myLatlng,
mapTypeId: google.maps.MapTypeId.ROADMAP,
}
map = new google.maps.Map(document.getElementById("map_canvas"), myOptions);
return map;
}
function add_marker(map, lat, long) {
var marker_image='/site_media/images/map_marker.png';
var myLatlng = new google.maps.LatLng(lat,long);
title='title';
var marker = new google.maps.Marker({
position: myLatlng,
map: map,
title:title,
icon:marker_image
});
//map.panTo(myLatlng);
}
//window.onload=initialize();
setTimeout('map=initialize();',2000);
$.getJSON("/ajax/get", function(data) {
$.each(data, function(i,val) {
latitude = val.fields.latitude;
longitude = val.fields.longitude;
add_marker(map, latitude, …文件指出:
该Q_OBJECT宏必须出现在类的定义,声明自己的信号和槽的私人部分或使用通过Qt的元对象系统提供的其他服务.
但究竟是什么意思呢？我可以安全地省略哪些QObject派生类？如果你在一个QObject派生类忽略Q_OBJECT会出现的问题,然后从一个继承？基本上我想了解更多关于何时可以从我的Qt课程中省略它的信息.
有没有办法获得一个"更漂亮"的例外而不是一个前言__main__MyExceptionTitle？
例:
>>> class BadThings(Exception):
... def __init__(self, msg):
... self.msg = msg
... return
...
>>> class BadThings(Exception):
... def __init__(self, msg):
... self.msg = msg
... return
... def __str__(self):
... return self.msg
...
>>> raise BadThings("This is really, really bad")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
__main__.BadThings: This is really, really bad
我想它只是说:
BadThings:这真的非常糟糕
就好像有一种类型:
>>> raise Exception("This, too, is really, really bad")
Traceback (most recent call last):
File "<stdin>", …我想在Rails 3中与以下字段联系我们:
发布的消息旨在转到我的电子邮件地址,因此我不一定必须将消息存储在数据库中.我必须使用ActionMailer任何宝石或插件吗？
编辑:好的,好的,我误读了.我不是将int与Integer进行比较.正好指出.
我的SCJP书说:
当==用于将基元与包装器进行比较时,包装器将被解包,并且比较将是原始的.
所以你认为这段代码会打印出来true:
Integer i1 = 1; //if this were int it'd be correct and behave as the book says.
Integer i2 = new Integer(1);
System.out.println(i1 == i2);
但它打印出来false.
另外,根据我的书,这应该打印true:
Integer i1 = 1000; //it does print `true` with i1 = 1000, but not i1 = 1, and one of the answers explained why.
Integer i2 = 1000;
System.out.println(i1 != i2);
不.是的false.
是什么赋予了？
http://msdn.microsoft.com/en-us/library/dd988458.aspx
UPD:
那么,让我们来讨论这篇文章:http://msdn.microsoft.com/en-us/library/dd997396.aspx
我已经改变了一点代码:
static void Main()
{
var tokenSource2 = new CancellationTokenSource();
CancellationToken ct = tokenSource2.Token;
var task = Task.Factory.StartNew(() =>
{
// Were we already canceled?
ct.ThrowIfCancellationRequested();
bool moreToDo = true;
Thread.Sleep(5000);
while (moreToDo)
{
// Poll on this property if you have to do
// other cleanup before throwing.
if (ct.IsCancellationRequested)
{
Console.WriteLine("exit");
// Clean up here, then...
ct.ThrowIfCancellationRequested();
}
}
}, tokenSource2.Token); // this parameter useless
Console.WriteLine("sleep");
Thread.Sleep(2000);
Console.WriteLine("cancel");
tokenSource2.Cancel();
// Just …在C++中,是一个const方法返回一个指向非const对象的指针被认为是不好的做法吗？例如.考虑以下方法:
// Makes perfect sense
bool isActive() const { return m_isActive; }
// Makes perfect sense, both consts ensure nothing is modified
const SomeObject* someObject() const { return m_someObject; }
// Makes perfect sense, no const because SomeObject is mutable,
// thus MyClass is indirectly mutable
SomeObject* someObject() { return m_someObject; }
// Makes no sense, if you're just returning a pointer to a const object,
// then the method should clearly be const since nothing can be modified
const …