我正在使用jQuery UI组件,但有一些问题.如果我尝试做一些简单的事情,如$("#mydiv").draggable()我收到一条错误消息"Microsoft JScript运行时错误:对象不支持此属性或方法".
似乎加载了jQuery UI,因为我在其包含的js文件中放置了一个alert()(请参阅代码)并显示警报.我真的很困惑.
;jQuery.ui || (function($) {
var _remove = $.fn.remove,
isFF2 = $.browser.mozilla && (parseFloat($.browser.version) < 1.9);
alert("jquery.ui.loading"); //
//Helper functions and ui object
$.ui = {
version: "1.7.2",
Run Code Online (Sandbox Code Playgroud) 我有一个'2009/05/13 19:19:30 -0400'形式的日期字符串.对于尾随时区规范,似乎以前版本的Python可能支持strptime中的%z格式标记,但2.6.x似乎已经删除了它.
将此字符串解析为日期时间对象的正确方法是什么?
我的应用是基于景观的.我想在横向上使用MFMailComposeViewController但是找不到任何关于方向的信息.在横向中,MFMailComposeViewController仅在横向显示邮件撰写窗口的顶部,从左侧进入.基本上它覆盖了一半的景观屏幕.有没有办法让邮件撰写窗口以横向而不是纵向显示?
---编辑---我想加载邮件撰写的视图是这样派生的:
//from first UIViewController
[self presentModalViewController:secondView animated:YES];
//from secondView (a UIViewController), I load the info view, which is where the mail compose shows
InfoController *infoController = [[InfoController alloc] initWithNibName:@"Info" bundle:[NSBundle mainBundle]];
[self.view addSubview:infoController.view];
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从上面可以看出,邮件撰写父视图是第三个加载的.在info.plist中,我这样做:
UIInterfaceOrientation = UIInterfaceOrientationLandscapeRight
Run Code Online (Sandbox Code Playgroud) 我想我想使用日历模块中内置的pythons来创建带有数据的HTML日历.我说我想是因为我可能会想到一个更好的方法,但现在它有点个人化了.我不知道这是否打算以这种方式使用,但如果你不能至少把这些日子变成一个,那似乎有点无意义<a hrefs>.
这将设置本月的日历,以星期日为第一天.
import calendar
myCal = calendar.HTMLCalendar(calendar.SUNDAY)
print myCal.formatmonth(2009, 7)
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它打印
<table border="0" cellpadding="0" cellspacing="0" class="month">\n<tr>
<th colspan="7" class="month">July 2009</th></tr>\n<tr><th class="sun">Sun</th>
<th class="mon">Mon</th><th class="tue">Tue</th><th class="wed">Wed</th>
<th class="thu">Thu</th><th class="fri">Fri</th><th class="sat">Sat</th></tr>\n
<tr><td class="noday"> </td><td class="noday"> </td>
<td class="noday"> </td><td class="wed">1</td><td class="thu">2</td><td class="fri">3</td>
<td class="sat">4</td></tr>\n<tr><td class="sun">5</td><td class="mon">6</td><td class="tue">7</td>
<td class="wed">8</td><td class="thu">9</td><td class="fri">10</td>
<td class="sat">11</td></tr>\n<tr><td class="sun">12</td><td class="mon">13</td>
<td class="tue">14</td><td class="wed">15</td><td class="thu">16</td><td class="fri">17</td>
<td class="sat">18</td></tr>\n<tr><td class="sun">19</td><td class="mon">20</td>
<td class="tue">21</td><td class="wed">22</td><td class="thu">23</td><td class="fri">24</td>
<td class="sat">25</td></tr>\n<tr><td class="sun">26</td><td class="mon">27</td>
<td class="tue">28</td><td class="wed">29</td><td class="thu">30</td><td class="fri">31</td>
<td class="noday"> </td></tr>\n</table>\n
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我想在呈现html字符串之前将一些数据插入到HTMLCalendar对象中.我只是无法弄清楚如何.
例如 …
我gprof在OS X上运行时遇到问题.该文件test.c是:
#include <stdio.h>
int main() {
printf("Hello, World!\n");
return 0;
}
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我的终端看起来像:
$ gcc -pg test.c
$ gcc -pg -o test test.c
$ ./test
Hello, World!
$ gprof test
gprof: file: test is not of the host architecture
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编辑:此外,它不会生成文件gmon.out.
这里发生了什么?
这是一个测试yield使用的示例脚本......我做错了吗?它总是返回'1'......
#!/usr/bin/python
def testGen():
for a in [1,2,3,4,5,6,7,8,9,10]:
yield a
w = 0
while w < 10:
print testGen().next()
w += 1
Run Code Online (Sandbox Code Playgroud) 我正在关注Boost入门文章.我已经用Bjam安装了它,我可以看到包含文件和库文件(.a,.so).
#include <boost/regex.hpp>
#include <iostream>
#include <string>
int main()
{
std::string line;
boost::regex pat( "^Subject: (Re: |Aw: )*(.*)" );
}
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如果我使用此命令构建上面的代码
g++ -I./boost/include -L./boost/lib -lboost_regex-gcc43-mt -static -o test_boost2 test_boost2.cc
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我收到此错误:
/tmp/ccJFVVid.o: In function `boost::basic_regex<char, boost::regex_traits<char, boost::cpp_regex_traits<char> > >::assign(char const*, char const*, unsigned int)':
test_boost2.cc:(.text._ZN5boost11basic_regexIcNS_12regex_traitsIcNS_16cpp_regex_traitsIcEEEEE6assignEPKcS7_j[boost::basic_regex<char, boost::regex_traits<char, boost::cpp_regex_traits<char> > >::assign(char const*, char const*, unsigned int)]+0x22): undefined reference to `boost::basic_regex<char, boost::regex_traits<char, boost::cpp_regex_traits<char> > >::do_assign(char const*, char const*, unsigned int)'
collect2: ld returned 1 exit status
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它在我的Windows机器下运行正常.
怎么了?
gcc版本4.3.2(Debian 4.3.2-1.1)
Linux xxxxxxxx …
我编写了一个相当简单的小型C#Web服务,通过WCF从独立的EXE托管.代码 - 有点简化 - 看起来像这样:
namespace VMProvisionEXE
{
class EXEWrapper
{
static void Main(string[] args)
{
WSHttpBinding myBinding = new WSHttpBinding();
myBinding.Security.Mode = SecurityMode.None;
Uri baseAddress = new Uri("http://bernard3:8000/VMWareProvisioning/Service");
ServiceHost selfHost = new ServiceHost(typeof(VMPService), baseAddress);
try
{
selfHost.AddServiceEndpoint(typeof(IVMProvisionCore), myBinding, "CoreServices");
ServiceMetadataBehavior smb = new ServiceMetadataBehavior();
smb.HttpGetEnabled = true;
smb.MetadataExporter.PolicyVersion = PolicyVersion.Policy12;
selfHost.Description.Behaviors.Add(smb);
// Add MEX endpoint
selfHost.AddServiceEndpoint(ServiceMetadataBehavior.MexContractName, MetadataExchangeBindings.CreateMexHttpBinding(), "mex");
selfHost.Open();
Console.WriteLine("The service is ready.");
Console.ReadLine();
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其余的C#代码; 上面的类VMPService实现了VMProvisionCore.IVMProvisionCore.
namespace VMProvisionCore
{
[ServiceContract(Namespace = "http://Cisco.VMProvision.Core", ProtectionLevel = System.Net.Security.ProtectionLevel.None)]
public interface IVMProvisionCore
{ …Run Code Online (Sandbox Code Playgroud) 我得到一个"对象不支持此属性或方法错误",有谁知道为什么?
我确实将值插入到userId,fname,lname,oname,sam,hasAccess中
function Employee(id, fname, lname, oname, sam, access) {
this.id = id;
this.fname = fname;
this.lname = lname;
this.oname = oname
this.sam = sam;
this.access = access;
}
var emp = new Employee(userId, fname, lname, oname, sam, hasAccess);
var jsonstuff = emp.toSource(); //Breaking here
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虽然这个链接说它可能是http://www.w3schools.com/jsref/jsref_toSource_date.asp
我many_to_many在ImageShells和用户之间有关系.我需要找到所有没有用户的ImageShells.
我有一个查找它的查询,但我如何把它放入named_scope?
SELECT * FROM image_shells INNER JOIN image_shells_users ON (image_shells_users.image_shell_id!=image_shells.id)
class ImageShell < ActiveRecord::Base
has_and_belongs_to_many :riders, :class_name => "User"
end
class User < ActiveRecord::Base
has_and_belongs_to_many :image_shells
end
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我可以使用find by sql但这很麻烦.
img_shells_with_out_users = ImageShell.find_by_sql 'SELECT * FROM image_shells INNER JOIN image_shells_users ON (image_shells_users.image_shell_id!=image_shells.id)'
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