我已经下载并在我的Windows中安装了Python 2.7.6,我也安装了最新版本的GAE SDK for Windows.
我的GAE上有以下配置:

我正在使用以下文件运行Hello World项目:
Main.py
import webapp2
class MainHandler(webapp2.RequestHandler):
def get(self):
self.response.write('Hello world!')
app = webapp2.WSGIApplication([('/', MainHandler)], debug=True)
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的app.yaml
application: HelloWorld
version: 1
runtime: python27
api_version: 1
threadsafe: yes
handlers:
- url: /favicon\.ico
static_files: favicon.ico
upload: favicon\.ico
- url: .*
script: main.app
libraries:
- name: webapp2
version: "2.5.2"
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但是,当我单击" 运行 "按钮时,为了破坏他的项目,我在LOG输出中得到以下错误:
2013-12-04 14:51:49 Running command: "['C:\\Python27\\pythonw.exe', 'C:\\Program Files (x86)\\Google\\google_appengine\\dev_appserver.py', '--skip_sdk_update_check=yes', '--port=8080', '--admin_port=8000', 'D:\\Users\\pedro\\Desktop\\HelloWorld']"
Traceback (most recent call last):
File "C:\Program Files …Run Code Online (Sandbox Code Playgroud) 我在Linux Mint 16上使用PhpStorm.我创建了一个Hello World,但我正在努力运行它.我收到错误"未安装PHP解释器.按'修复'编辑项目配置."

当我点击"修复"时,我得到:

任何想法如何解决这一问题?没有解释器选项.
我有几个divs .posts的attr data-id对应于mysql DB id.
<div class="posts" data-id="1"></div>
<div class="posts" data-id="2"></div>
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现在,如果我想滚动到一个div我只知道的特定的data-id.我将如何滚动到它?我的JSFiddle就在这里.任何人都可以和JSFiddle一起举个例子吗?
是否可以在不使用printf函数的情况下在C中编写"hello world"程序?(同时仍然保持程序相对几行)
这是一个简单的golang脚本T1.go:
package main
import "fmt"
func main() {
fmt.Println("Hello world")
}
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运行它go run T1.go,我得到:
T1.go:1:15: expected ';', found 'import'
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如果我添加;到行结束,该程序可以运行:
package main;
import "fmt";
func main() {
fmt.Println("Hello world")
}
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但是不是以golang结尾的分号冗余吗?
PS:我是在64位窗口7,golang版本devel +f4d1cb8d9a91或1.2rc1.该错误代码运行完美的http://play.golang.org/
更新我曾经dos2unix将源代码转换为unix行结尾,但它不会改变任何东西
Notes我的go安装在C:\go目录中,C:\go\bin并被添加到%PATH%环境变量中; 源代码T1.go放在C:\t\go目录中,这与go安装目录不同.不确定此配置是否会导致此问题.
我收到这个JSON响应:
{
"status": "Ok",
"total": 105373,
"result": [
{
"id": 668839,
"cn": "A-12157-68812",
"pD": "09-12-2012",
"cCD": "06-05-2012",
"cT": "PERM",
"fN": "COMCAST CABLE COMMUNICATIONS, LLC",
"s": "NJ",
"cR": "Certified",
"pT": "ENGINEER 3",
"vC": 6,
"cddt": "html"
},
{
"id": 725695,
"cn": "A-12361-25668",
"pD": "05-28-2013",
"cCD": "12-26-2012",
"cT": "PERM",
"fN": "L'ONVIE INC.",
"s": "NY",
"cR": "Certified",
"pT": "Biochemist",
"vC": 6,
"cddt": "html"
},
{
"id": 0,
"cn": "A-11199-93239",
"pD": "05-31-2012",
"cCD": "07-18-2011",
"cT": "PERM",
"fN": "ROCKWELL AUTOMATION, INC.",
"s": "PR",
"cR": "Certified", …Run Code Online (Sandbox Code Playgroud) 我有三个变量(asp.net中不同控件的宽度).他们的类型是Unit.
我想做以下行动.例:
Contol1.Width = Control2.Width - Control3.Width
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现在我收到一个错误:
无法将操作符" - "应用于"System.Web.UI.WebControls.Unit"类型的操作数.
我如何用这个值进行数学运算?
所以我认为我的问题是我有一个使用标准表单身份验证的MVC网站和一个更适合ASP.NET身份的MVC.
我要做的是配置我的Owin身份验证以读取和接受由标准MVC表单auth站点生成的auth cookie.一切都配置如下,但我似乎无法让Owin接受cookie.
app.SetDefaultSignInAsAuthenticationType(DefaultAuthenticationTypes.ExternalCookie);
app.UseCookieAuthentication(new CookieAuthenticationOptions
{
AuthenticationType = DefaultAuthenticationTypes.ExternalCookie,
AuthenticationMode = AuthenticationMode.Active,
LoginPath = new PathString("/Account/Login"),
CookieHttpOnly = true,
CookieName = "myAuthCookie",
CookieDomain = ".mydomain.com",
CookiePath = "/",
Provider = new CookieAuthenticationProvider
{
// Enables the application to validate the security stamp when the user logs in.
// This is a security feature which is used when you change a password or add an external login to your account.
OnValidateIdentity = SecurityStampValidator.OnValidateIdentity<ApplicationUserManager, ApplicationUser>(
validateInterval: TimeSpan.FromMinutes(30),
regenerateIdentity: (manager, user) => …Run Code Online (Sandbox Code Playgroud) 我想把@ Html.Actionlink作为按钮.
工作Html.Actionlink:
<li class="btn btn-sm"> @Html.ActionLink("Redeem Reward", "GetReward",
"Home", new { id = price.PriceId }, new { @class = "lnkGetReward"})</li>
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目前,当我点击它时看起来好像点击了一个链接,如果我们单击按钮边框,它就不起作用了.
我也在使用字体很棒的类.
我可以使用@Url.Action相同的id和@class?
我在Visual Studio中有一个asp.net核心应用程序.在自动创建的视图(Events/index.cshtml)中,我想从EventsController.cs中调用此方法
public Boolean IsInRole(string Role)
{
Boolean roleMembership = false;
if (HttpContext.Session.GetInt32("ID") != null)
{
if (HttpContext.Session.GetString("Role") == Role)
{
roleMembership = true;
}
}
return roleMembership;
}
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我的想法是在视图的顶部调用此方法
@if(IsInRole("Admin")) {
show some content
}
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我怎样才能完成这项任务?
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