小编Gan*_*rez的帖子

在Eclipse中的Preferences中打开Debug时出错

当我在Eclipse中的Preferences中打开Debug选项卡时出现此错误,有谁知道发生了什么？

PS:.这是在我做更新后发生的.

调试首选项编辑:

!SUBENTRY 1 org.eclipse.equinox.registry 4 1 2012-02-12 15:21:50.290
!MESSAGE Plug-in org.eclipse.jdt.debug.ui was unable to load class org.eclipse.jdt.debug.ui.launchConfigurations.JavaApplicationLaunchShortcut.
!STACK 0
org.eclipse.core.runtime.internal.adaptor.EclipseLazyStarter$TerminatingClassNotFoundException: An error occurred while automatically activating bundle org.eclipse.jdt.debug.ui (258).
    at org.eclipse.core.runtime.internal.adaptor.EclipseLazyStarter.postFindLocalClass(EclipseLazyStarter.java:122)
    at org.eclipse.osgi.baseadaptor.loader.ClasspathManager.findLocalClass(ClasspathManager.java:462)
    at org.eclipse.osgi.internal.baseadaptor.DefaultClassLoader.findLocalClass(DefaultClassLoader.java:216)
    at org.eclipse.osgi.internal.loader.BundleLoader.findLocalClass(BundleLoader.java:400)
    at org.eclipse.osgi.internal.loader.BundleLoader.findClassInternal(BundleLoader.java:476)
    at org.eclipse.osgi.internal.loader.BundleLoader.findClass(BundleLoader.java:429)
    at org.eclipse.osgi.internal.loader.BundleLoader.findClass(BundleLoader.java:417)
    at org.eclipse.osgi.internal.baseadaptor.DefaultClassLoader.loadClass(DefaultClassLoader.java:107)
    at java.lang.ClassLoader.loadClass(Unknown Source)
    at org.eclipse.osgi.internal.loader.BundleLoader.loadClass(BundleLoader.java:345)
    at org.eclipse.osgi.framework.internal.core.BundleHost.loadClass(BundleHost.java:229)
    at org.eclipse.osgi.framework.internal.core.AbstractBundle.loadClass(AbstractBundle.java:1207)
    at org.eclipse.core.internal.registry.osgi.RegistryStrategyOSGI.createExecutableExtension(RegistryStrategyOSGI.java:174)
    at org.eclipse.core.internal.registry.ExtensionRegistry.createExecutableExtension(ExtensionRegistry.java:905)
    at org.eclipse.core.internal.registry.ConfigurationElement.createExecutableExtension(ConfigurationElement.java:243)
    at org.eclipse.core.internal.registry.ConfigurationElementHandle.createExecutableExtension(ConfigurationElementHandle.java:55)
    at org.eclipse.debug.internal.ui.launchConfigurations.LaunchShortcutExtension.getDelegate(LaunchShortcutExtension.java:410)
    at org.eclipse.debug.internal.ui.launchConfigurations.LaunchShortcutExtension.getLaunchableResource(LaunchShortcutExtension.java:516)
    at org.eclipse.debug.internal.ui.contextlaunching.LaunchingResourceManager.getLaunchableResource(LaunchingResourceManager.java:417)
    at org.eclipse.debug.internal.ui.contextlaunching.ContextRunner.selectAndLaunch(ContextRunner.java:117)
    at org.eclipse.debug.internal.ui.contextlaunching.ContextRunner.selectAndLaunch(ContextRunner.java:164)
    at org.eclipse.debug.internal.ui.contextlaunching.ContextRunner.launch(ContextRunner.java:82)
    at org.eclipse.debug.ui.actions.AbstractLaunchToolbarAction.run(AbstractLaunchToolbarAction.java:79)
    at org.eclipse.debug.ui.actions.AbstractLaunchHistoryAction.runWithEvent(AbstractLaunchHistoryAction.java:404) …

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java eclipse

Gan*_*rez

2017 08-08

14
推荐指数

3
解决办法

2万
查看次数

指定编码XmlSerializer

我已经正确定义了一个类,并在将其序列化为XML后,我没有编码.

如何定义编码"ISO-8859-1"？

这是一个示例代码

var xml = new XmlSerializer(typeof(Transacao));
var file = new FileStream(Path.Combine(AppDomain.CurrentDomain.BaseDirectory, "transacao.xml"),FileMode.OpenOrCreate);            
xml.Serialize(file, transacao);            
file.Close();

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这是xml生成的开始

<?xml version="1.0"?>
<requisicao-transacao xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
  <dados-ec>
    <numero>1048664497</numero>

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c# xml export-to-xml c#-4.0

Gan*_*rez

lucky-day

9
推荐指数

1
解决办法

3万
查看次数

Linq - 相当于左连接中的BETWEEN

我在LINQ中看到了这个话题

我在SQL中的原始查询:

SELECT ISNULL(Tcar.name, '') FROM dbo.models model
LEFT JOIN cars Tcar on Tcar.model = model.id AND
                     Tcar.year between model.Start and model.End

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我需要在"左连接"内部之间实现,我试过这个:

我的课程:

public class car
{
    public string name { get; set; }
    public int model { get; set; }
    public DateTime year { get; set; }
}

public class model
{
    public int id { get; set; }
    public DateTime Start { get; set; }
    public DateTime End { get; set; }
}

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我的实施:

var …

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c# linq

Gan*_*rez

2017 05-23

6
推荐指数

1
解决办法

1839
查看次数

Linq to SQL Left Join,Order and Group By Count

我的查询完美无缺:

SELECT B.ID, B.NAME, COUNT(BU.ID) AS TOTAL
FROM Building B
LEFT JOIN BuildingUser BU ON BU.ID_BUILDING = B.ID    
GROUP BY B.ID, B.NAME
ORDER BY COUNT(BU.ID) DESC, B.NAME

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但是当我将它转换为Linq时,我没有得到预期的结果.当左连接返回null时,它返回count = 1.所以,我一直在尝试这个查询:

var list1 = (from building in db.GetTable<Building>()
             join entitybuildinguser in db.GetTable<BuildingUser>()
                 on building.ID equals entitybuildinguser.ID_BUILDING into tmpbuildinguser
                 from buildinguser in tmpbuildinguser.DefaultIfEmpty()                 
             group building by new
             {
                 building.ID,
                 building.NAME
             } into grpBuilding                                                
             orderby grpBuilding.Select(g => g.ID).Count() descending, grpBuilding.Key.NAME
             select new
             {
                 ID_BUILDING = grpBuilding.Key.ID,
                 NAME = grpBuilding.Key.NAME,
                 users = grpBuilding.Select(g => …

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c# sql linq linq-to-sql

Gan*_*rez

2013 07-31

6
推荐指数

1
解决办法

4982
查看次数

如何在Angular 2中实现等待连接的AuthGuard

我正在使用Angular 2(RC.3)和@ angular/router alpha.8创建一个Web应用程序.这个新路由器提供"Guard",它帮助我们实现处理授权重定向.

正式文档写了如何创建和使用Guard,但其示例代码没有考虑连接时间. https://angular.io/docs/ts/latest/guide/router.html#!#can-activate-guard

所以我想在其中使用Observable(或Promise).

export class ApiService {
  constructor(private _http: Http) {}
  head(url: string): Observable<any> {
    const req: any = { method: RequestMethod.Head, url: URI_SCHEME + url };

    return this._http.request(new Request(req));
  }
}

export class AuthGuard implements CanActivate {
  constructor(private _api: ApiService) {}

  canActivate(): Observable<boolean> {
    return this._api.head('/users/self').subscribe(
      (): Observable<boolean> => {
        // when gets 200 response status...
        return Observable.of(true);
      },
      (): Observable<boolean> => {
        // when gets 401 error response...
        // TODO: …

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rxjs typescript angular

Yoh*_*oda

2017 12-12

6
推荐指数

1
解决办法

1372
查看次数

Linq to SQL - Group By和Count

我正在尝试转换此查询(已经工作)

SELECT Building.NAME, COUNT([User].ID)
FROM BuildingUser
INNER JOIN Building ON Building.ID = BuildingUser.ID_BUILDING
INNER JOIN [User] ON [User].ID = BuildingUser.ID_USER
GROUP BY Building.NAME

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要Linq to SQL,但我不知道我做错了什么.看看我的尝试

from buildinguser in db.GetTable<BuildingUser>()
join building in db.GetTable<Building>()
on buildinguser.ID_BUILDING equals building.ID
join user in db.GetTable<User>()
on buildinguser.ID_USER equals user.ID
group building by building.NAME into grpBuilding
select new
{
    building = grpBuilding.Key,
    users = 
};

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我只需要对我的建筑物进行分组,并计算每个建筑物的用户数量.

c# sql linq linq-to-sql c#-4.0

Gan*_*rez

lucky-day

4
推荐指数

1
解决办法

1万
查看次数

Android中不同版面尺寸的最佳图像尺寸

这是Android中ldpi,mdpi,hdpi和xhdpi密度的最佳尺寸图像？

我问,因为我没有找到任何关于每个密度的完美图像大小的主题.

例如,中等密度是320X480,480X800,480X854,1280X800等.知道我需要根据更大的尺寸绘制图像？

android image android-layout

Gan*_*rez

2012 03-01

3
推荐指数

1
解决办法

2754
查看次数

如何为 320dp 和 360dp 创建布局？

我非常失望。我刚刚完成了基于 360dp 的“普通屏幕”项目，但是当我尝试在 Motorola Atrix 中运行时，我感到惊讶。Motorola Atrix 是 360dp 而不是 320dp，因为他的宽度是 540px。现在我正在努力寻找要解决的问题。如何为 360dp 创建布局？

我尝试了所有这些：

res/values-sw360dp
res/layout-sw360dp

主文件

<?xml version="1.0" encoding="utf-8"?>
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
    android:layout_width="fill_parent"
    android:layout_height="fill_parent"
    android:orientation="vertical"
    android:background="#DFEFF1">

    <Button
        android:background="#AAAA11"
        android:id="@+id/button1"
        android:layout_width="@dimen/default_width"
        android:layout_height="@dimen/default_height"
        android:text="SOME TEXT 1"         
        />


    <Button
        android:background="#FFAA11"
        android:id="@+id/button2"
        android:layout_width="@dimen/default_width"
        android:layout_height="@dimen/default_height"
        android:text="SOME TEXT 2"
        android:layout_alignParentRight="true" />
</RelativeLayout>

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res/values/strings.xml

<?xml version="1.0" encoding="utf-8"?>
<resources>

    <dimen name="default_width">160dp</dimen>
    <dimen name="default_height">160dp</dimen>

</resources>

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摩托罗拉Atrix

三星 Galaxy SII

android android-emulator android-layout

Gan*_*rez

2012 06-19

1
推荐指数

1
解决办法

2467
查看次数

使用正则表达式查找和替换 URL

我读过一些文章和帖子，但找不到从文本中删除特定网址的解决方案。我尝试过至少3种不同的表达方式。很高兴知道我正在以编程方式（C#）执行此操作。

\n\n

(?<!"")((www\\.|(http|https|ftp|news|file)+\\:\\/\\/)[_.a-z0-9-]+\\.[a-z0-9\\/_:@=.+?,##%&amp;~-]*[^.|\\\'|\\# |!|\\(|?|,| |>|<|;|\\)])(?!"")\n\n((([A-Za-z]{3,9}:(?:\\/\\/)?)(?:[-;:&=\\+\\$,\\w]+@)?[A-Za-z0-9.-]+|(?:www.|[-;:&=\\+\\$,\\w]+@)[A-Za-z0-9.-]+)((?:\\/[\\+~%\\/.\\w-_]*)?\\??(?:[-\\+=&;%@.\\w_]*)#?(?:[\\w]*))?)\n\n(?<!\\w?="")(((http|https|ftp|news|file)+://)[_.a-z0-9-]+\\.[a-z0-9\\/_:@=.+?,##%&\xe2\x80\x8c\xe2\x80\x8bamp;~-]*[^.|\\\'|\\# |!|\\(|?|,| |>|<|;|\\)])\n

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示例网址如下。请注意，并非所有网址都完全相同。这是最糟糕的。

\n\n

http://api.ning.com/files/tHd uipNJAW29nIL5jzWbrYON5NhTKYVXmzollyMEHasDDvnq1454BrFkXFfmzj1gH7Ye3xw03C9ulNTICYAdueKg6vaIeu /PauloSkaf.png

\n\n

有人可以指导我找到解决方案吗？

\n\n

更新

\n\n

正如hwnd提到的，最好的模式是

\n\n