问题列表 - 第41750页

如何禁用第 3 方库（例如 AdWhirl 和 AdMob）的 logcat 输出？

有时他们提供禁用日志输出的方法，有时他们不提供。对于那些没有应用程序范围的设置，我可以禁止某些类或标签的任何日志输出或日志输出吗？

我试图在终端中使用以下命令在我的第一个Rails应用程序中运行我的测试:

rspec spec/

但它只会导致以下错误:

Exception encountered: #<NameError: uninitialized constant PagesController>
backtrace:
/Users/rich/.rvm/gems/ruby-1.9.2-p0@abunchofletters/gems/rspec-core-2.3.0/lib/rspec/core/backward_compatibility.rb:20:in `const_missing'
/Users/rich/.rvm/gems/ruby-1.9.2-p0@abunchofletters/gems/rspec-expectations-2.3.0/lib/rspec/expectations/backward_compatibility.rb:6:in `const_missing'
/Users/rich/dev/rails_projects/abunchofletters/spec/controllers/pages_controller_spec.rb:3:in `<top (required)>'
/Users/rich/.rvm/gems/ruby-1.9.2-p0@abunchofletters/gems/rspec-core-2.3.0/lib/rspec/core/configuration.rb:388:in `block in load_spec_files'
/Users/rich/.rvm/gems/ruby-1.9.2-p0@abunchofletters/gems/rspec-core-2.3.0/lib/rspec/core/configuration.rb:388:in `map'
/Users/rich/.rvm/gems/ruby-1.9.2-p0@abunchofletters/gems/rspec-core-2.3.0/lib/rspec/core/configuration.rb:388:in `load_spec_files'
/Users/rich/.rvm/gems/ruby-1.9.2-p0@abunchofletters/gems/rspec-core-2.3.0/lib/rspec/core/command_line.rb:18:in `run'
/Users/rich/.rvm/gems/ruby-1.9.2-p0@abunchofletters/gems/rspec-core-2.3.0/lib/rspec/monkey/spork/test_framework/rspec.rb:4:in `run_tests'
/Users/rich/.rvm/gems/ruby-1.9.2-p0@abunchofletters/gems/spork-0.9.0.rc2/lib/spork/run_strategy/forking.rb:13:in `block in run'
/Users/rich/.rvm/gems/ruby-1.9.2-p0@abunchofletters/gems/spork-0.9.0.rc2/lib/spork/forker.rb:21:in `block in initialize'
/Users/rich/.rvm/gems/ruby-1.9.2-p0@abunchofletters/gems/spork-0.9.0.rc2/lib/spork/forker.rb:18:in `fork'
/Users/rich/.rvm/gems/ruby-1.9.2-p0@abunchofletters/gems/spork-0.9.0.rc2/lib/spork/forker.rb:18:in `initialize'
/Users/rich/.rvm/gems/ruby-1.9.2-p0@abunchofletters/gems/spork-0.9.0.rc2/lib/spork/run_strategy/forking.rb:9:in `new'
/Users/rich/.rvm/gems/ruby-1.9.2-p0@abunchofletters/gems/spork-0.9.0.rc2/lib/spork/run_strategy/forking.rb:9:in `run'
/Users/rich/.rvm/gems/ruby-1.9.2-p0@abunchofletters/gems/spork-0.9.0.rc2/lib/spork/server.rb:47:in `run'
/Users/rich/.rvm/rubies/ruby-1.9.2-p0/lib/ruby/1.9.1/drb/drb.rb:1558:in `perform_without_block'
/Users/rich/.rvm/rubies/ruby-1.9.2-p0/lib/ruby/1.9.1/drb/drb.rb:1518:in `perform'
/Users/rich/.rvm/rubies/ruby-1.9.2-p0/lib/ruby/1.9.1/drb/drb.rb:1592:in `block (2 levels) in main_loop'
/Users/rich/.rvm/rubies/ruby-1.9.2-p0/lib/ruby/1.9.1/drb/drb.rb:1588:in `loop'

我可以在浏览器中成功运行该站点,并且"Spork准备就绪并且正在倾听",这让我觉得我正在运行的Gems,RSpec,Spork还是......有什么问题？而不是我作为PagesController和RSpec测试编写的测试是基于RailsTutorial.org的书.我很茫然.

这是我的Gemfile:

source 'http://rubygems.org'

gem 'rails', '3.0.3'
gem 'sqlite3-ruby', '1.2.5', :require => 'sqlite3'
gem 'gravatar_image_tag', '0.1.0'
gem "nokogiri", '1.4.4'
gem …

刚从大学毕业,我将很快成为一家财富500强公司的软件工程师.该公司使用Java和Spring MVC作为其在线服务.

我知道如何用Java编写类,但我发现Java还有很多东西(例如Java Enterprise Edition,Java Beans,JavaBeans,JSP,Servlets等等).我想在几周内开始工作之前至少熟悉"大规模"/"公司"Java的使用.

Java如何在企业界使用？关于如何将我的"小规模"Java知识带到"大规模",有什么建议吗？

我有多个目标的项目有多个客户端但是大多数应用程序是非常相同的,到目前为止,我能够控制不同的程序流与运行时读取的属性列表.

一个客户有一个特定的视图,我需要先显示所有其他人.

我的问题是我遇到了构建错误(实际上是链接错误),因为控制器类不在其他客户端目标中,我也不想包含它.所以我一直在寻找一些编译时控件.

我正在寻找类似的东西

#ifdef client1target
     ... do something
#else
     ... do something else
#endif    

我遇到麻烦的程序部分就是这样看的

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions {    

    NSString *client = [myEnvVariables getShortName]; // In this class method I read the plist

    if ([client isEqualToString:@"CLIENT1"]) {

        Client1SpecificController *mm = [[Client1SpecificController alloc] initWithNibName:@"Client1SpecificView" bundle:nil];
// here happens the compile error because Client1SpecificController is not known at other targets

        mm.view.frame = CGRectMake(0,20,320,460);

        [window addSubview:mm.view];
        [window makeKeyAndVisible]; 

    } else {

        [window addSubview:navigationController.view];
        [window makeKeyAndVisible]; 
    }

    return YES;
}

错误看起来像这样 …

这个PHP数组在JavaScript中会是什么样子？

什么是最接近的翻译？

$nursery_rhyme = array(“mary”, “had”, “a”, “little”, “lamb”);

此函数使用范围对象返回用户选择并将其包装为粗体标记.有没有一种方法可以删除标签？如在<b>text<b> = text.


我实际上需要一个切换功能,它将选择包装在标签中,如果已经包含标签,则将其解包.与切换粗体按钮时文本编辑器的操作类似.

if "text" then "<b>text</b>"
else "<b>text</b>" then "text"  

...

function makeBold() {

    //create variable from selection
    var selection = window.getSelection();
        if (selection.rangeCount) {
        var range = selection.getRangeAt(0).cloneRange();
        var newNode = document.createElement("b");

            //wrap selection in tags
        range.surroundContents(newNode);

            //return the user selection
        selection.removeAllRanges();
        selection.addRange(range);
    } 
}

随着编程语言的不断发展,我们看到这些编程语言位于Java或.NET等框架之上,这很好.

在您看来,在框架之上构建语言的最佳理由是什么？

免责声明:我不是想证明一些事情,我认为这是一个很好的讨论话题

使用此示例,如何在输入字段中右对齐数字？

<html>
<body>

<form name="input" action="html_form_action.asp" method="get">
Taxes: <input type="number" name="Taxes" value="" /><br />
Shipping: <input type="number" name="shipping" value="" align="right" /><br />
<input type="submit" value="Submit" />
</form> 

<p>Click the "Submit" button, the form-data will be sent to a page called "html_form_action.asp".</p>

</body>
</html>

我使用vs 2008 ..我在vb.net中创建一个Windows窗体应用程序我想要帮助,其中.........如果我退出子*check_fill_for_New()*使用 EXIT SUB然后在*bt_Ok_Click*sub not发一个msgbox ......但它也会退出一半

Public Sub check_fill_for_New()     
    If tb_UserName.Text = "" Then         
        MsgBox("Please Insert User Name Field", MsgBoxStyle.OkOnly, "Error")          
        tb_UserName.Focus()          
        Exit Sub      
     End If
End Sub    

Private Sub bt_Ok_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles bt_Ok.Click                
    If maintain_department = "Admin" Then                
        Call check_fill_for_New()                            
        MsgBox("nooooooooo")        
    End If
End Sub

如果我将列表定义为list=['Ford','Mustang','1966','red'] 并尝试打印它,我的输出将如下所示:Ford Mustang 1966 red

但是我如何实现一个标题,以便输出看起来像:

Brand Modell Year Color
Ford Mustang 1966 Red

那看起来会更专业.