小编Son*_*bby的帖子

在执行长任务时,在Swift IOS应用程序中加载叠加层的示例是什么.从远程服务器加载数据的示例.我用谷歌搜索,但没有找到任何答案.

更新:

感谢@Sebastian Dressler这是一个简单的方法.我更新了我的代码并且运行很酷

public class LoadingOverlay{

var overlayView = UIView()
var activityIndicator = UIActivityIndicatorView()

class var shared: LoadingOverlay {
    struct Static {
        static let instance: LoadingOverlay = LoadingOverlay()
    }
    return Static.instance
}

    public func showOverlay(view: UIView) {

        overlayView.frame = CGRectMake(0, 0, 80, 80)
        overlayView.center = view.center
        overlayView.backgroundColor = UIColor(hex: 0x444444, alpha: 0.7)
        overlayView.clipsToBounds = true
        overlayView.layer.cornerRadius = 10

        activityIndicator.frame = CGRectMake(0, 0, 40, 40)
        activityIndicator.activityIndicatorViewStyle = .WhiteLarge
        activityIndicator.center = CGPointMake(overlayView.bounds.width / 2, overlayView.bounds.height / 2)

        overlayView.addSubview(activityIndicator)
        view.addSubview(overlayView)

        activityIndicator.startAnimating() …

在Swift中,我无法通过以下方式将Int转换为String:

var iString:Int = 100
var strString = String(iString)

但我在Int中的变量？,有错误:Cant invoke 'init' with type '@Ivalue Int?'

例

let myString : String = "42"
let x : Int? = myString.toInt()

if (x != null) {
    // Successfully converted String to Int
    //And how do can i convert x to string???
}

当我启动eclipse时,在启动加载窗口时,eclise自动退出并显示错误:

        !SESSION 2013-09-05 14:52:04.771 -----------------------------------------------
    eclipse.buildId=4.3.0.I20130605-2000
    java.version=1.7.0_25
    java.vendor=Oracle Corporation
    BootLoader constants: OS=win32, ARCH=x86, WS=win32, NL=en_US
    Framework arguments:  -product org.eclipse.epp.package.jee.product
    Command-line arguments:  -os win32 -ws win32 -arch x86 -product org.eclipse.epp.package.jee.product

    !ENTRY org.eclipse.ui.workbench 4 2 2013-09-05 14:52:09.260
    !MESSAGE Problems occurred when invoking code from plug-in: "org.eclipse.ui.workbench".
    !STACK 0
    java.lang.NullPointerException
        at org.eclipse.equinox.internal.p2.core.helpers.ServiceHelper.getService(ServiceHelper.java:74)
        at org.eclipse.equinox.internal.p2.engine.SimpleProfileRegistry.updateRoamingProfile(SimpleProfileRegistry.java:156)
        at org.eclipse.equinox.internal.p2.engine.SimpleProfileRegistry.updateSelfProfile(SimpleProfileRegistry.java:147)
        at org.eclipse.equinox.internal.p2.engine.SimpleProfileRegistry.getProfileMap(SimpleProfileRegistry.java:344)
        at org.eclipse.equinox.internal.p2.engine.SimpleProfileRegistry.internalGetProfile(SimpleProfileRegistry.java:248)
        at org.eclipse.equinox.internal.p2.engine.SimpleProfileRegistry.getProfile(SimpleProfileRegistry.java:178)
        at org.eclipse.equinox.internal.p2.ui.sdk.scheduler.AutomaticUpdateScheduler.earlyStartup(AutomaticUpdateScheduler.java:88)
        at org.eclipse.ui.internal.EarlyStartupRunnable.runEarlyStartup(EarlyStartupRunnable.java:87)
        at org.eclipse.ui.internal.EarlyStartupRunnable.run(EarlyStartupRunnable.java:66)
        at org.eclipse.core.runtime.SafeRunner.run(SafeRunner.java:42)
        at org.eclipse.ui.internal.Workbench$55.run(Workbench.java:2552)
        at org.eclipse.core.internal.jobs.Worker.run(Worker.java:53)

    !ENTRY org.eclipse.ui.workbench 4 2 2013-09-05 14:52:09.264
    !MESSAGE Problems occurred …

我有Pojo对象,使用getAsJson函数返回此对象的Json字符串.我使用JsonProperty来定义此对象中的json属性.使用ObjectMapper的writeValueAsString为此对象编写json字符串.

import org.codehaus.jackson.JsonGenerationException;
import org.codehaus.jackson.annotate.JsonIgnore;
import org.codehaus.jackson.annotate.JsonIgnoreProperties;
import org.codehaus.jackson.annotate.JsonProperty;
import org.codehaus.jackson.map.JsonMappingException;
import org.codehaus.jackson.map.ObjectMapper;

@JsonIgnoreProperties(ignoreUnknown=true)
public class LogLikeArticleDetail extends BaseObject {
    private static final long serialVersionUID = -2018373118257019033L;
    @JsonProperty("LikeArticleGUId")
    private String likeArticleGUId;
    @JsonProperty("UserId")
    private String userID;
    @JsonProperty("UserName")
    private String userName;
    @JsonProperty("IP") 
    private String ip;
    @JsonProperty("OS") 
    private String os;
    @JsonProperty("UserAgent") 
    private String userAgent;
    @JsonProperty("WebsiteCode") 
    private String websiteCode;
    @JsonProperty("ArticleId") 
    private String articleID;
    @JsonProperty("ATitle") 
    private String aTitle;
    @JsonProperty("CateAlias") 
    private String cateAlias;
    @JsonProperty("LikeStatus") 
    private String likeStatus;
    @JsonProperty("TimeStamp") 
    private Date timeStamp;
        //get, set....
        //....
        @JsonIgnore
    public …

我用C#,.NET Framework 4.0编写了一个控制台应用程序.

这个项目参考Microsoft.Office.Interop.Excel.dll.

当我构建项目成功,但在build文件夹中不包含Microsoft.Office.Interop.Excel.dll.

当我改为.NET 2.0并构建它时,Microsoft.Office.Interop.Excel.dll包括在内!

但我想在.NET 4.0中构建和运行我的程序.

我使用VS 2010.

告诉我如何解决.

我用Log4j写一些日志我的程序.我在这个网站上找到并阅读了很多问题和答案,但我无法解决我的问题.

这是我的代码:

1. log4j.xml

<appender name="rollingfileAppender" class="org.apache.log4j.DailyRollingFileAppender">
    <param name="append" value="true"/>
    <param name="file" value="logs/process.log"/>
    <param name="DatePattern" value="'.'yyyy-MM-dd-HH"/>
    <layout class="org.apache.log4j.PatternLayout">
       <param name="ConversionPattern" value="%d{yyyy-MM-dd HH:mm:ss:SSS} %-5p [%c{1}] %m%n"/>
    </layout>
</appender>
<root>
    <level value="DEBUG"/>
    <appender-ref ref="rollingfileAppender"/>
    <appender-ref ref="stdout"/>
</root>

2.我的java代码

package TestPacket;

import org.apache.log4j.Logger;
import org.apache.log4j.xml.DOMConfigurator;


public class TestLog4jXML {
    static Logger logger = org.apache.log4j.Logger.getLogger(TestLog4jXML.class.getName());
    public TestLog4jXML() {
    }
    public static void main(String[] args) {
        try {
            DOMConfigurator.configure("log4j1.xml");
            logger.trace("Entering application.");
            logger.debug("Debug");
            logger.info("info");
            logger.warn("warn");
            logger.error("error");
            logger.fatal("fatal");
            lungtng();
        } catch (Exception e) {
            e.printStackTrace();
        }
    } …

我有2个班:

public class Vote
{
    public string VoteId { get; set; }
    public string Question { get; set; }
    public List<VoteAnswer> AnswerList { get; set; }
}

和:

public class VoteOption
{
    public string OptionId { get; set; }
    public string OptionName { get; set; }
    public double VoteCount { get; set; }
}

如何更新/删除VoteOption在一个Vote地方VoteId = voteId和OptionId = optionId？使用C#驱动程序

首先,我通过以下方式获得VoteOption:

        var v = col.FindOneAs<Vote>(Query.EQ("VoteID", voteId));
        VoteOption vo = v.AnswerList.Find(x => …

在 Swift 中，我使用 Decoding HTML NSAttributedString，如下所示：

let encodedString = "Ph\xe1\xba\xa3i c\xc3\xb4ng nh\xe1\xba\xadn r\xe1\xba\xb1ng k\xe1\xbb\x83 t\xe1\xbb\xab l\xc3\xbac \xc3\xb4ng Th\xc4\x83ng\xc2\xa0l\xc3\xa0m b\xe1\xbb\x99 tr\xc6\xb0\xe1\xbb\x9fng"\nlet encodedData = encodedString.dataUsingEncoding(NSUTF8StringEncoding)\nlet attributedOptions = [NSDocumentTypeDocumentAttribute: NSHTMLTextDocumentType]\nlet attributedString = NSAttributedString(data: encodedData, options: attributedOptions, documentAttributes: nil, error: nil)         \nlet decodedString = attributedString.string\nprintln(decodedString)\n

但结果是这样的：

\n

Ph\xc3\xa1\xc2\xba\xc2\xa3i c\xc3\x83\xc2\xb4ng nh\xc3\xa1\xc2\xba\xc2\xadn r\xc3\xa1\xc2\xba\xc2\xb1ng k\ xc3\xa1\xc2\xbb\xc6\x92 t\xc3\xa1\xc2\xbb\xc2\xab l\xc3\x83\xc2\xbac \xc3\x83\xc2\xb4ng Th\xc3\x84\xc6\x92ng \xc3\x82\xc2\xa0l\xc3\x83\xc2\xa0m b\xc3\xa1\xc2\xbb\xe2\x84\xa2\n tr\xc3\x86\xc2\xb0\xc3\xa1\xc2\xbb \xc5\xb8ng

\n

真实结果一定与encodedString

这个方法有什么问题吗？

当我选择主从应用程序时，我想在 XCode 故事板设计中删除图片中的这个箭头。

我有一个List<Object>ñ Objects.在n对象中有一些它们具有相同的对象Id.例List<T> = {t1, t2, t3, t4, t5....,t10}.

假如说:

t1.Id = t5.id

所以,我想删除t5从List.

谁能帮助我PLZ.

在这个问题中: 使用LINQ删除List中的对象你知道该对象重复(例如:"bob"),但在我的问题中,在这个List中,我不知道之前有任何对象重复.