小编Ken*_*ter的帖子

我正在尝试执行以下代码:

import com.xuggle.mediatool.IMediaReader;
import com.xuggle.mediatool.IMediaWriter;
import com.xuggle.mediatool.ToolFactory;
public class vidcon {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        IMediaReader reader = ToolFactory.makeReader("test.m4v");
        IMediaWriter writer = ToolFactory.makeWriter("out.flv");

        reader.addListener(writer);
        while(reader.readPacket() == null)
            ;
    }
}

但这是给我这个错误:

Exception in thread "main" java.lang.NoClassDefFoundError: org/slf4j/LoggerFactory
    at com.xuggle.ferry.JNILibrary.<clinit>(JNILibrary.java:42)
    at com.xuggle.ferry.FerryJNI.<clinit>(FerryJNI.java:14)
    at com.xuggle.ferry.Ferry.<clinit>(Ferry.java:25)
    at com.xuggle.xuggler.XugglerJNI.<clinit>(XugglerJNI.java:19)
    at com.xuggle.xuggler.IContainer.<clinit>(IContainer.java:1622)
    at com.xuggle.mediatool.MediaReader.<init>(MediaReader.java:137)
    at com.xuggle.mediatool.ToolFactory.makeReader(ToolFactory.java:77)
    at vidcon.main(vidcon.java:11)
Caused by: java.lang.ClassNotFoundException: org.slf4j.LoggerFactory
    at java.net.URLClassLoader$1.run(Unknown Source)
    at java.net.URLClassLoader$1.run(Unknown Source)
    at java.security.AccessController.doPrivileged(Native Method)
    at java.net.URLClassLoader.findClass(Unknown …

鉴于此文本文件:

Received: from unknown (HELO aws-bacon-delivery-svc-iad-1007.vdc.g.com) ([10.146.157.151])
  by na-mm-outgoing-6102-bacon.iad6.g.com with ESMTP; 12 Apr 2011 14:30:47 +0000
Return-Path: 0000012f4a2a0037-528dbafb-e773-44be-bef5-07d8f63e6aee-000000@email-bounces.g.com
Date: Tue, 12 Apr 2011 14:42:37 +0000
From: xxx@xxx.example.com
To: yyy@yyy.example.com
Message-ID: <0000012f4a2a0037-528dbafb-e773-44be-bef5-07d8f63e6aee-000000@email.g.com>
Subject: test
Mime-Version: 1.0
Content-Type: text/plain;
 charset=UTF-8
Content-Transfer-Encoding: 7bit
X-AWS-Outgoing: 199.255.192.79

testing123

我想获得每个领域(Return-path,Date,From,To等)以及body("testing123").

我尝试过匹配使用:

    var bodyRegex = /[\n]Subject: (.+)[\n](.+)/

但我得到空值.

在iphone中加载UIWebView时是否有任何方法可以检测到网址重定向.当我加载www.example.com网址时,它会重定向到其他网址.我需要获取那些重定向网址,实际上我们处理重定向的位置.

当我运行Xcode 6并使用iOS 8模拟器时,我正在尝试加载MFMailComposeViewController,它没有出现.相反,我得到这个错误......

"警告:尝试在ViewController上显示MFMailComposeViewController:0x7c30c400:0x7baf6000已经呈现(null)"

这个完全相同的代码已经工作了将近一年,没有变化.可以肯定的是,我使用iOS 7.1模拟器在Xcode 6中运行我的应用程序,它完全像以前一样工作.

有谁知道如何在iOS 8中显示MFMailComposeViewController视图？这似乎是一个简单的任务,我确信这是一个简单的解决方案.

PS.不需要发布,因为它是标准代码,但无论如何......

MFMailComposeViewController *picker = [[MFMailComposeViewController alloc] init];

picker.mailComposeDelegate = self;

//Current Date
NSDateFormatter *formatter = [[NSDateFormatter alloc]  init];
NSDate  *date = [NSDate date];
//Ensure UTC DATE
[formatter setTimeZone:[NSTimeZone timeZoneWithAbbreviation:@"UTC"]];
[formatter setDateFormat:@"MM/dd/yyyy 'at' HH:mm:ss UTC"];
NSString *myString = [formatter stringFromDate:date];

NSString *emailBody = [NSString stringWithFormat:@"Feedback time at %@.<br><br>Name:", myString];
[picker setMessageBody:emailBody isHTML:YES];

NSString *subject, *emailAddress;

[formatter setTimeZone:[NSTimeZone timeZoneWithAbbreviation:@"UTC"]];
[formatter setDateFormat:@"MM/dd/yyyy"];
subject = [NSString stringWithFormat:@"Feedback"];

//Set the Subject
[picker setSubject:subject];

emailAddress = @"xxx@example.com"; …

我有一个Visual Studio项目,它取决于我们公司的feed提供的一些NuGet包.可以从网上访问该Feed.该项目存储在Git存储库中的Visual Studio Online中.但那些NuGet包不是!我想使用VSO的自动构建和持续集成功能.我如何"告诉"VSO不应该从nuget.org下载那些NuGet包,应该从www.example.com获取？有办法吗？

我正尝试使用HTML5音频在Safari中播放语音.但它不适用于https,http很好.

var audio = new Audio();
audio.src = "https://example.com/xxxx.wav";
audio.play();

是不是可以通过https播放音频？如果不使用html5音频,还有其他替代方案可以在Safari下播放https语音吗？

谢谢

我们使用远程身份验证到SharePoint Online来访问目标O365 sharepoint站点并检索FedAuth和rtFA cookie.这适用于具有xxx.sharepoint.com域的目标URL.现在,如果已将此域自定义为example.com,则身份验证机制将无法抛出错误 - "无法找到登录请求中使用的伙伴DNS".有没有办法我们可以对其域名已被定制为不反映.sharepoint.com的sharepoint在线站点进行身份验证.？

我正在使用JavaScript来查询基于Mongo DB的API.

我需要根据LIKE运算符过滤结果,类似于

select * from playlist where title like '%a%'

目前我称这个URL

var assetUrl = 'https://example.com/playlist?oauth_token=' + accessToken + '&account=XXX'+ '&fields={"title":true,"splash":true,"description":true,"source":true}'+ '&criteria={"title":/.*a.*/}';

没有成功(返回0个对象).

我想知道我是否应该使用正则表达式以及如何在这种情况下使用它们.谢谢

我刚刚将TortoiseGit更新为1.8,从那以后就不可能从服务器推送或拉出.在他们要求输入我的密码之前,他们不再这样做了.相反,他们告诉我:

致命:无法从远程存储库读取.

请确保您具有正确的访问权限并且存储库存在.

git没有干净地退出(退出代码128)(1326 ms @ 2013-01-29 10:34:32)

我在设置中没有更改任何内容,我使用ssh Url作为远程源(Vincent@example.com/git/...)

我在Windows 8中.

设置SPF记录后,我仍然在这里验证为什么我无法收到此电子邮件到hotmail帐户.我可以毫无问题地将它发送到Gmail.请确认代码是否正确,SPF记录是否正确:

<?php 

require_once 'PHPmailer/class.phpmailer.php';
$mail = new PHPMailer();


$body = "Thankyou for your Purchase. <br/><br/> Here is your Policy! You are now Protected during your Travels.";

$mail->AddAddress('$payer_email');
$mail->From = "noreply@example.com";
$mail->FromName = "Name";
$mail->Subject = "Thankyou for Your Purchase";
$mail->MsgHTML($body);
$mail->AddAttachment("tosend/xxx.pdf");
if(!$mail->Send()) {
echo "There was an error sending the message";
    $sql = "UPDATE purchases SET policy_sent = 'Not Sent' WHERE id = '$lastid' ";
    $stmt = $mysqli->query($sql);
    $mysqli->close();
exit;
}
echo "Message was sent successfully";
$sql = "UPDATE purchases …