有没有人知道维护的gem通过现有的Rails 3应用程序处理Zendesk API的用户身份验证?
我问了Zendesk IT并被发送到https://github.com/tobias/zendesk_remote_auth,但它看起来并不兼容rails 3并且自2009年以来一直没有更新.
当请求进入页面时,例如app.get("/")我想从amazon s3返回静态HTML页面.我知道我可以从S3请求它然后发送它,但这似乎很慢.反正有没有告诉请求者直接从s3获取文件而不更改URL?
谢谢.
如果做不到这一点,从s3提供文件的最快方式是什么?
本教程首先介绍如何编写文件
http://www.hacksparrow.com/node-js-amazon-s3-how-to-get-started.html
// We need the fs module so that we can write the stream to a file
var fs = require('fs');
// Set the file name for WriteStream
var file = fs.createWriteStream('slash-s3.jpg');
knox.getFile('slash.jpg', function(err, res) {
res.on('data', function(data) { file.write(data); });
res.on('end', function(chunk) { file.end(); });
});
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有没有办法发送文件而不先写它?写它看起来非常慢.
我有一个类,所有方法都需要同步(没有静态方法).其中一种方法将每50 ms调用一次.
我想知道在哪里放置synchronized关键字有最短的执行时间?
即关于执行时间的2个选项之间是否有任何差异?
选项1(同步方法)
public class MyNativeObject{
private boolean released = false;
public synchronized String read(){
if(released) return null;
return Read();
}
public synchronized void release(){
if(released) return;
released = true;
Release();
}
private native String Read();
private native void Release();
}
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选项2(同步块)
public class MyNativeObject{
private Boolean released = false;
public String read(){
synchronized(released){
if(released) return null;
return Read();
}
}
public void release(){
synchronized(released){
if(released) return;
released = true;
Release();
}
}
private native String Read();
private …Run Code Online (Sandbox Code Playgroud) 这是一个脑筋急转弯的问题,我实际上知道答案.我正在抛出一笔赏金,因为它代表了一个有价值的节点编程安全提示(这是第一个提示).
我正在使用
var knox = require('knox');
var s3 = knox.createClient({
key: ...,
secret: ...,
bucket: ...
});
// The bug is below:
var stringVal = JSON.stringify(<2d javascript array from a large spreadsheet>)
var req = s3.put(path + filename, {
'Content-Length': stringVal.length,
'Content-Type': 'application/json'
});
req.end(stringVal);
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生成的上载被截断或以其他方式损坏.我们有stringVal.length === 322889,并且生成的S3项目大小与之匹配.但是下载并重新加载文件会产生一个长度为的字符串322140.在尝试使用JSON.parse字符串(可预测地)导致语法错误之前,没有错误出现.
这是怎么回事?
javascript file-upload amazon-s3 node.js knox-amazon-s3-client
我改变了我的应用程序以使用Hibernate EntityManager(来自Hibernate会话),
但我有一些旧代码(我无法更改这些代码),在下面的代码中使用:
getSessionFactory().getCurrentSession() .
我有sessionFactory的bean,所以上面的代码应该工作,但在运行时我有
HibernateException("No Session found for current thread"),甚至上面的代码是在事务块中执行的.
仅供参考:我检查了事务资源(在调试模式下)并且使用密钥EntityManagerFactory,会话存在但不在SessionFactory下
我使用eclipse juno创建了一个maven项目并编辑了pom文件.当我使用命令提示符"mvn clean install"运行pom文件时,我收到此错误.
[ERROR] The build could not read 1 project -> [Help 1]
[ERROR]
[ERROR] The project se.cambio.platform.sdk:sdk-documentation-samples-laki:1 (F:\newWorkspace\sdk-documentation-samples-laki\pom.xml) has 2 errors
[ERROR] 'dependencies.dependency.version' for se.cambio.platform.sdk:sdk-client:jar must be a valid version but is '${version.sdk.client}'. @ line 38, column 16
[ERROR] 'dependencies.dependency.version' for se.cambio.platform.sdk:sdk-common:jar must be a valid version but is '${version.sdk.common}'. @ line 44, column 16
[ERROR]
[ERROR] To see the full stack trace of the errors, re-run Maven with the -e switch.
[ERROR] Re-run Maven using the …Run Code Online (Sandbox Code Playgroud) 我目前正在开发一个带有Hadoop集群的未来项目.我需要找到有关集群安全性的信息.
我发现API Apache Knox Gateway似乎是我们需要的.我们现在与Cloudera 4.5合作.将来,我们将升级到Cloudera 5.
我的问题是Knox似乎与Cloudera 4.5不兼容(http://www.cloudera.com/content/cloudera-content/cloudera-docs/CDH4/latest/CDH-Version-and-Packaging-Information/cdhvd_topic_3. HTML).
WebHDFS 2.4.0
WebHCat/Templeton 0.13.0 0.12.0
Ozzie 4.0.0
HBase/Stargate 0.98.0
Hive (via WebHCat) 0.13.0
Hive (via JDBC) 0.13.0
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(http://knox.apache.org/books/knox-0-4-0/knox-0-4-0.html#Quick+Start)
根据该网站,这是唯一测试的版本.有没有人有这个API和Cloudera 4.5的经验?
如果它无法工作,是否有任何解决方案可以保证集群安全?
我有这段代码可以生成一个 XML 文件并保存到一个外部目录。
try //database structure
{
DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder docBuilder = docFactory.newDocumentBuilder();
//test report elements, does not reflect the real database in the future
//mandatory/ username, location, caption, media, time(Actually, it's best if the server determines recieve time)
//optional/
//reportdata elements
Document doc = docBuilder.newDocument();
Element rootElement = doc.createElement("reportdata");
doc.appendChild(rootElement);
//username elements
Element username = doc.createElement("username");
username.appendChild(doc.createTextNode("testusername"));
rootElement.appendChild(username);
//Location ELEMENTS
//latitude elements
Element lat = doc.createElement("latitude");
lat.appendChild(doc.createTextNode(String.valueOf(latitude)));
rootElement.appendChild(lat);
//longitude
Element longi = doc.createElement("longitude");
longi.appendChild(doc.createTextNode(String.valueOf(longitude)));
rootElement.appendChild(longi);
//caption text …Run Code Online (Sandbox Code Playgroud) 我想在我的 Android 应用程序中显示 html 图像地图。为此,我使用 WebView,并且我希望当<area>单击 a 时,单击的区域执行某些操作。例如,如果我使用踢或触摸“北卡罗来纳州”,我会在祝酒词中显示它的名称。
我的代码是:
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>map_usa_856</title>
</head>
<body>
<img src="usamap.png" width="320" height="266" style="margin:0 auto; display:block;" usemap="#Map" border="0" />
<map name="Map" id="Map">
<area shape="poly" href="North Carolina" alt="North Carolina" title="North Carolina" coords="266,109,267,107,270,107,270,105,273,102,278,100,277,98,275,98,276,96,278,96,280,93,278,91,277,91,278,90,277,88,245,93,244,97,242,97,239,100,236,102,233,102,232,104,231,104,231,105,237,104,240,102,250,102,250,102,252,104,258,102">
<area shape="poly" href="North Dakota" alt="North Dakota" title="North Dakota" coords="119,17,152,17,155,38,118,36">
<area shape="poly" href="Rhode Island" alt="Rhode Island" title="Rhode Island" coords="289,48,291,53,292,52,302,54,302,57,311,57,311,51,302,51,302,53,292,51,291,48">
<area shape="poly" href="Illinois" alt="Illinois" title="Illinois" coords="208,79,207,80,208,82,207,85,205,87,205,91,203,93,203,93,202,93,200,93,200,95,198,93,199,90,192,87,194,82,191,82,190,82,190,80,185,77,185,75,187,69,187,67,190,67,192,62,189,60,204,59,204,60,206,62">
<area shape="poly" href="New Mexico" alt="New Mexico" title="New Mexico" coords="117,96,83,93,78,130,83,131,84,127,92,129,92,127,115,129"> <area shape="poly" alt="Georgia" …Run Code Online (Sandbox Code Playgroud) <ToggleButton
android:id="@+id/toggle"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:background="@drawable/check"
android:layout_margin="10dp"
android:textOn=""
android:textOff=""
android:focusable="false"
android:focusableInTouchMode="false"
android:layout_centerVertical="true"/>
<?xml version="1.0" encoding="utf-8"?>
<selector xmlns:android="http://schemas.android.com/apk/res/android">
<!-- When selected, use grey -->
<item android:drawable="@drawable/selected_image"
android:state_checked="true" />
<!-- When not selected, use white-->
<item android:drawable="@drawable/unselected_image"
android:state_checked="false"/>
</selector>
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如何使用颜色代替可绘制的颜色?
我试过这个:
<?xml version="1.0" encoding="utf-8"?>
<selector xmlns:android="http://schemas.android.com/apk/res/android">
<item android:state_checked="true" android:color="@color/bg_blue"/>
<item android:state_checked="false" android:color="@color/bg_light_grey"/>
</selector>
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在名称为bg_toggle.xml的颜色文件夹中定义了上述选择器
和
android:background="@color/bg_toggle"
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这给出了一个例外:
android.view.InflateException:二进制XML文件行#50:错误膨胀类android.widget.ToggleButton
org.xmlpull.v1.XmlPullParserException:二进制XML文件行#4:标记需要一个'drawable'属性或定义drawable的子标记
谢谢
java ×4
android ×3
amazon-s3 ×2
javascript ×2
node.js ×2
xml ×2
apache ×1
cloudera ×1
eclipse ×1
exception ×1
express ×1
file-upload ×1
hadoop ×1
hibernate ×1
jpa ×1
maven ×1
performance ×1
pom.xml ×1
security ×1
selector ×1
synchronized ×1
togglebutton ×1
zendesk ×1