我有以下XML配置我想转换为java属性文件.
我收到了以下错误
log4j:WARN Failed to set property [filter] to value "org.apache.log4j.varia.DenyAllFilter".
log4j:ERROR Could not instantiate class [true].
java.lang.ClassNotFoundException: true
at java.net.URLClassLoader$1.run(URLClassLoader.java:202)
at java.security.AccessController.doPrivileged(Native Method)
at java.net.URLClassLoader.findClass(URLClassLoader.java:190)
at java.lang.ClassLoader.loadClass(ClassLoader.java:307)
at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:301)
at java.lang.ClassLoader.loadClass(ClassLoader.java:248)
at java.lang.Class.forName0(Native Method)
at java.lang.Class.forName(Class.java:169)
at org.apache.log4j.helpers.Loader.loadClass(Loader.java:198)
at org.apache.log4j.helpers.OptionConverter.instantiateByClassName(OptionConverter.java:326)
at org.apache.log4j.PropertyConfigurator.parseAppenderFilters(PropertyConfigurator.java:881)
at org.apache.log4j.PropertyConfigurator.parseAppender(PropertyConfigurator.java:812)
at org.apache.log4j.PropertyConfigurator.parseCategory(PropertyConfigurator.java:735)
at org.apache.log4j.PropertyConfigurator.configureRootCategory(PropertyConfigurator.java:615)
at org.apache.log4j.PropertyConfigurator.doConfigure(PropertyConfigurator.java:502)
at org.apache.log4j.PropertyConfigurator.doConfigure(PropertyConfigurator.java:395)
at org.apache.log4j.PropertyConfigurator.configure(PropertyConfigurator.java:403)
at simpandfile.main(simpandfile.java:10)
我的XML配置
<filter class="org.apache.log4j.varia.StringMatchFilter">
<param name="StringToMatch" value="FileNotfound" />
<param name="AcceptOnMatch" value="true" />
</filter>
<filter class="org.apache.log4j.varia.LevelRangeFilter">
<param name="LevelMin" value="ERROR"/>
<param name="LevelMax" value="FATAL"/>
</filter>
<filter class="org.apache.log4j.varia.DenyAllFilter"/> …我从这边使用SSL Socket和Trustmanager 自签名SSL
但我不断收到以下错误:
09-28 19:52:41.942:WARN/System.err(10101):javax.net.ssl.SSLHandshakeException:org.bouncycastle.jce.exception.ExtCertPathValidatorException:无法验证证书签名.
怎么了？我已经在stackoverflow上检查了不同的帖子,但我似乎无法让它工作.
我的代码:
SchemeRegistry schemeRegistry = new SchemeRegistry();
// http scheme
schemeRegistry.register(new Scheme("http", PlainSocketFactory.getSocketFactory(), 80));
// https scheme
schemeRegistry.register(new Scheme("https", new EasySSLSocketFactory(), 443));
params = new BasicHttpParams();
params.setParameter(ConnManagerPNames.MAX_TOTAL_CONNECTIONS, 1);
params.setParameter(ConnManagerPNames.MAX_CONNECTIONS_PER_ROUTE, new ConnPerRouteBean(1));
params.setParameter(HttpProtocolParams.USE_EXPECT_CONTINUE, false);
HttpProtocolParams.setVersion(params, HttpVersion.HTTP_1_1);
HttpProtocolParams.setContentCharset(params, "utf8");
CredentialsProvider credentialsProvider = new BasicCredentialsProvider();
credentialsProvider.setCredentials(new AuthScope("www.example.com", AuthScope.ANY_PORT),
new UsernamePasswordCredentials("user", "password"));
clientConnectionManager = new ThreadSafeClientConnManager(params, schemeRegistry);
context = new BasicHttpContext();
context.setAttribute("http.auth.credentials-provider", credentialsProvider);
DefaultHttpClient client = new DefaultHttpClient(clientConnectionManager, params);
HttpGet get = new HttpGet("https://www.example.com/web/restricted/form/formelement=512663");
HttpResponse response = client.execute(get, …我尝试从使用PasswordText WSS类型的Web服务获取信息.首先,我使用soapUI测试它并成功获取数据.然后我在Java上实现了身份验证,编写了SecurityHandler:
public final class SecurityHandler implements SOAPHandler<SOAPMessageContext> {
...
@Override
public boolean handleMessage(SOAPMessageContext messageContext) {
boolean outInd = (Boolean) messageContext.get(MessageContext.MESSAGE_OUTBOUND_PROPERTY);
if (outInd) {
try {
WSSecUsernameToken builder = new WSSecUsernameToken();
builder.setPasswordType(WSConstants.PASSWORD_TEXT);
builder.setUserInfo(_username, _password);
builder.addNonce();
builder.addCreated();
Document doc = messageContext.getMessage().getSOAPPart().getEnvelope().getOwnerDocument();
WSSecHeader secHeader = new WSSecHeader();
secHeader.insertSecurityHeader(doc);
builder.build(doc, secHeader);
} catch (Exception e) {
LOGGER.error("Unable to handle SOAP message", e);
return false;
}
}
return true;
}
...
}
我查看了doc对象,XMLUtils.PrettyDocumentToString(doc)看到它看起来像是由soupUI发送的XML - 所有认证信息(登录,密码,现时和创建时间)都已到位,标签的mustUnderstand属性Security为true.
然后我遇到了错误:
javax.xml.ws.soap.SOAPFaultException:MustUnderstand标题:[{ http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-secext-1.0.xsd …
我已经坚持了一段时间,似乎并没有解决这个问题.
我试图从URL读取URL的内容作为字符串,但我得到一个奇怪的
错误 - >错误域= NSCocoaErrorDomain代码= 256"操作无法完成.(Cocoa错误256.)"
我的代码:
fetchedString = [[NSString alloc] initWithContentsOfURL:[NSURL URLWithString:@"www.example.com/iphone"] encoding:NSUTF8StringEncoding error:&error];
NSLog(@"%@",fetchedString);
// if there is something wrong with the URL
if (error) {
NSLog(@"Error -> %@", error);
return ;
}
我究竟做错了什么？我尝试使用获取作为NSData,但我得到null.
我在阅读SOA记录时遇到了一些麻烦.在dig用于获取我控制的域的SOA记录后,我得到以下响应.
SOA ns1.example.com. xxx.gmail.com. 2013041357 14400 14400 1209600 86400
我怎么知道哪个值与每个字段相关联？根据这篇文章,SOA记录可以有九个字段.
我想象每个空格表示值的结束,我可以看到第二个值是电子邮件地址.但是,我如何确定哪个是刷新值,哪个是重试值？它们只是旁边没有字段名称的数字!
我在这里错过了什么吗？提前致谢.
我已经分叉了某人的GIT存储库:
https://github.com/nippysaurus/toodledo-objc
将其克隆到我的本地计算机,显示原点并提供以下信息:
* remote origin
Fetch URL: https://nippysaurus@github.com/nippysaurus/toodledo-objc.git
Push URL: https://nippysaurus@github.com/nippysaurus/toodledo-objc.git
HEAD branch: master
Remote branch:
master tracked
Local branch configured for 'git pull':
master merges with remote master
Local ref configured for 'git push':
master pushes to master (up to date)
当我将我的更改推送到"origin master"时,git会打印出"最新的一切",但在我的GitHub仓库中没有更新任何内容.
这里发生了什么？
编辑:
有人建议我检查文件是否实际提交...文件被提交,我向你保证.
commit 0d3a21616d82c8e5a89baaf85d745fc2cfdf614f
Author: nippysaurus <nippysaurus@example.com>
Date: Wed Jun 1 13:19:14 2011 +1000
updated readme
这是更新的文件:
commit 0d3a21616d82c8e5a89baaf85d745fc2cfdf614f
Author: nippysaurus <nippysaurus@example.com>
Date: Wed Jun 1 13:19:14 2011 +1000
updated readme
diff --git a/README.mdown …我的网站上有一个表单,其中有一个页面,我可以编辑/删除/添加邮箱:
每当我对邮箱做某事(更新,销毁)时,我都会收到此错误:
重定向到http://example.com/ 已完成406在64毫秒内无法接受
但数据会更新.
这是控制器代码:
# PUT /mailboxes/1
def update
@mailbox = Mailbox.find(params[:id])
if @mailbox.update_attributes(params[:mailbox])
redirect_to(root_path, :notice => 'Mailbox was successfully updated.')
else
render :action => "edit"
end
end
# DELETE /mailboxes/1
def destroy
@mailbox = Mailbox.find(params[:id])
@mailbox.destroy
redirect_to(root_path)
end
这是routes.rb信息:
match 'settings.js' => 'settings#javascript', :via => :get, :format => :js
scope '/settings' do
# Directs /settings/mailboxes/* to Settings::MailboxesController
# (app/controllers/settings/mailboxes_controller.rb)
resources :mailboxes
end
我究竟做错了什么？这是日志显示的内容:
if @mailbox.update_attributes(params[:mailbox])
(rdb:2) response.status
200
(rdb:2) next
/Users/Fallen/Projects/support-app/trunk/app/controllers/mailboxes_controller.rb:65
redirect_to(mailboxes_path, …我有一个JAX-WS Web服务应用程序,它作为Tomcat 7的WAR文件进行部署.它使用最新版本的Metro库,我将其包含在WAR文件中,并且工作正常.
我正在尝试简化部署包.据我所知,太阳JDK包括地铁的复印件(请参阅这个问题和这个为例),但出于某种原因,这显然是强制性的,同时一个从GlassFish网站下载,以取代地铁的这个副本.我试图了解是否有可能只使用Tomcat和JDK附带的城域网实现,或者如果不是为什么不这样做.
WAR内容如下(删除了类文件):
META-INF/MANIFEST.MF
WEB-INF/classes/
WEB-INF/classes/com/[et cetera]
WEB-INF/ibm-web-ext.xml
WEB-INF/lib/
WEB-INF/lib/stax-api.jar
WEB-INF/lib/webservices-api.jar
WEB-INF/lib/webservices-extra-api.jar
WEB-INF/lib/webservices-extra.jar
WEB-INF/lib/webservices-rt.jar
WEB-INF/lib/webservices-tools.jar
WEB-INF/sun-jaxws.xml
WEB-INF/web.xml
wsdl/
wsdl/MyService.wsdl
web.xml 包含,部分:
<servlet>
<servlet-name>MyService</servlet-name>
<servlet-class>
com.sun.xml.ws.transport.http.servlet.WSServlet
</servlet-class>
</servlet>
当我从WAR中删除webservices-*jars(Metro jar)时,Web服务失败并显示错误"Wrapper找不到servlet类com.sun.xml.ws.transport.http.servlet.WSServlet或类这取决于".这并不奇怪,因为我无法在Java 7 SE附带的罐子里找到该类.
那么,如果你必须下载另一份Metro来制作像这样的工作,那么说Java 7附带Metro是什么意思呢？是否可以仅使用Java附带的jar在Tomcat中运行JAX-WS Web服务？
我正在尝试设置一些cookie WebView来打开一个浏览器,其浏览器与我在我的应用程序上的会话相同.
我读了很多答案,但它们对我不起作用.我发现的唯一解决方案是loadUrl对cookie数据进行硬编码extraHeaders,但正如预期的那样,这仅适用于此请求,并且不会维护会话.
我的代码是:
CookieSyncManager cookieSyncManager = CookieSyncManager.createInstance(mWebView.getContext());
CookieManager cookieManager = CookieManager.getInstance();
cookieManager.setAcceptCookie(true);
cookieManager.removeSessionCookie();
cookieManager.setCookie("http://xx.xxx.example.com","mid="+MySession.GetSession().sessionId+" ; Domain=.example.com");
cookieSyncManager.sync();
String cookie = cookieManager.getCookie("http://xx.xxx.example.com");
Log.d(LOGTAG, "cookie ------>"+cookie);
mWebView.getSettings().setJavaScriptEnabled(true);
mWebView.setWebViewClient(new TuWebViewClient());
mWebView.loadUrl("http://xx.xx.example.com");
getCookie()返回正确的数据,但是当我从服务器读取cookie时,这些是空的.怎么了？请指教.谢谢!!!
我有WSDL以下格式的网址,似乎是正确的.使用后,wsimport在终端解析,我得到一个错误:
从终端运行:
tux-world@alachiq:~/Desktop/Project/java > wsimport -keep -s wsdl/ example.com/wsdl/wsdl.php?wsdl
parsing WSDL...
[ERROR] "Use of SOAP Encoding is not supported.
SOAP extension element on line 59 in example.com/wsdl/wsdl.php?wsdl has use="encoded" "
Failed to parse the WSDL.
解析的WSDL:
<?xml version="1.0" encoding="ISO-8859-1"?>
<definitions xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:SOAP-ENC="http://schemas.xmlsoap.org/soap/encoding/" xmlns:tns="urn:tsmswsdl" xmlns:soap="http://schemas.xmlsoap.org/wsdl/soap/" xmlns:wsdl="http://schemas.xmlsoap.org/wsdl/" xmlns="http://schemas.xmlsoap.org/wsdl/" targetNamespace="urn:tsmswsdl">
<types>
<xsd:schema targetNamespace="urn:tsmswsdl"
>
<xsd:import namespace="http://schemas.xmlsoap.org/soap/encoding/" />
<xsd:import namespace="http://schemas.xmlsoap.org/wsdl/" />
<xsd:complexType name="outGetMessages">
<xsd:complexContent>
<xsd:restriction base="SOAP-ENC:Array">
<xsd:attribute ref="SOAP-ENC:arrayType" wsdl:arrayType="xsd:string[]"/>
</xsd:restriction>
</xsd:complexContent>
</xsd:complexType>
</xsd:schema>
</types>
<message name="GetMessagesRequest"> …