小编use*_*509的帖子

如何删除xml文档中的独立属性声明?

我目前正在使用Java创建一个xml,然后将其转换为String.xml声明如下:

DocumentBuilderFactory dbfac = DocumentBuilderFactory.newInstance();
DocumentBuilder docBuilder = dbfac.newDocumentBuilder();
Document doc = docBuilder.newDocument();
doc.setXmlVersion("1.0");
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为了将文档转换为String,我包含以下声明:

TransformerFactory transfac = TransformerFactory.newInstance();
Transformer trans = transfac.newTransformer();
trans.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "no");
trans.setOutputProperty(OutputKeys.VERSION, "1.0");
trans.setOutputProperty(OutputKeys.ENCODING,"UTF-8");
trans.setOutputProperty(OutputKeys.INDENT, "yes");
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然后我做了转型:

StringWriter sw = new StringWriter();
StreamResult result = new StreamResult(sw);
DOMSource source = new DOMSource(doc);
trans.transform(source, result);
String xmlString = sw.toString();
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问题是在XML声明属性中,包含独立属性,我不希望这样,但我希望显示版本和编码属性:

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
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有没有可以指定的财产?

java xml dom

33
推荐指数
1
解决办法
3万
查看次数

如何在location.href中使用target

我目前正在开发一个Web应用程序,我需要打开一个弹出窗口来显示报告.问题是某些版本的资源管理器不支持window.open javascript函数,所以在这种情况下我抓住错误并用location.href打开新的url.这里的代码:

try {
    window.open(url, "","width=1002,height=700,location=0,menubar=0,scrollbars=1,status=1,resizable=0")
} catch(e) {
    location.target = "_blank";
    location.href = url;
}
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问题是location.target无法正常工作,我想知道是否有办法指定location.href的目标,以便可以在新选项卡中打开它.

javascript window.open

16
推荐指数
3
解决办法
17万
查看次数

使用DER格式从String base64编码创建PrivateKey和PublicKey

我在Base64中的String中有我的私钥和公钥,其中使用ANS1 DER进行编码.我尝试创建java的实例PrivateKeyPublicKey:

byte [] llave2 = DatatypeConverter.parseBase64Binary(key);
PKCS8Key pkcs8 = new PKCS8Key( llave2, password.toCharArray()); //line 2
llave2 = pkcs8.getDecryptedBytes();                             //line 3
certificado = DatatypeConverter.parseBase64Binary(cer);

KeyFactory kf = KeyFactory.getInstance("RSA");  
PKCS8EncodedKeySpec ks = new PKCS8EncodedKeySpec(llave2);
PrivateKey privateKey = kf.generatePrivate(ks);
X509EncodedKeySpec x = new X509EncodedKeySpec(certificado);
PublicKey publicKey = kf.generatePublic(x);
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我收到以下错误PublicKey publicKey = kf.generatePublic(x).

    java.security.spec.InvalidKeySpecException: java.security.InvalidKeyException:     IOException: ObjectIdentifier() -- data isn't an object ID (tag = -96)
    at sun.security.rsa.RSAKeyFactory.engineGeneratePublic(Unknown Source)
    at java.security.KeyFactory.generatePublic(Unknown Source)
    at vital.cancelaciones.GeneraXMLCancelacion.main(GeneraXMLCancelacion.java:118)
Caused by: java.security.InvalidKeyException: …
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java encryption der private-key

14
推荐指数
1
解决办法
4万
查看次数

WSDL中的参数名称具有重要名称

我正在使用JAXWS RI在Java中创建WebService.在自动部署应用程序WAR时会创建WSDL文件.问题是我希望WSDL文件中的参数(每个操作都收到)具有重要的名称,但是它们显示为arg0,arg1,arg2 ...有没有办法为这些参数定义名称而不是使用默认名称?

我已经实现了以下内容:

WebService接口

@WebService
@SOAPBinding(style = Style.RPC)
public interface WS2 {
    @WebMethod String confirmaXML(String lrt_id);
}
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WebService接口实现

@WebService(endpointInterface = "vital.tde.ws2.WS2")
public class WS2Imp implements WS2{
    public String confirmaXML(String lrt_id) {
        String respuesta = null;
        //CODE
        return respuesta;
    }
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太阳jaxws.xml

<?xml version="1.0" encoding="UTF-8"?>
<endpoints
    xmlns="http://java.sun.com/xml/ns/jax-ws/ri/runtime"
    version="2.0">
    <endpoint name="WS2" implementation="vital.tde.ws2.WS2Imp" url-pattern="/WS2" />
</endpoints>
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web.xml中

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" version="3.0">
  <display-name>WS2</display-name>
  <listener>
    <listener-class>
                com.sun.xml.ws.transport.http.servlet.WSServletContextListener
        </listener-class>
  </listener>
  <servlet>
    <servlet-name>WS2</servlet-name>
    <servlet-class>
            com.sun.xml.ws.transport.http.servlet.WSServlet
        </servlet-class>
    <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>WS2</servlet-name>
    <url-pattern>/WS2</url-pattern> …
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java wsdl web-services jax-ws

13
推荐指数
1
解决办法
1万
查看次数

由于缺少java中的证书而导致Web服务出错(基于XML SOAP)

我需要使用在Java中使用Https构建的Web服务.使用Eclipse生成Web服务客户端并调用它我使用以下代码:

ServicioTimbradoPruebasLocator ServicioTimbradoLocator = new ServicioTimbradoPruebasLocator();
                            ServicioTimbradoPruebasSoap ServicioTimbrado = ServicioTimbradoLocator.getServicioTimbradoPruebasSoap();
                            javax.xml.rpc.Stub s =((javax.xml.rpc.Stub)ServicioTimbrado);
                            s._setProperty(javax.xml.rpc.Stub.USERNAME_PROPERTY, "XXXXXXXX");
                            s._setProperty(javax.xml.rpc.Stub.PASSWORD_PROPERTY, "psswd");          
                            String resultado = ServicioTimbrado.generaTimbre(xml.getBytes());
                            System.out.println("resultado: " +resultado);
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在这一行String resultado = ServicioTimbrado.generaTimbre(xml.getBytes()); 我收到以下错误:

AxisFault
     [java]  faultCode: {http://schemas.xmlsoap.org/soap/envelope/}Server.userException
     [java]  faultSubcode: 
     [java]  faultString: javax.net.ssl.SSLHandshakeException: sun.security.validator.ValidatorException: PKIX path building failed: sun.security.provider.certpath.SunCertPathBuilderException: unable to find valid certification path to requested target
     [java]  faultActor: 
     [java]  faultNode: 
     [java]  faultDetail: 
     [java]     {http://xml.apache.org/axis/}stackTrace:javax.net.ssl.SSLHandshakeException: sun.security.validator.ValidatorException: PKIX path building failed: sun.security.provider.certpath.SunCertPathBuilderException: unable to find valid certification path to requested target
     [java]     at sun.security.ssl.Alerts.getSSLException(Unknown …
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java soap web-services certificate webservice-client

7
推荐指数
1
解决办法
3万
查看次数

设置WebLogic Server 11g时出错(10.3.6):找不到本机库从命令行安全地读取密码

我试图在Windows中为开发人员10.3.6(11g)配置我的Weblogic Server.从read me文件中,我在以下步骤中收到错误:

  1. 创建一个新的WLS域并启动WLS.(建议您在MW_HOME之外创建域)Windows $ mkdir C:\ home\myhome\mydomain $ cd C:\ home\myhome\mydomain $%JAVA_HOME%\ bin\java.exe -Xmx1024m -XX:MaxPermSize = 128m weblogic.Server

我在(C:/ WebLogic)中成功创建了目录,并在运行命令时:$ %JAVA_HOME%\bin\java.exe -Xmx1024m -XX:MaxPermSize=128m weblogic.Server,我被要求输入WebLogic的用户名,输入后我收到以下错误:<Error><Security><BEA-090783><Server is Running in Development Mode and Native Library<terminalio> to read the password securely from commandline is not found.>

我该怎么做才能解决这个问题?

java oracle oracle11g weblogic11g server

6
推荐指数
1
解决办法
6491
查看次数

从Visual Basic 6调用Java方法

我需要使用visual basic的java类6.我应该如何声明java类?我需要收到4个字符串,我该如何收到它们?如何从Visual Basic 6中调用此类?我该如何处理返回值?

当你意识到我对此很新,并感谢你的帮助.

java vb6

5
推荐指数
1
解决办法
1526
查看次数

Java中的XMLSignature中的DigestValue与C#不同

我有一个在C#中运行的程序,它将XMLSignature应用于xml文档.我在两种情况下(C#和Java)都有相同的XML文档,但我没有得到相同的摘要和签名值.我知道我的C程序的结果是正确的,但我不能在Java中正确地得到它们.

这里是C#代码:

      public void SignXml(XmlDocument xmlDoc, RSA Key)
        {
            // Check arguments.
            if (xmlDoc == null)
                throw new ArgumentException("xmlDoc");
            if (Key == null)
                throw new ArgumentException("Key");

            // Create a SignedXml object.
            SignedXml signedXml = new SignedXml(xmlDoc);

            // Add the key to the SignedXml document.
            signedXml.SigningKey = Key;

            // Create a reference to be signed.
            Reference reference = new Reference();
            reference.Uri = "";

            // Add an enveloped transformation to the reference.
            XmlDsigEnvelopedSignatureTransform env = new XmlDsigEnvelopedSignatureTransform();
            reference.AddTransform(env);

            // Add the reference …
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c# java xml-signature

5
推荐指数
1
解决办法
4954
查看次数

无法将X509证书转换为Base64

我有一个X509 Certificate,我想将其存储在一个MySQL DB而不是保存文件.因此,我认为最好将证书转换为base64并存储它.我完成了转换的前进和后退,但是我没有得到真正的结果.

我正在使用javax.security.cert.X509Certificate和导入org.apache.commons.codec.binary.Base64如下:

X509Certificate cert = X509Certificate.getInstance(new FileInputStream(certFile));
System.out.println("Vigencia: "+cert.getNotAfter());
System.out.println("Inicio: "+cert.getNotBefore());
String cert64 = bytes2String(Base64.encodeBase64(cert.toString().getBytes("UTF-8")));
System.out.println("Cert 64: "+ cert64);
String certRegreso = bytes2String(Base64.decodeBase64(cert64.getBytes()));
System.out.println("Cert Regreso: "+ certRegreso);
X509Certificate certNuevo = X509Certificate.getInstance(certRegreso.getBytes());
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java base64 x509certificate

3
推荐指数
1
解决办法
5140
查看次数

创建具有以base 64编码的字节数组的PrivateKey和PublicKey

我有byte[]我的私钥(llave)和byte[]我的公钥(certificado =.

我需要在a PrivateKeyPublicKeyobject(分别)中转换这两个值,以便稍后使用它们以数字方式签署xml文档.

我试过了KeyStore,我做不到.我也尝试过使用KeyFactory以下代码:

import java.io.StringWriter;
import java.security.KeyFactory;
import java.security.PrivateKey;
import java.security.PublicKey;
import java.security.spec.PKCS8EncodedKeySpec;
import java.security.spec.X509EncodedKeySpec;
import java.util.Collections;

import javax.xml.crypto.dsig.CanonicalizationMethod;
import javax.xml.crypto.dsig.DigestMethod;
import javax.xml.crypto.dsig.Reference;
import javax.xml.crypto.dsig.SignatureMethod;
import javax.xml.crypto.dsig.SignedInfo;
import javax.xml.crypto.dsig.Transform;
import javax.xml.crypto.dsig.XMLSignature;
import javax.xml.crypto.dsig.XMLSignatureFactory;
import javax.xml.crypto.dsig.dom.DOMSignContext;
import javax.xml.crypto.dsig.keyinfo.KeyInfo;
import javax.xml.crypto.dsig.keyinfo.KeyInfoFactory;
import javax.xml.crypto.dsig.keyinfo.KeyValue;
import javax.xml.crypto.dsig.spec.C14NMethodParameterSpec;
import javax.xml.crypto.dsig.spec.TransformParameterSpec;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.transform.OutputKeys;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerConfigurationException;
import javax.xml.transform.TransformerException;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;

import org.w3c.dom.DOMImplementation; …
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java encryption public-key private-key

2
推荐指数
1
解决办法
7754
查看次数