您好,我正在尝试编组一个SAMLResponse对象&我得到了以下错误。有什么想法吗?
java.lang.StringIndexOutOfBoundsException: String index out of range: -4
at java.lang.String.substring(Unknown Source)
at org.apache.xml.security.transforms.params.InclusiveNamespaces.<init>(Unknown Source)
at org.opensaml.common.impl
我正在使用Security扩展来集成ACA医疗保健(又名奥巴马医改)网站。它使用IDP启动的SSO。SAML握手失败,输出如下
org.opensaml.xml.security.keyinfo.provider.InlineX509DataProvider] Single certificate was present, treating as end-entity certificate
org.opensaml.xml.security.keyinfo.BasicProviderKeyInfoCredentialResolver] Credentials successfully ext
在尝试使用OpenSAML完成单点登录时,我们看到以下错误。这是工作的,然后似乎停止当我们升级到Java 7(尽管这很可能是一个转移注意力的问题)。我们确实安装了JCE,所以不要认为它与此相关。以前有没有人遇到过这种情况?
java.lang.NoClassDefFoundError: Could not initialize class org.apache.commons.ssl.TrustMaterial
org.opensaml.xml.security.x509.X509Util.decodeCertificate(X509Util.java:319)
org.ope
错误:
Exception in thread "main" org.opensaml.xml.io.UnmarshallingException: the assertion does not appear to be encrypted
at AssertionDecrypter.unmarshallEncryptedAssertion(AssertionDecrypter.java:162)
at AssertionDecrypter.decryptAssertion(AssertionDecrypter.java:119)
at Assertio
我正在尝试用OpenSAML2 (2.6.6)来编写一个Java应用程序来解密加密的断言,但是我得到了:
[main] ERROR org.opensaml.xml.encryption.Decrypter - Failed to decrypt EncryptedKey, valid decryption key could not be resolved
[main] ERROR org.opensaml.xml.encryption.Decrypter - Failed to decrypt EncryptedData using either EncryptedData KeyInfo
我是WSO2 API Manager版本1.9.1的新手。我希望x-jwt-assertion可以使用OpenSAML库()解码。我希望同样的x-jwt-断言(在这个链接中显示)被解码/解析,但是当实现下面的代码时,我看到接下来会出现错误。请指点。
这是否可以使用x-jwt-assertion (API管理器)解码WSO2的APIM (API管理器)?
供参考的守则:
Exception in thread "main" org.xml.sax.SAXParseException; lineNumber: 1; columnNumber: 1; Content is not a
我一直试图让Spring示例应用程序启动并运行,但几天来我一直在苦苦挣扎,在互联网上搜索却没有成功。我遵循了“快速启动指南”中的所有步骤。当我单击'Start单登录‘按钮时,我被重定向到SSOCircle,我登录,然后被重定向回示例应用程序,但是它返回以下错误:
Message:
Could not initialize class org.apache.commons.ssl.TrustMaterial
StackTrace:
java.lang.NoClassDefFoundError: Could not initialize class org.apache.common
考虑一下下面的例子,似乎它在索引0中运行得很好:
$ a1=(1 2 3)
$ a2=(a b c)
$ for x in a1 a2; do echo "${!x}"; done
1
a
$ for x in a1 a2; do echo "${!x[0]}"; done
1
a
但是,对于索引1,它不会输出任何内容:
$ for x in a1 a2; do echo "${!x[1]}"; done
数组本身是可以的:
$ echo "${a1[1]} ${a2[1]}"
2 b
编辑-一个基于ilkkachu答案