小编npa*_*ace的帖子

我们正在使用一个oracle数据库,其中设置它的人"已经不在了",因此不知道sysdba密码,但需要它.我们有root访问权限(在Linux上).有没有办法恢复或更改系统密码？

我试图通过Bluemix部署Java应用程序.当我在本地部署它时Liberty Server(Liberty 16.0.0.2),它工作正常,我能够从SOAP客户端点击URL .

但是当我在Bluemix上尝试它时,它给了我这个错误:

应用程序类'com.sun.xml.messaging.saaj.soap.SOAPPartImpl.:119'抛出的异常java.lang.VerifyError:JVMVRFY012堆栈形状不一致; class = com/sun/xml/messaging/saaj/soap/SOAPDocumentImpl,method = createDocumentFragment()Lorg/w3c/dom/DocumentFragment;,pc = 5; 在签名com/sun/xml/messaging/saaj/soap/SOAPDocumentFragment中键入Mismatch,参数0.:( Lcom/sun/org/apache/xerces/internal/dom/CoreDocumentImpl;)V与Exception Details不匹配:Location:com /sun/xml/messaging/saaj/soap/SOAPDocumentImpl.createDocumentFragment()Lorg/w3c/dom/DocumentFragment; @ 5:JBinvokespecial Reason:键入'com/sun/xml/messaging/saaj/soap/SOAPDocumentImpl'(当前帧,堆栈[2])不能分配给'com/sun/org/apache/xerces/internal/dom/CoreDocumentImpl'当前帧:bci:@ 5 flags:{} locals:{'com/sun/xml/messaging/saaj/soap/SOAPDocumentImpl'} stack:{'uninitialized','uninitialized','com/sun/xml/com.sun.xml.messaging.saaj.soap.SOAPPartImpl.(SOAPPartImpl.java:119)中的messaging/saaj/soap/SOAPDocumentImpl'},位于com.sun.xml.messaging.saaj.soap.ver1_1.SOAPPart1_1Impl.(SOAPPart1_1Impl的.java:89)在com.sun.xml.messaging.saaj.soap.ver1_1.Message1_1Impl.getSOAPPart(Message1_1Impl.java:109)在org.apache.cxf.binding.soap.saaj.SAAJInInterceptor.handleMessage(SAAJInInterceptor.java :101)在org.apache.cxf.jaxws.handler.soap.SOAPMessageContextImpl.getMessage(SOAPMessageContextImpl.java:78)在org.apache.cxf.jaxws.handler.soap.SOAPHandlerInterceptor.getOpQName(SOAPHandlerInterceptor.java:294)在org.apache.cxf.jaxws.handler.AbstractJAXWSHandlerInterceptor.s etupBindingOperationInfo(AbstractJAXWSHandlerInterceptor.java:111)org.apache.cxf.jaxws.handler.soap.SOAPHandlerInterceptor.createProtocolMessageContext(SOAPHandlerInterceptor.java:235)at org.apache.cxf.jaxws.handler.soap.SOAPHandlerInterceptor.handleMessageInternal(SOAPHandlerInterceptor.的java:144)在org.apache.cxf.jaxws.handler.soap.SOAPHandlerInterceptor.handleMessage(SOAPHandlerInterceptor.java:119)在org.apache.cxf.jaxws.handler.soap.SOAPHandlerInterceptor.handleMessage(SOAPHandlerInterceptor.java:69)在org.apache.cxf.phase.PhaseInterceptorChain.doIntercept(PhaseInterceptorChain.java:236)在org.apache.cxf.transport.ChainInitiationObserver.onMessage(ChainInitiationObserver.java:89)在org.apache.cxf.transport.servlet.ServletDestination .invoke(ServletDestination.java:99)在org.apache.cxf.transport.servlet.ServletController.invokeDestination(ServletController.java:368)在org.apache.cxf.transport.servlet.ServletController.invoke(ServletController.java:183 )org.apache.cxf.t ransport.servlet.AbstractCXFServlet.invoke(AbstractCXFServlet.java:163)在org.apache.cxf.transport.servlet.AbstractCXFServlet.doGet(AbstractCXFServlet.java:145)在javax.servlet.http.HttpServlet.service(HttpServlet.java: 687)在[内部类]的com.ibm.ws.webcontainer.servlet.ServletWrapper.service(ServletWrapper.java:1290)的javax.servlet.http.HttpServlet.service(HttpServlet.java:790)

我已经尝试部署打包的自由服务器和服务器目录选项.但它不起作用.我也试着用ibm-web-ext.xml方法改变 fileServingEnabled="false".但它给出了同样的错误.

我正在研究填字游戏算法来开发一个文字应用程序。经过大量谷歌搜索或在 StackOverflow 上搜索后，我终于达到了这一点。但我还无法理解 Java 中算法的正确实现。下面是我使用的类。

public class Crosswords {

    char[][] cross;
    int rows;
    int cols;
    char[][] numberGrid;
    boolean startword;
    final char DEFAULT = ' ';

    public Crosswords() {
        rows = 50;
        cols = 50;
        cross = new char[rows][cols];
        numberGrid = new char [rows][cols];
        for (int i = 0; i < cross.length;i++){
            for (int j = 0; j < cross[i].length;j++){
                cross[i][j] = DEFAULT;
            }
        }
    }

    public Crosswords(int ros, int colls) {
        rows = ros;
        cols = colls;
        cross = new …

有人能告诉我将数组转换为树状结构的最有效方法吗？

var array= [
        {id: "1",     name: "header1"},
        {id: "2",     name: "header2"},
        {id: "1.1",   name: "subheader1.1"},
        {id: "1.2",   name: "subheader1.2"},
        {id: "2.1",   name: "subheader2.1"},
        {id: "2.2",   name: "subheader2.2"},
        {id: "1.1.1", name: "subheader1detail1"},
        {id: "2.1.1", name: "subheader2detail2"}
];

结果数组必须如下:

var array = [{
    id: "1",
    name: "header1",
    items: [{
        id: "1.1",
        name: "subheader1.1",
        items: [{
            id: "1.1.1",
            name: "subheader1detail1",
        }]
    }, {
        id: "1.2",
        name: "subheader1.2"
    }]
}, {
    id: "2",
    name: "header2",
    items: [{
        id: "2.1",
        name: "subheader2.1",
        items: [{ …

假设我有一个类Parent,它有四个字段A,B,C和D,这样C和D可选地使用默认实现传递或初始化:

open class Parent(val a: A, val b: B, val c: C, val d: D) {
    constructor(a: A, b: B, c: C): this(a, b, c, DImpl()){}
    constructor(a: A, b: B): this(a, b, CImpl(), DImpl()){}
}

我需要扩展这个类并向子类添加另一个字段:

class Child: Parent {
     val e: E // How do I initialize this?
}

传递val给辅助构造函数不起作用,也不使用init{}块.

传递val给主构造函数可以工作,但后来我失去了对Parent类中辅助构造函数的委托- 我需要使用所有Parent构造函数和所有参数,或者复制辅助构造函数,将实现细节泄露给Child类.

这应该很简单,我在这里遗漏了什么？

我正在处理以瑞典格式编写的数据集.在瑞典使用逗号代替十进制数字的点.

我的数据集是这样的:

1,188,1,250,0,757,0,946,8,960

1,257,1,300,0,802,1,002,9,485

1,328,1,350,0,846,1,058,10,021

1,381,1,400,0,880,1,100,10,418

我希望将其他所有逗号更改为指向并输出如下:

1.188,1.250,0.757,0.946,8.960

1.257,1.300,0.802,1.002,9.485

1.328,1.350,0.846,1.058,10.021

1.381,1.400,0.880,1.100,10.418

有关如何使用简单的shell脚本执行此操作的任何想法.这很好如果我分多步完成.我的意思是如果我首先更改逗号的第一个实例,然后更改第三个实例和...

非常感谢您的帮助.