无法找到与from-view-id'/ index.xhtml'匹配的导航案例以便采取行动:JSF

Question

我收到以下错误:无法找到与from-view-id'/index.xhtml'匹配的导航案例,用于行动'#{medcontroller.getMedGeneric}',结果为'javax.faces.model.ListDataModel@7a652236'

我是jsf的新手,我对解决这个错误真的很无能为力.我有一个ManagedBean,代码如下:

MedController.java

@ManagedBean(name = "medcontroller")
@SessionScoped

public class MedController implements Serializable {
    int startId;
    String gName;
    int endId;
    DataModel medNames;
    //DataModel medGeneric;
    MedicineHelper helper;
    private int recordCount = 1000;
    private int pageSize = 10;
    private Medicine current;
    private int selectedItemIndex;

    public MedController() {
        helper = new MedicineHelper();
        startId = 1;
        endId = 10;
    }

    public MedController(int startId, int endId) {
        helper = new MedicineHelper();
        this.startId = startId;
        this.endId = endId;
    }

    public Medicine getSelected() {
        if (current == null) {
            current = new Medicine();
            selectedItemIndex = -1;
        }
        return current;
    }

    public DataModel getMedNames() {
        if (medNames == null) {
            medNames = new ListDataModel(helper.getMedNames(startId, endId));
        }
        return medNames;
    }

    public String getgName()
        {
        return gName;
    }

    public void setgName(String gName)
        {
        this.gName = gName;
    }

    public DataModel getMedGeneric() {
        if (medNames == null) {
            medNames= new ListDataModel(helper.getMedGeneric(gName));
        }
        return medNames;
    }

    void recreateModel() {
        medNames = null;
    }

    public boolean isHasNextPage() {
        if (endId + pageSize <= recordCount) {
            return true;
        }
        return false;
    }

    public boolean isHasPreviousPage() {
        if (startId-pageSize > 0) {
            return true;
        }
        return false;
    }

    public String next() {
        startId = endId+1;
        endId = endId + pageSize;
        recreateModel();
        return "index";
    }

    public String previous() {
        startId = startId - pageSize;
        endId = endId - pageSize;
        recreateModel();
        return "index";
    }

    public int getPageSize() {
        return pageSize;
    }

   public String prepareView(){
        current = (Medicine) getMedNames().getRowData();
        return "browse";
    }
   public String prepareList(){
        recreateModel();
        return "index";
    }

}

这是我的JSF文件

的index.xhtml

<ui:define name="body">
    <h:form styleClass="jsfcrud_list_form">
    <h:commandLink action="#{medcontroller.previous}" value="Previous #{medcontroller.pageSize}" rendered="#{medcontroller.hasPreviousPage}"/> 
    <h:commandLink action="#{medcontroller.next}" value="Next #{medcontroller.pageSize}" rendered="#{medcontroller.hasNextPage}"/> 
    <h:dataTable value="#{medcontroller.medNames}" var="item" border="1" cellpadding="15" cellspacing="10" rowClasses="jsfcrud_odd_row,jsfcrud_even_row" rules="all" style="border:solid 1px">
    <h:column>
        <f:facet name="header">
            <h:outputText value="BrandName"/>
        </f:facet>
            <h:outputText value="#{item.brandName}"/>
    </h:column>
    <h:column>
        <f:facet name="header">
            <h:outputText value="Generic"/>
        </f:facet>
            <h:outputText value="#{item.generic}"/>
    </h:column>
    <h:column>
        <f:facet name="header">
            <h:outputText value=" "/>
        </f:facet>
            <h:commandLink action="#{medcontroller.prepareView}" value="View"/>
    </h:column>
    </h:dataTable>

    <h:inputText value="#{medcontroller.gName}" />
    <h:commandButton value="Submit" action="#{medcontroller.getMedGeneric}" />
    </h:form>
</ui:define>

请帮我解决错误.另外,我没有faces-config.xml文件.我正在使用netbeans ide 7.1.2 web应用程序与jsf和hibernate框架.

先感谢您.

Answer 1

在<h:commandButton action>必须指向一个方法,调用一些业务逻辑,再返回void或String代表你想要的目标页面(重新)显示.但是你返回了一个ListDataModel对JSF导航处理程序没有任何意义的整体,因此出现了这个错误.

这样的事情应该做:

public String getMedGeneric(){ 
    // Do your business logic here.

    return "someViewId";
}

这将导航到someViewId.xhtml.但是,如果您打算坚持使用相同的视图,只需让它返回void(或null),它将重新显示相同的视图.

public void getMedGeneric(){ 
    // Do your business logic here.
}

顺便说一下,为操作方法名称添加前缀实际上是一个糟糕的命名约定get.这令人困惑,使您的代码无法自我记录.而是将它命名为loadMedGeneric()左右.我只是猜测,因为你没有告诉任何具体的功能要求,究竟该按钮应该做什么.