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java - JXtreeTable 忽略树节点的 getValueAt

转载 作者:行者123 更新时间:2023-12-01 12:22:15 25 4
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我有一个 JXTreeTable,它的左侧有一棵节点树(不同的数据类型),右侧有一个网格(默认)。

现在我希望所有显示的值(还有树标题)均取自 getValueAt(row,column) 方法,而不是通过其 toString() 方法呈现树节点

如何强制 JXTreeTable 不采用特定于类的 toString() 并使用 getValueAt(...) 返回的值?

谢谢!

这是我的模型:

public class TreeTableGroupTreeTableModel extends AbstractTreeTableModel {

private final static String[] COLUMN_NAMES = { "Tree",
"Add/Edit", "Remove" };

//Group-Hierarchy:
private JXGroup groupRootObject;

public TreeTableGroupTreeTableModel(JXGroup groupRoot) {

this.groupRootObject = groupRoot;

}

// Here is my structure:
//GROUP 1-1 ModuleAggregation 1-* Modules 1-1 JXGroupAccessMemberAggregation 1-* JXGroupAccessSimpleUserObjectLeaf

/**
* Returns the value for the node at columnIndex. The node must be managed
* by this model. Unamanaged nodes should throw an IllegalArgumentException.
*
*
* node - the node whose value is to be queried column - the column whose
* value is to be queried
*/
@Override
public Object getValueAt(Object node, int column) {

System.out.println(node.getClass());

if (node instanceof JXGroup) {
switch (column) {
case 0:
return "this should render this string instead of the class tostring()";//((JXGroup) node);
case 1:
case 2:
return "";
}
} else if (node instanceof JXGroupAccessModuleAggregation) {
switch (column) {
case 0:
return "Modules";
case 1:
return "ADD";
case 2:
return "REMOVE ALL";
}
} else if (node instanceof JXGroupAccessModule) {
switch (column) {
case 0:
return ((JXGroupAccessModule) node).getModule();
case 1:
return "ADD";
case 2:
return "REMOVE";
}
} else if (node instanceof JXGroupAccessMemberAggregation) {
switch (column) {
case 0:
return "Members";
case 1:
return "ADD";
case 2:
return "REMOVE ALL";
}
} else if (node instanceof JXGroupAccessSimpleUserObjectLeaf) {
switch (column) {
case 0:
SimpleUserObject c = FunctionsUser.getUser(
metaDataPackage.getUsers(),
((JXGroupAccessSimpleUserObjectLeaf) node).getUserId().getId());
return c.getShowAs();
case 1:
return "";
case 2:
return "REMOVE";
}
}

return null;
}

/**
* Sets the value for the node at columnIndex to value. The node must be
* managed by this model. Unamanaged nodes should throw an
* IllegalArgumentException.
*
* value - the new value node - the node whose value is to be changed column
* - the column whose value is to be changed
*/
@Override
public void setValueAt(Object value, Object node, int column) {

return; // changes are not allowed

}

@Override
public Object getRoot() {
return groupRootObject;
}

@Override
public int getColumnCount() {
return COLUMN_NAMES.length;
}

@Override
public String getColumnName(int column) {
return COLUMN_NAMES[column];
}

@Override
public boolean isCellEditable(Object node, int column) {
return false;
}

@Override
public boolean isLeaf(Object node) {
return node instanceof JXGroupAccessCategoryLeaf
|| node instanceof JXGroupAccessChapterLeaf
|| node instanceof JXGroupAccessEntryLeaf
|| node instanceof JXGroupAccessSimpleUserObjectLeaf;
}

@Override
public int getChildCount(Object obj) {

if (!(obj instanceof JXICountable)) {
System.out.println("parsing error. should be JXCountable (but is "
+ obj.getClass() + ")");
return 0;
}

JXICountable obj2 = (JXICountable) obj;

return obj2.getChildrenCount();

}

@Override
public Object getChild(Object parent, int index) {
if (parent instanceof JXITreeNodeAble) {
return ((JXITreeNodeAble) parent).getChild(index);
}
throw new IllegalArgumentException(
"parent is not a member of JXITreeNodeAble");
}

@Override
public int getIndexOfChild(Object parent, Object child) {
if (parent instanceof JXITreeNodeAble) {
return ((JXITreeNodeAble) parent).getIndexOfChild(child);
}
throw new IllegalArgumentException(
"parent is not a member of JXITreeNodeAble");
}
}

最佳答案

首先请注意,树表是最复杂的组件之一,因为节点必须保存大量数据:

  • 层级:赋予父子关系。
  • 用户数据:能够通过 setUserObject(..)getUserObject() 方法保留和检索用户特定的对象。
  • 列数据:每个节点负责检索与每个列索引关联的数据,以便提供以表格形式表示的能力。

此外,所有节点可能保存不同的数据对象,因此问题变得更加复杂。

更新

现在您已经发布了代码和 source从它的基础来看,请注意,实现旨在没有根(您遇到问题的节点)并直接与域对象一起工作,以避免处理TreeTableNodes。有一个subsequent post其中作者修改了它的树表模型实现以包含根节点。话虽如此,您至少应该修改两件事:

  1. 不要重写 getRoot() 方法并让实现继承自 AbstractTreeTableModel
  2. 更改类构造函数,调用适当的super构造函数。

在代码中:

public class TreeTableGroupTreeTableModel extends AbstractTreeTableModel {
...
// private JXGroup groupRootObject; Not needed!
...
public TreeTableGroupTreeTableModel(JXGroup groupRoot) {
super(groupRoot);
}
...
// @Override
// public Object getRoot() {
// return groupRootObject;
// }
...
}

虽然这可能有效(如果没有您的所有类(class),我无法测试它),但我想再次强调,这种实现树表模型的方法对我来说并不合适。我一开始就不会搞乱树表模型,而是使用 DefaultTreeTableModel 来代替。我还将为需要在树表中表示的每个域类实现一个树表节点。例如:

class JXGroupAccessModuleTreeTableNode extends AbstractMutableTreeTableNode {

public JXGroupAccessModuleTreeTableNode(JXGroupAccessModule module) {
super(module);
}

public JXGroupAccessModuleTreeTableNode(JXGroupAccessModule module, boolean allowsChildren) {
super(module, allowsChildren);
}

@Override
public Object getValueAt(int column) {
JXGroupAccessModule module = (JXGroupAccessModule)getUserObject();
switch (column) {
case 0: return module.getModule();
case 1: return "ADD";
case 2: return "REMOVE";
default: throw new ArrayIndexOutOfBoundsException(column);
}
}

@Override
public int getColumnCount() {
return 3;
}
}

要允许单元格可编辑,您还必须覆盖 setValueAt(Object value, int column)isEditable(column) 方法:默认情况下 AbstractMutableTreeTableNodes 不可编辑。

按照这种方法,您将获得以下好处:

  • 根本不需要搞乱树表模型。
  • 如果需要更改树结构,您只需根据需要重新排序节点即可。再次无需修改任何模型的代码。
  • 每个节点的表示与模型分离,因此您可以用不同的方式表示同一域对象,只需覆盖树表节点(而非模型)中的几个方法即可。

关于java - JXtreeTable 忽略树节点的 getValueAt,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26591255/

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