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With a ZK6 tree model (like DefaultTreeModel/DefaultTreeNode) how do I setup the model so that all the parent nodes are open at start?
I've tried the following and it doesn't work.
<tree id="planningTree" width="100%"
sizedByContent="true" model="@load(vm.planningMenuModel)"
selectedItem="@bind(vm.planningMenuItem)"
onSelect="@global-command('setDataArea',DataViewCommand=vm.planningMenuItem.data)" >
<treecols>
<treecol label="" />
</treecols>
<template name="model">
<treeitem open="true" >
<treerow>
<treecell
label="${each.data.commandName}" />
</treerow>
</treeitem>
</template>
</tree>
Hi davout
I think you implemented tree view with MVVM. Can you please share your example code
@davout Can you please share your example code?
Thank you!
Some code from an example MVVM class:
setPlanningMenuModel(new DefaultTreeModel(aParent)); // set the model with the tree structure
int[] aPath = {0};
getPlanningMenuModel().addOpenPath(aPath);
Hi davout.
I'm not sure if the following code fits your needs... just try
public void doCollapseExpandAll(Component component, boolean open) {
if (component!=null) {
if (component instanceof Treeitem) {
Treeitem treeitem = (Treeitem) component;
treeitem.setOpen(open);
}
Collection<?> com = component.getChildren();
if (com != null) {
for (Iterator<?> iterator = com.iterator(); iterator.hasNext();) {
doCollapseExpandAll((Component) iterator.next(), open);
}
}
}
}
version zk 6.5
private void doExpandAll(Component component, boolean aufklappen) {
if (component instanceof Treeitem) {
Treeitem treeitem = (Treeitem) component;
this.treeModel.addOpenPath(this.treeModel.getPath((DefaultTreeNode) treeitem.getValue()));
}
Collection<?> com = component.getChildren();
if (com != null) {
for (Iterator<?> iterator = com.iterator(); iterator.hasNext();) {
doExpandAll((Component) iterator.next(), aufklappen);
}
}
}
Asked: 2012-04-18 18:34:37 +0800
Seen: 468 times
Last updated: May 27 '16
