问题
I have a domain
class Node {
String nodeId
String label
Node parent
}
In the GSP page I want to start with the root and print all its children (note I have a reference to parent and not the child). Any idea how to do this ? I am sorry but I am new and really confused. I want to print everything going down from root(not root) and root has no parent(its null). So i have written
<g:each in="${nodes}" var="node">
<g:if test="${node.parent!=null}">
${node.label}
<g:render template="node" model="[nodes:Node.findAllByParent(node)]" />
</g:if>
</g:each>
In the above code not sure what parent_node_intance should be. The nodes list starts with root. I dont want to print that but start from everything else that has root as parent.
node.gsp
<g:if test="${nodes}">
<ul>
<g:each in="${nodes}" var="node">
<li>
${node.label}
<g:render template="node" model="[nodes:Node.findAllByParent(node)]" />
</li>
</g:each>
</ul>
</g:if>
Getting the following error which I am sure is caused by the root
2014-10-02 12:28:21,693 [http-bio-8080-exec-1] ERROR errors.GrailsExceptionResolver - NullPointerException occurred when processing request: [GET] /taxonomy - parameters:
outputFormat: concept
hierarchyId: lp
Cannot invoke method findAllByParent() on null object. Stacktrace follows:
Message: Error evaluating expression [[nodes:Node.findAllByParent(node)]] on line [11]: Cannot invoke method findAllByParent() on null object
Line | Method
->> 11 | run in C:/Users/U6021072/Documents/workspace-ggts-3.6.0.RELEASE/ae-and-sdx-analysis-ui/target/work/plugins/taxonomy-toolkit-for-grails-0.02-SNAPSHOT/grails-app/views/taxonomy/concept.gsp
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Caused by NullPointerException: Cannot invoke method findAllByParent() on null object
->> 11 | doCall in C__Users_U6021072_Documents_workspace_ggts_3_6_0_RELEASE_ae_and_sdx_analysis_ui_target_work_plugins_taxonomy_toolkit_for_grails_0_02_SNAPSHOT_grails_app_views_taxonomy_concept_gsp$_run_closure4
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
| 198 | doFilter in grails.plugin.cache.web.filter.PageFragmentCachingFilter
| 63 | doFilter in grails.plugin.cache.web.filter.AbstractFilter
| 1145 | runWorker in java.util.concurrent.ThreadPoolExecutor
| 615 | run . . . in java.util.concurrent.ThreadPoolExecutor$Worker
^ 745 | run in java.lang.Thread
回答1:
So, you want to print all the nodes that have the known parent?
What about something like:
<g:each in="${Node.findAllByParent(parent_node_instance)}" var="node">
${node}<br>
</g:each>
But since you are talking about recursion, I think you also want to print all descendants.
Create a _node.gsp
template:
<g:each in="${nodes}" var="node">
${node}<br>
<g:render template="node" model="[nodes:Node.findAllByParent(node)]" />
</g:each>
And call with:
<g:render template="node" model="[nodes:Node.findAllByParent(parent_node_instance)]" />
You can easily add a depth variable to the model to indent each generation properly, or use the <ul>
and <li>
elements in the template.
parent_node_instance
is the root you want to start printing your tree from, it can be the absolute root or any node in the tree.
findAllByParent()
is a dynamic finder function. See http://grails.org/doc/latest/guide/GORM.html#finders for details.
回答2:
you can also use the @collection
/@bean
:
_node.gsp
<div class="fancy nested">
id:${it.nodeId} - label:${it.label}
<g:render template="node" collection="${Node.findAllByParent(it.node)}" />
</div>
main.gsp
<g:render template="node" bean="${rootNode}" />
来源:https://stackoverflow.com/questions/26129073/recursion-in-gsp-page