我在尝试从一个节点获取所有祖先时遇到了一点麻烦;
这是我的计划。yml:
Constante:
connection: doctrine
tableName: constante
actAs:
NestedSet:
hasManyRoots: true
rootColumnName: parent_id
columns:
id:
type: integer(8)
fixed: false
unsigned: false
primary: true
autoincrement: true
parent_id:
type: integer(8)
fixed: false
unsigned: false
primary: false
notnull: true
autoincrement: false
lft:
type: integer(8)
fixed: false
unsigned: false
primary: false
notnull: true
autoincrement: false
rgt:
type: integer(8)
fixed: false
unsigned: false
primary: false
notnull: true
autoincrement: false
level:
type: integer(8)
fixed: false
unsigned: false
primary: false
notnull: true
autoincrement: false
cod_interno:
type: string(5)
fixed: false
unsigned: false
primary: false
notnull: false
autoincrement: false
nombre:
type: string(64)
fixed: false
unsigned: false
primary: false
notnull: true
autoincrement: false
这就是我试图从一个节点(不是根)获得所有祖先的方法
$path = Doctrine_Core::getTable('Constante')->find($condicion); // $condicion = 57
$node = $path->getNode();
$isLeaf = $node->isLeaf(); //var_dump throws true
$ancestors = $node->getAncestors(); //var_dump throws false
$isValidNode = $node->isValidNode(); //var_dump throws true
作为$ancestors == false
,我无法对其进行迭代并获得所有祖先(我正在尝试构建一个简单的面包屑)
这是我在数据库中存储的,这是真实的数据(仅用于测试小狗)
+---------++---------++---------++---------++----------++---------+
|ID ||PARENT_ID||LFT ||RGT ||LEVEL ||NOMBRE |
|---------||---------||---------||---------||----------||---------|
|56 ||56 ||1 ||4 ||0 ||COUNTRY | --> this is root
|57 ||56 ||2 ||3 ||1 ||CANADA | --> child of root
+---------++---------++---------++---------++----------++---------+
据此,如果Ancestors返回false,则表示所选节点是根节点。
我花了几个小时寻找解决方案,但没有成功。
如果您需要进一步的信息,请随时询问!
编辑:我在键入表中的内容时犯了一个错误,多亏了olivierw提醒我这一点。
表中的rgt
字段似乎有错误。如果id 56
是根,则它应该具有rgt = 4
,并且id 57
应该具有rgt = 3
。所以你的表格应该是:
+---------++---------++---------++---------++----------++---------+
|ID ||PARENT_ID||LFT ||RGT ||LEVEL ||NOMBRE |
|---------||---------||---------||---------||----------||---------|
|56 ||56 ||1 ||4 ||0 ||COUNTRY |
|57 ||56 ||2 ||3 ||1 ||CANADA |
+---------++---------++---------++---------++----------++---------+
所以你会得到正确的祖先。
我想分享我的解决方案,希望它对其他人有用:
行动:
//action.class.php
public function executeIndex(sfWebRequest $request) {
$condicion = $request->getParameter('id') ? $request->getParameter('id') : 0;
if ($condicion > 0) {
$path = $tree = Doctrine_Core::getTable('Constante')->find($condicion);
} else {
$tree = Doctrine_Core::getTable('Constante');
$path = null;
}
$this->constantes = $tree;
$this->path = $path;
$this->setTemplate('index');
}
视图:
//indexSuccess.php
<?php
$var = (is_null($sf_request->getParameter('id'))) ? $constantes->getTree()->fetchRoots() : $constantes->getNode()->getChildren();
?>
<?php if ($var): foreach ($var as $constante): ?>
<tr>
<td><?php echo $constante->getNombre() ?></td>
</tr>
<?php
endforeach;
endif;
?>
部分:
//_breadcrumb.php
<?php
if ($path) {
echo '<ul class="breadcrumb">';
$node = $path->getNode();
$ancestors = $node->getAncestors();
if ($ancestors)
foreach ($ancestors AS $ancestor) {
echo '<li>' . link_to($ancestor->getNombre(), 'constantes/more?id=' . $ancestor->getId()) . '<span class="divider">/</span></li>';
}
echo '</ul>';
}
?>
我认为代码是不言自明的,但如果你有任何问题,请告诉我!