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View Code? Open in Web Editor NEWA closure table tree package for laravel eloquent
License: MIT License
A closure table tree package for laravel eloquent
License: MIT License
$menu->getAncestorsAndSelf(); 如何逐级顺序输出
$multiTree = Menu::all()->toTree();
var_dump($multiTree);
返回[]
跟我另外那个问题非常相似,执行返回true,但是没有建立任何数据,但perfectNode就可以的。
建议支持 id 可以为 UUID 或者 ULID
$t = $root->getTree([], 'children'); //ok, 所有字段都返回
$t = $root->getTree([], 'children', ['id', 'name']); //返回空数组
PHP7 Laravel5.8 自定义parent为parent_id:
使用$menu->getRoot();返回为空,但是改成getRoots就可以返回多节点,搞不懂是为什么,我想通过子节点获取它的root然后生成树,但是始终返回为空。
应用场景如下:存在大量节点的时候,考虑性能不能一次性展开所有节点,但是如果有初始访问节点,我希望展开以这个节点为脉络的所有父节点,其他节点默认保持关闭,等用户点击再展开。
同样的代码mysql没问题,pgsql出问题了
"message": "SQLSTATE[22P02]: Invalid text representation: 7 ERROR: invalid input syntax for type bigint: \"tree\" (SQL: select * from \"categories\" where \"categories\".\"id\" = tree and \"categories\".\"deleted_at\" is null limit 1)"
层级越深
每插入一个node就会增加很多条数据
单表很快就会满了
如果分表,那么按什么维度拆、怎样才能均匀分布?
if (!isset($data[0][$parent])) {
return [];
}
换成
if (!key_exists($parent, $data[0])) {
return [];
}
如何看这个节点是一级还是二级呢?我想把这个扩展用作分销绑定里面
如果在laravel中设置了表前缀,
$query = " INSERT INTO {$table} ({$ancestorColumn}, {$descendantColumn}, {$distance}) SELECT tbl.{$ancestorColumn}, {$descendantId}, tbl.{$distance}+1 FROM {$table} AS tbl WHERE tbl.{$descendantColumn} = {$ancestorId} UNION ALL SELECT {$descendantId}, {$descendantId}, 0 ON DUPLICATE KEY UPDATE {$distance} = VALUES ({$distance}) ";
和
public function getRoot(array $columns = ['*']) { $parentColumn = $this->getParentColumn(); return $this ->joinRelationBy('ancestor') ->where($parentColumn, 0) ->orWhere(function ($query) use ($parentColumn) { $query->whereNull($parentColumn); }) ->first($columns); }
2种操作数据库方法会冲突,望处理下,3qs
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