我正在从dropbox api中检索文件树。在api中,每个文件夹都使用单独的api调用来读取,因此我将遍历整个文件树以获取所有文件夹。这是通过cron-job完成的。
从dropbox检索数据的函数如下所示:
function renderFolderTree($myobject, $path){
$entry = $myobject->getMetadataWithChildren($path);
foreach ($entry['contents'] as $child) {
if ($child['is_dir']){
$folderpath = $child['path'];
//this will retrieve the child-folder
renderFolderTree($myobject, $folderpath, $filetree);
//here I need something that saves the folder
}else{
print_r($child);
//here I need something that saves the file
}
}
}
我想将文件树保存到postgres数据库,以便以后可以将其作为代表它的json对象输出。
我是数据库设计的新手,不知道保存数据的方法。我假设每个文件和文件夹都应该有自己的数据库条目。我可以让每个孩子都引用它的父母ID,或者我可以让每个父母都包含它的孩子列表。
我是新手,我想要一个相当简单的解决方案,阅读速度比写作更重要!
答案 0 :(得分:5)
在关系数据库中存储树有几种选择。 为了更好地概述,我推荐Bill Karwin的slides。
由于您提到读取速度最重要,因此闭包表将是一种适当的强大编码。闭包表是多对多关系,其为每个路径(例如,/ a / b / c)存储所有父/子(传递)。这样,可以使用一个SQL查询(非递归)完成对树的许多查询。
那看起来像
create table nodes (
path varchar primary key
/* your other attributes here, can be null */
);
create table parents_children (
parent_path varchar,
child_path varchar,
primary key(parent_path,child_path),
foreign key (parent_path) references nodes (path),
foreign key (child_path) references nodes (path)
);
要在目录/ a / b /下插入新文件/ a / b / c,您可以这样做:
insert into nodes values ('/a/b/c');
insert into parents_children
select parent_path, '/a/b/c' from parents_children where child_path = '/a/b/'
union all select '/a/b/c','/a/b/c';
例如,查询来自' / a'的所有儿童。在递归上,你会这样做:
select *
from nodes join parents_children on path = child_path
where parent_path = '/a';
更详尽的示例,它存储以下文件树:
/
/a/
/a/b/
/a/b/d
/a/c
/b
要插入数据:
insert into nodes values ('/');
insert into parents_children values ('/','/');
insert into nodes values ('/a/');
insert into parents_children
select parent_path, '/a/' from parents_children where child_path = '/'
union all select '/a/','/a/';
insert into nodes values ('/a/b/');
insert into parents_children
select parent_path, '/a/b/' from parents_children where child_path = '/a/'
union all select '/a/b/','/a/b/';
insert into nodes values ('/a/c');
insert into parents_children
select parent_path, '/a/c' from parents_children where child_path = '/a/'
union all select '/a/c','/a/c';
insert into nodes values ('/a/b/d');
insert into parents_children
select parent_path, '/a/b/d' from parents_children where child_path = '/a/b/'
union all select '/a/b/d','/a/b/d';
insert into nodes values ('/b');
insert into parents_children
select parent_path, '/b' from parents_children where child_path = '/'
union all select '/b','/b';
查询/ a /
的所有子项select node.*
from nodes join parents_children on path = child_path
where parent_path = '/a/';
path
----------
/a/
/a/b/
/a/b/d
/a/c