我试图浏览一堆带有其他对象链接的对象。我想从最低的id号（根对象）开始，并根据连接的链接浏览每个对象。一些对象链接将循环回到以前的对象，所以我想确保只查看每个对象链接，否则我将陷入无限循环。我还希望能够通过从第一个链接开始浏览链接来判断哪些对象无法访问。
我数据库中的表格如下所示：
对象表：
+----+---------+
| id | title |
+----+---------+
| 1 | Apple |
| 3 | Carrot |
| 4 | Dill |
| 5 | Egg |
| 6 | Fred |
| 7 | Goat |
| 8 | Harry |
| 9 | Igloo |
| 10 | Jason |
| 11 | Klaus |
| 12 | Banana |
| 15 | Oyster1 |
| 16 | Oyster2 |
+----+---------+
Object_Links表：
+----+---------+--------------+
| id | obj_id | obj_link_id |
+----+---------+--------------+
| 1 | 1 | 12 |
| 2 | 1 | 5 |
| 3 | 3 | 1 |
| 4 | 3 | 12 |
| 5 | 3 | 3 |
| 6 | 4 | 1 |
| 7 | 4 | 5 |
| 8 | 5 | 6 |
| 9 | 6 | 7 |
| 10 | 7 | 7 |
| 11 | 7 | 8 |
| 12 | 9 | 12 |
| 13 | 9 | 5 |
| 14 | 10 | 1 |
| 15 | 10 | 5 |
| 16 | 10 | 8 |
| 17 | 11 | 1 |
| 18 | 11 | 5 |
| 19 | 11 | 10 |
| 20 | 12 | 3 |
| 21 | 15 | 16 |
| 22 | 16 | 15 |
+----+---------+--------------+
因此，从表中可以看到对象1具有指向对象12和5的链接。
我的SQL查询如下所示：
select object.id, title, obj_link_id
from object
left join object_links ON object.id = object_links.object_id
order by object.id
给出了表格：
+----+---------+--------------+
| id | title | obj_link_id |
+----+---------+--------------+
| 1 | Apple | 12 |
| 1 | Apple | 5 |
| 3 | Carrot | 1 |
| 3 | Carrot | 12 |
| 3 | Carrot | 3 |
| 4 | Dill | 1 |
| 4 | Dill | 5 |
| 5 | Egg | 6 |
| 6 | Fred | 7 |
| 7 | Goat | 7 |
| 7 | Goat | 8 |
| 8 | Harry | NULL |
| 9 | Igloo | 12 |
| 9 | Igloo | 5 |
| 10 | Jason | 1 |
| 10 | Jason | 5 |
| 10 | Jason | 8 |
| 11 | Klaus | 1 |
| 11 | Klaus | 5 |
| 11 | Klaus | 10 |
| 12 | Banana | 3 |
| 15 | Oyster1 | 16 |
| 16 | Oyster2 | 15 |
+----+---------+--------------+
在PHP中我使用：
$objects = $stmt->fetchAll(PDO::FETCH_CLASS);
我不确定是否有更好的方法来获取这些用于我的目的，所以我愿意接受建议。
A print_r($objects)产生：
Array
(
[0] => stdClass Object
(
[id] => 1
[title] => Apple
[obj_link_id] => 12
)
[1] => stdClass Object
(
[id] => 1
[title] => Apple
[obj_link_id] => 5
)
[2] => stdClass Object
(
[id] => 3
[title] => Carrot
[obj_link_id] => 1
)
[3] => stdClass Object
(
[id] => 3
[title] => Carrot
[obj_link_id] => 12
)
[4] => stdClass Object
(
[id] => 3
[title] => Carrot
[obj_link_id] => 3
)
[5] => stdClass Object
(
[id] => 4
[title] => Dill
[obj_link_id] => 1
)
[6] => stdClass Object
(
[id] => 4
[title] => Dill
[obj_link_id] => 5
)
[7] => stdClass Object
(
[id] => 5
[title] => Egg
[obj_link_id] => 6
)
[8] => stdClass Object
(
[id] => 6
[title] => Fred
[obj_link_id] => 7
)
[9] => stdClass Object
(
[id] => 7
[title] => Goat
[obj_link_id] => 7
)
[10] => stdClass Object
(
[id] => 7
[title] => Goat
[obj_link_id] => 8
)
[11] => stdClass Object
(
[id] => 8
[title] => Harry
[obj_link_id] =>
)
[12] => stdClass Object
(
[id] => 9
[title] => Igloo
[obj_link_id] => 12
)
[13] => stdClass Object
(
[id] => 9
[title] => Igloo
[obj_link_id] => 5
)
[14] => stdClass Object
(
[id] => 10
[title] => Jason
[obj_link_id] => 1
)
[15] => stdClass Object
(
[id] => 10
[title] => Jason
[obj_link_id] => 5
)
[16] => stdClass Object
(
[id] => 10
[title] => Jason
[obj_link_id] => 8
)
[17] => stdClass Object
(
[id] => 11
[title] => Klaus
[obj_link_id] => 1
)
[18] => stdClass Object
(
[id] => 11
[title] => Klaus
[obj_link_id] => 5
)
[19] => stdClass Object
(
[id] => 11
[title] => Klaus
[obj_link_id] => 10
)
[20] => stdClass Object
(
[id] => 12
[title] => Banana
[obj_link_id] => 3
)
[21] => stdClass Object
(
[id] => 15
[title] => Oyster1
[obj_link_id] => 16
)
[22] => stdClass Object
(
[id] => 16
[title] => Oyster2
[obj_link_id] => 15
)
)
请注意，括号中的数字只是数组索引，而不是对象ID号，所以不要让索引让你失望。
我正在尝试找到一种方法来确定哪些是链接的，哪些是未链接的对象。基于上述场景，对象应按如下方式分开：
**Linked:**
Apple
Banana
Carrot
Egg
Fred
Goat
Harry
**Not Linked:**
Dill
Igloo
Jason
Klaus
Oyster1
Oyster2
我的主要问题：
如何在PHP中创建循环来循环遍历这样的结构，尤其是当每个对象可以有多个链接时？最后，我想生成两个对象集合，一个包含链接对象，另一个包含未链接对象。样本集可能如下所示：
stdClass Object
(
[LinkedElements] => stdClass Object
(
[1] => stdClass Object
(
[id] => 1
[name] => Apple
[link] => Array
(
[0] => 14
[1] => 5
)
)
[14] => stdClass Object
(
[id] => 14
[name] => Banana
[link] => Array
(
[0] => 3
)
)
[3] => stdClass Object
(
[id] => 3
[name] => Carrot
[link] => Array
(
[0] => 1
[1] => 14
[2] => 3
)
)
[5] => stdClass Object
(
[id] => 5
[name] => Egg
[link] => Array
(
[0] => 6
)
)
[6] => stdClass Object
(
[id] => 6
[name] => Fred
[link] => Array
(
[0] => 7
)
)
[7] => stdClass Object
(
[id] => 7
[name] => Goat
[link] => Array
(
[0] => 7
[1] => 8
)
)
[8] => stdClass Object
(
[id] => 8
[name] => Harry
)
)
[UnLinkedElements] => stdClass Object
(
[4] => stdClass Object
(
[id] => 4
[name] => Dill
[link] => Array
(
[0] => 1
[1] => 5
)
)
[9] => stdClass Object
(
[id] => 9
[name] => Igloo
[link] => Array
(
[0] => 14
[1] => 5
)
)
[10] => stdClass Object
(
[id] => 10
[name] => Jason
[link] => Array
(
[0] => 1
[1] => 5
[2] => 8
)
)
[11] => stdClass Object
(
[id] => 11
[name] => Klaus
[link] => Array
(
[0] => 1
[1] => 5
[2] => 10
)
)
[15] => stdClass Object
(
[id] => 15
[name] => Oyster1
[link] => Array
(
[0] => 16
)
)
[16] => stdClass Object
(
[id] => 16
[name] => Oyster2
[link] => Array
(
[0] => 15
)
)
)
)
请注意：
答案 0 :(得分：1)
这是一个graph traversal问题。从节点（根）开始，您希望遍历图形，跟踪沿途访问的每个节点。遍历结束后，访问过的已连接。广度优先搜索可以通过以下方式完成：
//To form a graph fetch all objects from the database (sorted by id) and
//index them in a hash map
$objects = $stmt->fetchAll(PDO::FETCH_OBJ);
$nodes = [];
foreach ($objects as $object) {
$nodes[$object->id] = new Node($object);
}
//fetch all connections from the database and link the objects
$links = $stmt->fetchAll(PDO::FETCH_OBJ);
foreach ($links as $link) {
$nodes[$link->obj_id]->addLink($nodes[$link->obj_link_id]);
}
//let's assume root is the first node (sorted by id),
//traverse the graph starting from root
$root = reset($nodes);
$root->traverse();
//now if a node can be reached by the root it is marked as visited
$linked = [];
$notLinked = [];
foreach ($nodes as $node) {
if ($node->isVisited()) {
$linked[] = $node;
} else {
$notLinked[] = $node;
}
}
节点类：
class Node
{
/**
* List of neighbor nodes.
*
* @var Node[]
*/
private $links = [];
/**
* id, title info
*
* @var array
*/
private $data = [];
/**
* To track visited nodes.
*
* @var bool
*/
private $visited = false;
/**
* Node constructor.
* @param array $data
*/
public function __construct($data)
{
$this->data = $data;
}
/**
* Add a link to this node.
*
* @param Node $node
* @return void
*/
public function addLink(Node $node)
{
$this->links[] = $node;
}
/**
* Traverse the graph in a Breadth-First-Search fashion marking
* every visited node.
* @return void
*/
public function traverse()
{
//initialize queue
$q = new SplQueue();
//add root to queue and mark as visited
$q->enqueue($this);
$this->visited = true;
while (!$q->isEmpty()) {
/** @var Node $cur */
$cur = $q->dequeue();
foreach ($cur->links as $link) {
//if link not visited already add it to queue and mark visited
if (!$link->visited) {
$link->visited = true;
$q->enqueue($link);
}
}
}
}
/**
* Checks if node has been visited.
*
* @return bool
*/
public function isVisited()
{
return $this->visited;
}
}
答案 1 :(得分：1)
将数据作为两个独立的数组处理更容易（恕我直言）。一组对象及其链接。此外，作为第一部分，我将对象转换为ID作为键，这允许我直接使用它，而不是每次都必须搜索ID。
另外为了使解决方案更加简单，我在访问它时在对象数组中创建了一个标志，这样当我再次尝试引用它时，我可以检查它是否已经存在访问。
<?php
error_reporting ( E_ALL );
ini_set ( 'display_errors', 1 );
$objects =[[1,'apple'],
[3, 'Carrot'],
[4, 'Dill'],
[5, 'Egg '],
[6, 'Fred'],
[7, 'Goat'],
[8, 'Harry'],
[9, 'Igloo'],
[10, 'Jason'],
[11, 'Klaus'],
[12, 'Banana'],
[15, 'Oyster1'],
[16, 'Oyster2' ]];
$links =[[1,12],
[1,5],
[3,1],
[3,12],
[3,3],
[4,1],
[4,5],
[5,6],
[6,7],
[7,7],
[7,8],
[8,null],
[9,12],
[9,5],
[10,1],
[10,5],
[10,8],
[11,1],
[11,5],
[11,10],
[12,3],
[15,16],
[16,15 ]];
function buildTree ( $id, &$objects, $links ) {
foreach ( $links as $testNode ) {
if ( $testNode[0] == $id &&
$testNode[1] != $id &&
$testNode[1] != null &&
!isset($objects[$testNode[1]]['visited']) ) {
$objects[$testNode[1]]['visited'] = true;
buildTree ( $testNode[1], $objects, $links);
}
}
}
// Convert array to have the ID as key
$objects = array_combine(array_column($objects, 0), $objects);
// Fetch ID of first item
reset($objects);
$startID = key($objects);
// Mark as visited
$objects[$startID]['visited'] = true;
// Process
buildTree ( $startID, $objects, $links);
$linked = [];
$notLinked = [];
foreach ( $objects as $object) {
if ( isset($object['visited']) ) {
$linked[] = $object[1];
}
else {
$notLinked[] = $object[1];
}
}
echo "Linked...".PHP_EOL;
print_r($linked);
echo "Not linked...".PHP_EOL;
print_r($notLinked);
如您所见，核心是递归buildTree函数。这使用&$objects，因为这意味着对函数的所有调用都将使用相同的数组。由于我想建立项目的所有用途，这很重要。
buildTree中的条件，检查它是否是我们想要的节点，它不是指同一个节点（浪费时间再看），不是null（不知道为什么你链接为null，但又不值得再查看了）并且该节点还没有被访问过。如果这些条件没有问题，它会将下一个节点标记为已访问并进入下一级别。
输出是......
Linked...
Array
(
[0] => apple
[1] => Carrot
[2] => Egg
[3] => Fred
[4] => Goat
[5] => Harry
[6] => Banana
)
Not linked...
Array
(
[0] => Dill
[1] => Igloo
[2] => Jason
[3] => Klaus
[4] => Oyster1
[5] => Oyster2
)
答案 2 :(得分：0)
假设“root”是Student.last.grade。
这是一种有点蛮力的算法，但它利用了SQL对'set'操作的喜好。
obj_id 1更接近SQL：
Insert into table1 the root (1)
Loop
Create table2 with all nodes linked to any node in table1
Exit if number of rows in table2 = num rows in table1
table1 := table2
输出是所有'连接'的ID。如果你想要未连接的那些：
# Initialize:
CREATE TABLE table1 (
obj_id ...
PRIMARY KEY(obj_id)
)
SELECT 1; # assuming root is 1
start of loop:
CREATE TABLE table2 (
obj_id ...
PRIMARY KEY(obj_id)
)
SELECT DISTINCT obj_link_id
FROM table1
JOIN object_links USING(obj_id);
SELECT @fini := ( SELECT COUNT(*) FROM table1 ) =
( SELECT COUNT(*) FROM table2 ) # will give true/false
DROP TABLE table1;
RENAME TABLE table2 TO table1;
loop if @fini=0
答案 3 :(得分：0)
以下是获取所有链接ID的简短方法：
$pdo = new PDO('mysql:host=localhost;dbname=test_obj_link', 'testread', 'testread');
$links = $pdo
->query('select obj_id, obj_link_id from object_links')
->fetchAll(PDO::FETCH_GROUP|PDO::FETCH_COLUMN);
function getLinks($objId, $links, $indirectNodes = []) {
$directNodes = isset($links[$objId]) ? $links[$objId] : [];
foreach($directNodes as $linkedNode) {
if (!in_array($linkedNode, $indirectNodes)) {
$indirectNodes[] = $linkedNode;
$indirectNodes = array_unique(array_merge(
$indirectNodes,
getLinks($linkedNode, $links, $indirectNodes)
));
}
}
return $indirectNodes;
}
$linkedObjectIds = getLinks(1, $links);
fetchAll(PDO::FETCH_GROUP|PDO::FETCH_COLUMN)将返回一个结构化数组，其中每个对象的链接由objectIDs索引，如下所示：
$links = [
1 => ['5', '12'],
3 => ['1', '3', '12'],
4 => ['1', '5'],
5 => ['6'],
6 => ['7'],
7 => ['7', '8'],
9 => ['5', '12'],
10 => ['1', '5', '8'],
11 => ['1', '5', '10'],
12 => ['3'],
15 => ['16'],
16 => ['15'],
];
getLinks函数将递归地“遍历”$links数组并合并在路上找到的所有子数组。由于PHP似乎没有array_union函数 - 而是使用array_unique(array_merge(..))。
结果：
$linkedObjectIds = array (
0 => '5',
1 => '6',
2 => '7',
3 => '8',
4 => '12',
10 => '3',
11 => '1',
)
请注意，这里的索引没有任何意义。
要获取相应的对象，您可以执行以下操作：
$objects = $pdo
->query('select id, title from object')
->fetchAll(PDO::FETCH_KEY_PAIR);
$linkedObjects = array_intersect_key($objects, array_flip($linkedObjectIds));
$notLinkedObjects = array_diff_key($objects, $linkedObjects);
变量将包含：
$objects = array (
1 => 'Apple',
3 => 'Carrot',
4 => 'Dill',
5 => 'Egg',
6 => 'Fred',
7 => 'Goat',
8 => 'Harry',
9 => 'Igloo',
10 => 'Jason',
11 => 'Klaus',
12 => 'Banana',
15 => 'Oyster1',
16 => 'Oyster2',
);
$linkedObjects = array (
1 => 'Apple',
3 => 'Carrot',
5 => 'Egg',
6 => 'Fred',
7 => 'Goat',
8 => 'Harry',
12 => 'Banana',
);
$notLinkedObjects = array (
4 => 'Dill',
9 => 'Igloo',
10 => 'Jason',
11 => 'Klaus',
15 => 'Oyster1',
16 => 'Oyster2',
);
演示：http://rextester.com/ZQQGE35352
请注意，in_array()和array_unique()可能很慢，因为他们必须搜索未排序的值。这可能会导致某些数据集出现性能问题。假设PHP可以更快地搜索密钥，我们可以使用array_key_exists()而不是in_array()和数组运算符+（按键联合）而不是array_unique(array_merge())。代码甚至会更短：
function getLinks($objId, $links, $indirectNodes = []) {
$directNodes = isset($links[$objId]) ? $links[$objId] : [];
foreach($directNodes as $linkedNode) {
if (!array_key_exists($linkedNode, $indirectNodes)) {
$indirectNodes[$linkedNode] = 0;
$indirectNodes = $indirectNodes + getLinks($linkedNode, $links, $indirectNodes);
}
}
return $indirectNodes;
}
但是 - 由于函数现在将所需结果作为键返回，我们需要使用array_keys()来提取它们：
$linkedObjectIds = array_keys(getLinks(1, $links));