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我決定關注http://www.artfulsoftware.com/mysqlbook/sampler/mysqled1ch20.html在MySQL中處理嵌套集？
所以現在我正在尋找一些幫助的代碼。
我使用他們的數據，我的測試， 所以，我想象的樹是像這樣：
array('value' => 'Richard Shakespeare',
array('value' => 'Henry',
array('value' => 'Joan'),
array('value' => 'Margaret'),
array('value' => 'William',
array('value' => 'Susana',
array('value' => 'Elizabeth Hall',
array('value' => 'John Bernard'))),
array('value' => 'Hamnet'),
array('value' => 'Judith',
array('value' => 'Shakespeare Quiney'),
array('value' => 'Richard Quiney'),
array('value' => 'Thomas Quiney'))),
array('value' => 'Gilbert'),
array('value' => 'Joan',
array('value' => 'William Hart'),
array('value' => 'Mary Hart'),
array('value' => 'Thomas Hart'),
array('value' => 'Micheal Hart')),
array('value' => 'Anne'),
array('value' => 'Richard'),
array('value' => 'Edmond')),
array('value' => 'John'));
所以，如果我們要插入到我們希望與
落得數據庫Array
(
[0] => Array
(
[value] => Richard Shakespeare
[left] => 1
[right] => 46
)
[1] => Array
(
[value] => Henry
[left] => 2
[right] => 43
)
[2] => Array
(
[value] => Joan
[left] => 3
[right] => 4
)
[3] => Array
(
[value] => Margaret
[left] => 5
[right] => 6
)
[4] => Array
(
[value] => William
[left] => 7
[right] => 24
)
[5] => Array
(
[value] => Susana
[left] => 8
[right] => 13
)
[6] => Array
(
[value] => Elizabeth Hall
[left] => 9
[right] => 12
)
[7] => Array
(
[value] => John Bernard
[left] => 10
[right] => 11
)
[8] => Array
(
[value] => Hamnet
[left] => 14
[right] => 15
)
[9] => Array
(
[value] => Judith
[left] => 16
[right] => 23
)
[10] => Array
(
[value] => Shakespeare Quiney
[left] => 17
[right] => 18
)
[11] => Array
(
[value] => Richard Quiney
[left] => 19
[right] => 20
)
[12] => Array
(
[value] => Thomas Quiney
[left] => 21
[right] => 22
)
[13] => Array
(
[value] => Gilbert
[left] => 25
[right] => 26
)
[14] => Array
(
[value] => Joan
[left] => 27
[right] => 36
)
[15] => Array
(
[value] => William Hart
[left] => 28
[right] => 29
)
[16] => Array
(
[value] => Mary Hart
[left] => 30
[right] => 31
)
[17] => Array
(
[value] => Thomas Hart
[left] => 32
[right] => 33
)
[18] => Array
(
[value] => Micheal Hart
[left] => 34
[right] => 35
)
[19] => Array
(
[value] => Anne
[left] => 37
[right] => 38
)
[20] => Array
(
[value] => Richard
[left] => 39
[right] => 40
)
[21] => Array
(
[value] => Edmond
[left] => 41
[right] => 42
)
[22] => Array
(
[value] => John
[left] => 44
[right] => 45
)
)
所以這個問題想起來，如何最好地做到這一點？
我的解決辦法是：
$container = array();
function children($item){
$children = 0;
foreach($item as $node)
if(is_array($node))
$children += children($node)+1;
return $children;
}
function calculate($item, &$container, $data = array(0,0)){
//althought this one is actually of no use, it could be useful as it contains a count
$data[0]++; //$left
$right = ($data[0]+(children($item)*2))+1;
//store the values in the passed container
$container[] = array(
'value' => $item['value'],
'left' => $data[0],
'right' => $right,
);
//continue looping
$level = $data[1]++;
foreach($item as &$node)
if(is_array($node))
$data = calculate($node, $container, $data);
$data[1] = $level;
$data[0]++;
return $data;
}
calculate($tree, $container);
其效率如何，我不知道。
但現在進入查詢。
選擇節點的所有後代，我們可以使用
SELECT child.value AS 'Descendants of William', COUNT(*) AS `Level`
FROM tester AS parent
JOIN tester AS child ON child.`left` BETWEEN parent.`left` AND parent.`right`
WHERE parent.`left` > 7 AND parent.`right` < 24
GROUP BY child.value ORDER BY `level`;
選擇節點的所有後代，到特定的深度，我們可以使用
注意，我們選擇威廉的後裔的深度的2
威廉姆斯左：7，威廉姆斯右：24，電平：2
SELECT child.value AS 'Descendants of William', COUNT(*) AS `Level`
FROM tester AS parent
JOIN tester AS child ON child.`left` BETWEEN parent.`left` AND parent.`right`
WHERE parent.`left` > 7 AND parent.`right` < 24
GROUP BY child.value HAVING `level` <= 2 ORDER BY `level`;
所以這很簡單。
但現在我想知道一些事情，
注意的是，在實際的數據庫，以及左/右的所有行都有一個唯一的ID，幷包含其inviteers ID的「父母」一欄，則返回null沒有邀請
- 可以說，我想插入
David作爲Judith一個孩子，我該怎麼辦呢？ - 可以說我想要得到
Mary Hart's父母和父母父母（array('Henery', 'Joan', 'Mary Hart')），我該怎麼做？ - 可以說我想從
Joan刪除William Hart我該怎麼做？
你試過了什麼？我討厭問，但嵌套很複雜，你的問題聽起來像是你的工作羣衆採購。另外，你甚至還沒有開始撇清這個問題的難度。如：「如果我想顛倒大衛和朱迪思，那麼朱迪思現在就把大衛的地方當作孩子，反之亦然，我怎麼做到這一點[沒有重新索引樹兩次]？」 – 2011-06-09 22:18:06