我有一张像
这样的表格id catagory suboff
1 software 0
2 programming 1
3 Testing 1
4 Designing 1
5 Hospital 0
6 Doctor 5
7 Nurses 5
9 Teaching 0
10 php programming 2
11 .net programming 2
如何编写代码以基于suboff在多维数组中获取所有这些信息，如下所示，
-software
--programming
---php programming
--- .net programming
--testing
--designing
-hospital
--doctor
--nurses
-teaching
答案 0 :(得分：3)
假设MySQL是您的数据库引擎：
// We'll need two arrays for this
$temp = $result = array();
// Get the data from the DB
$table = mysql_query("SELECT * FROM table");
// Put it into one dimensional array with the row id as the index
while ($row = mysql_fetch_assoc($table)) {
$temp[$row['id']] = $row;
}
// Loop the 1D array and create the multi-dimensional array
for ($i = 1; isset($temp[$i]); $i++) {
if ($temp[$i]['suboff'] > 0) {
// This row has a parent
if (isset($temp[$temp[$i]['suboff']])) {
// The parent row exists, add this row to the 'children' key of the parent
$temp[$temp[$i]['suboff']]['children'][] =& $temp[$i];
} else {
// The parent row doesn't exist - handle that case here
// For the purposes of this example, we'll treat it as a root node
$result[] =& $temp[$i];
}
} else {
// This row is a root node
$result[] =& $temp[$i];
}
}
// unset the 1D array
unset($temp);
// Here is the result
print_r($result);
使用references来完成这样的工作。
答案 1 :(得分：2)
$array = array(
array('1','software','0'),
array('2','programming','1'),
array('3','Testing','1'),
array('4','Designing','1'),
array('5','Hospital','0'),
array('6','Doctor','5'),
array('7','Nurses','5'),
array('9','Teaching','0'),
array('10','php programming','2'),
array('11','.net programming','2')
);
function menu_sort($results, $master = 0)
{
$open = array();
$return = NULL;
foreach($results as $result)
{
if($result[2] == $master){
if(!$open){
$return .= '<ul>';
$open = true;
}
$return .= '<li>'.$result[1];
$return .= menu_sort($results, $result[0]);
$return .= '</li>';
}
}
if($open)
$return .= '</ul>';
return $return;
}
echo menu_sort($array);
结果...
software programming php programming .net programming Testing Designing Hospital Doctor Nurses Teaching
答案 2 :(得分：1)
我会这样做的方式：
首先，您需要解析此表。我想你可以自己做;如果没有，谷歌“正则表达”，他们是你的朋友。
您使用的数据结构是经典的tree。您将需要两个数组来处理它。首先是节点数组$nodes，其中键是节点ID，值是节点名称，$links，其中每个键是父节点，每个值都是子节点数组（{每个元素{1}}就足够了。
现在你必须以递归的方式下降你拥有的树。您引入了一个带有如下签名的函数：
$links[$id][] = $suboff此函数应使用$ level padding破折号打印节点本身（存储在$ nodes中的信息）并调用自身以呈现所有子节点。它们将反过来渲染所有子节点等。您只需要为顶级节点调用此函数。
答案 3 :(得分：1)
此类将平面类别数组转换为结构化树数组：
<?php
/**
* Creates a structured tree out of a flat category list
*/
class CategoryTree {
/**
*
* @var array
*/
protected $categories = array();
/**
*
* @var array
*/
protected $tree = array();
/**
* Default constructor
* @param array $categories
*/
function __construct(array $categories) {
$this->categories = $categories;
}
/**
* Process a subtree
* @param array $categories
* @param integer $parentId
* @return array
*/
protected function getSubtree(array $categories, $parentId = 0) {
$tree = array();
foreach($categories as $category) {
if($category['suboff'] == $parentId) {
$tree[$category['id']] = $category;
$tree[$category['id']]['children'] = $this->getSubtree($categories, $category['id']);
}
}
return $tree;
}
/**
* Get the category tree as structured array
* @return array
*/
public function getTree() {
if(empty($this->tree)) {
$this->tree = $this->getSubtree($this->categories, 0);
}
return $this->tree;
}
/**
* Get the category tree as string representation
* @return string
*/
public function __toString() {
return "<pre>" . print_r($this->getTree(), true) . "</pre>";
}
}
// Now, use the class with the givven data:
$categories = array(
array(
'id' => 1,
'category' => 'software',
'suboff' => 0
),
array(
'id' => 2,
'category' => 'programming',
'suboff' => 1
),
array(
'id' => 3,
'category' => 'Testing',
'suboff' => 1
),
array(
'id' => 4,
'category' => 'Designing',
'suboff' => 1
),
array(
'id' => 5,
'category' => 'Hospital',
'suboff' => 0
),
array(
'id' => 6,
'category' => 'Doctor',
'suboff' => 5
),
array(
'id' => 7,
'category' => 'Nurses',
'suboff' => 5
),
array(
'id' => 9,
'category' => 'Teaching',
'suboff' => 0
),
array(
'id' => 10,
'category' => 'php programming',
'suboff' => 2
),
array(
'id' => 11,
'category' => '.net programming',
'suboff' => 2
)
);
$myTree = new CategoryTree($categories);
echo $myTree;
?>
答案 4 :(得分：1)
这就是我刚为我的应用所写的内容，它就像一个魅力：）
$array = [
'i' => ['key' => 'i', 'name' => 'php programming', 'parent' => 'b'],
'g' => ['key' => 'g', 'name' => 'Nurses', 'parent' => 'e'],
'j' => ['key' => 'j', 'name' => '.net programming', 'parent' => 'b'],
'b' => ['key' => 'b', 'name' => 'programming', 'parent' => 'a'],
'a' => ['key' => 'a', 'name' => 'software', 'parent' => 'asd'],
'c' => ['key' => 'c', 'name' => 'Testing', 'parent' => 'a'],
'd' => ['key' => 'd', 'name' => 'Designing', 'parent' => 'a'],
'e' => ['key' => 'e', 'name' => 'Hospital', 'parent' => 'asd'],
'f' => ['key' => 'f', 'name' => 'Doctor', 'parent' => 'e'],
'h' => ['key' => 'h', 'name' => 'Teaching'],
];
function getAsTree(array &$array)
{
foreach ($array as $key => $item) {
if (isset($item['parent']) && isset($array[$item['parent']])) {
$array[$item['parent']]['children'][] = $item;
unset($array[$key]);
return getAsTree($array);
}
}
return $array;
}
结果如下：
--- a: software ------ b: programming --------- i: php programming --------- j: .net programming ------ c: Testing ------ d: Designing --- e: Hospital ------ g: Nurses ------ f: Doctor --- h: Teaching
答案 5 :(得分：0)
恕我直言，逻辑是：
答案 6 :(得分：0)
您需要将整个表读入内存并将其转换为树，其中每个节点都可以使用其对应的ID号进行标识。然后对树进行预先遍历以将其打印出来。
答案 7 :(得分：0)
在PHP中，我从数据库中获取数据：
"SELECT* FROM Table WHERE suboff LIKE 0"
foreach(item..)
"SELECT* FROM Table WHERE suboff LIKE item.ID"
foreach(item2..)
$result[item][item2]
答案 8 :(得分：0)
这是一种非常容易理解的不同方法。它要求您按suboff排序表，即
SELECT * FROM table ORDER BY suboff
假设结果存储在$table中，您可以使用这个非常简洁的PHP代码：
// this will hold the result
$tree = array();
// used to find an entry using its id
$lookup = array();
foreach($table as $row){
if($row['suboff'] === 0){
// this has no parent, add it at base level
$tree[$row['category']] = array();
// store a reference
$lookup[$row['id']] =& $tree[$row['category']];
}else{
// find the right parent, add the category
$lookup[$row['suboff']][$row['category']] = array();
// store a reference
$lookup[$row['id']] =& $lookup[$row['suboff']][$row['category']];
}
}
答案 9 :(得分：0)
此解决方案适用于我。
Sub StateFill()
Dim x As Long, N As Long
With Sheet1
N = .Cells(Rows.Count, 1).End(xlUp).Row
For x = 1 To N
If Right(.Range("$A$" & x), 4) Like ", [A-Z][A-Z]" Then
.Range("$B$" & x) = .Range("$B$" & x) & Right(.Range("$A$" & x), 2)
End If
Next
End With
End Sub