以下是我要做的事情： - 我需要一个函数，当作为参数传递时，ID（对于一类事物）将提供所有子类别和子子类别以及sub-sub-sub..etc。 - 我正在考虑使用递归函数，因为我不知道它们的子子类别的子类别的数量等等 所以这就是我到目前为止所做的事情
function categoryChild($id) {
$s = "SELECT * FROM PLD_CATEGORY WHERE PARENT_ID = $id";
$r = mysql_query($s);
if(mysql_num_rows($r) > 0) {
while($row = mysql_fetch_array($r))
echo $row['ID'].",".categoryChild($row['ID']);
}
else {
$row = mysql_fetch_array($r);
return $row['ID'];
}
}
如果我使用return而不是echo，我将得不到相同的结果。我需要一些帮助来解决这个问题或从头开始重写
答案 0 :(得分：13)
我很难弄清楚你的功能。我想这会做你想要的。它获取ID $ id类别的所有子项，以及它们的子项（从而获得您想要的整个子类别，子子类别效果）。
function categoryChild($id) {
$s = "SELECT ID FROM PLD_CATEGORY WHERE PARENT_ID = $id";
$r = mysql_query($s);
$children = array();
if(mysql_num_rows($r) > 0) {
# It has children, let's get them.
while($row = mysql_fetch_array($r)) {
# Add the child to the list of children, and get its subchildren
$children[$row['ID']] = categoryChild($row['ID']);
}
}
return $children;
}
此功能将返回：
$var = array(
'categoryChild ID' => array(
'subcategoryChild ID' => array(
'subcategoryChild child 1' => array(),
'subcategoryChild child 2' => array()
)
),
'anotherCategoryChild ID' => array() # This child has no children of its own
);
它基本上返回一个包含子ID的数组和一个包含其子ID的数组。我希望这有任何帮助。
答案 1 :(得分：5)
database tree to multidimensional array
<?php
function getTree($rootid)
{
$arr = array();
$result = mysql_query("select * from PLD_CATEGORY where PARENT_ID='$rootid'");
while ($row = mysql_fetch_array($result)) {
$arr[] = array(
"Title" => $row["Title"],
"Children" => getTree($row["id"])
);
}
return $arr;
}
?>
答案 2 :(得分：4)
由于这是由@Pawan Sharma提出的，我认为我也可以给出一些答案。
所有给定的解决方案都存在常见问题 - 它们为每个孩子执行SQL查询。例如，如果第二级有100个孩子，那么将完成100个查询，而实际上可以使用where parent_id in (<list_of_ids>)在单个查询中完成。
示例DB：
create table category (
id int auto_increment primary key,
parent_id int default null,
title tinytext,
foreign key (parent_id) references category (id)
) engine = InnoDB;
insert into category (id, parent_id, title) values
(1, null, '1'),
(2, null, '2'),
(3, null, '3'),
(4, 1 , '1.1'),
(5, 1 , '1.2'),
(6, 1 , '1.3'),
(7, 4 , '1.1.1'),
(8, 4 , '1.1.2'),
(9, 7 , '1.1.1.1');
这是我的解决方案：
/**
* @param null|int|array $parentID
*/
function getTree($parentID) {
$sql = "select id, parent_id, title from category where ";
if ( is_null($parentID) ) {
$sql .= "parent_id is null";
}
elseif ( is_array($parentID) ) {
$parentID = implode(',', $parentID);
$sql .= "parent_id in ({$parentID})";
}
else {
$sql .= "parent_id = {$parentID}";
}
$tree = array();
$idList = array();
$res = mysql_query($sql);
while ( $row = mysql_fetch_assoc($res) ) {
$row['children'] = array();
$tree[$row['id']] = $row;
$idList[] = $row['id'];
}
mysql_free_result($res);
if ( $idList ) {
$children = getTree($idList);
foreach ( $children as $child ) {
$tree[$child['parent_id']]['children'][] = $child;
}
}
return $tree;
}
使用提供的示例数据，当调用getTree(null)时（对于所有条目），它最多可执行5次查询：
select id, parent_id, title from category where parent_id is null
select id, parent_id, title from category where parent_id in (1,2,3)
select id, parent_id, title from category where parent_id in (4,5,6)
select id, parent_id, title from category where parent_id in (7,8)
select id, parent_id, title from category where parent_id in (9)
当调用getTree(4)时，会执行3个查询：
select id, parent_id, title from category where parent_id = 4
select id, parent_id, title from category where parent_id in (7,8)
select id, parent_id, title from category where parent_id in (9)
答案 3 :(得分：1)
function categoryChild($id)
{
$s = "SELECT category_id,name FROM proads_categories WHERE parent_id =".$id;
$r = mysql_query($s);
$children = array();
if(mysql_num_rows($r) > 0)
{
#It has children, let's get them.
while($row = mysql_fetch_array($r))
{
#Add the child to the list of children, and get its subchildren
$children[$row['category_id']]['nam'] = $row['name'];
$arr = categoryChild($row['category_id']);
if(count($arr) > 0)
{
$children[$row['category_id']]['child'] = categoryChild($row['category_id']);
}
}
}
return $children;
}
这很完美。如果您需要，请试试这个
答案 4 :(得分：1)
function breadCrumb($id)
{
$ar = array();
$result = mysql_query("SELECT * FROM groups WHERE ParentID = '$id'");
if(mysql_num_rows($result) > 0)
{
while($row = mysql_fetch_object($result))
{
$ar[] = $row->DBGroupID;
$r = mysql_query("SELECT * FROM groups WHERE ParentID = '".$row->GroupID."'");
if(mysql_num_rows($r) > 0)
$ar = array_merge($ar, breadCrumb($row->GroupID, 1));
}
}
return $ar;
}
答案 5 :(得分：1)
<?php
require('db/dbconnect.php');
$user_id='triD-100';
$sql="select * from ajent_joining where sponser_id='".$user_id."'";
$qR=mysql_query($sql);
while($rowD=mysql_fetch_assoc($qR)){
echo $childId=$rowD["user_id"];
echo "<br/>";
categoryChild($childId);
}
function categoryChild($childId) {
$s = "select user_id from ajent_joining where sponser_id='".$childId."'";
$r = mysql_query($s);
if(mysql_num_rows($r) > 0) {
while($row = mysql_fetch_array($r)) {
echo $childId=$row["user_id"];
echo "<br/>";
categoryChild($childId);
}
}
}
?>
答案 6 :(得分：0)
使用Prestashop功能：
public function getRecursiveChildren() {
$subCategories = $this->recurseLiteCategTree();
//print_r($subCategories);
$my_tab = array();
foreach ($subCategories['children'] as $subc) {
$my_tab[] = $subc['id'];
foreach ($subc['children'] as $subc2) {
$my_tab[] = $subc2['id'];
foreach ($subc2['children'] as $subc3) {
$my_tab[] = $subc3['id'];
foreach ($subc3['children'] as $subc4) {
$my_tab[] = $subc4['id'];
}
}
}
}
$my_tab [] = $this->id;
return $my_tab;
}
这可以通过使用递归来改善，但今天没有时间：'（