我正在尝试将使用树遍历模型分层设置的数据转换为< UL>为了在我的网站上显示。
这是我的代码:
function getCats($) {
// retrieve all children of $parent
$query = "SELECT max(rght) as max from t_categories";
$row = C_DB::fetchSingleRow($query);
$max = $row["max"];
$result ="<ul>";
$query = "SELECT * from t_categories where lft >=0 and rght <= $max";
if($rs = C_DB::fetchRecordset($query)){
$p_right ="";
$p_left ="";
$p_diff="";
while($row = C_DB::fetchRow($rs)){
$diff = $row["rght"] -$row["lft"];
if($diff == $p_diff){
$result.= "<li>".$row['title']."</li>";
}elseif (($row["rght"] - $row["lft"] > 1) && ($row["rght"] > $p_right)){
$result. "<ul>";
$result.= "<li>".$row['title']."</li>";
}else{
$result.= "<li>".$row['title']."</li>";
}
$p_right = $row["rght"];
$p_left = $row["lft"];
$p_diff = $diff;
}
}
$result.= "</ul>";
return $result;
}
这是我的示例表:
|ID | TITLE | lft| rght |
|1 | Cat 1 | 1 | 16 |
|18 | Cat 2 | 3 | 4 |
|22 | Cat 3 | 5 | 6 |
|28 | Cat 4 | 7 | 8 |
|34 | Cat 5 | 9 | 9 |
|46 | Cat 6 | 11 | 10 |
|47 | Cat 7 | 13 | 12 |
|49 | Cat 8 | 15 | 14 |
现在输出如下内容:
<ul>
<li>Cat 1</li>
<li>Cat 2</li>
<li>Cat 3</li>
<li>Cat 4</li>
<li>Cat 5</li>
<li>Cat 6</li>
<li>Cat 7</li>
<li>Cat 8</li>
</ul>
任何人都可以告诉我为什么或如何在分层结构中输出列表?
答案 0 :(得分:51)
好的,让我们做一些赏金狩猎;)
第0步 - 清理示例:
如前所述,您的示例数据已被破坏,因为它没有定义有效的嵌套集。如果从应用程序中获取此数据,则应检查插入/删除逻辑。
因此,对于测试,我使用了一个像这样的消毒版本:
(MySQL在这里,因为它是第一个在手边)
CREATE TABLE t_categories`(
`id` INTEGER UNSIGNED NOT NULL AUTO_INCREMENT,
`title` VARCHAR(45) NOT NULL,
`lft` INTEGER UNSIGNED NOT NULL,
`rght` INTEGER UNSIGNED NOT NULL,
PRIMARY KEY (`id`)
);
INSERT INTO t_categories (title, lft, rght) VALUES ('Cat 1',1,16);
INSERT INTO t_categories (title, lft, rght) VALUES ('Cat 2',2,3);
INSERT INTO t_categories (title, lft, rght) VALUES ('Cat 3',4,7);
INSERT INTO t_categories (title, lft, rght) VALUES ('Cat 4',5,6);
INSERT INTO t_categories (title, lft, rght) VALUES ('Cat 5',8,13);
INSERT INTO t_categories (title, lft, rght) VALUES ('Cat 6',9,12);
INSERT INTO t_categories (title, lft, rght) VALUES ('Cat 7',10,11);
INSERT INTO t_categories (title, lft, rght) VALUES ('Cat 8',14,15);
第1步 - 让数据库进行排序
嵌套集主要发明为在数据库中存储树的便捷方式,因为它们可以很容易地查询子树,父关系,特别是在这种情况下有趣的顺序和深度:
SELECT node.title, (COUNT(parent.title) - 1) AS depth
FROM t_categories AS node
CROSS JOIN t_categories AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rght
GROUP BY node.title
ORDER BY node.lft
这将返回整齐排序的集合,从根节点开始并按预订继续到最后。最重要的是,它会将每个节点的深度添加为正整数,表示节点在根目录下的级别(级别0)。对于上面的示例数据,结果将是:
title, depth
'Cat 1', 0
'Cat 2', 1
'Cat 3', 1
'Cat 4', 2
'Cat 5', 1
'Cat 6', 2
'Cat 7', 3
'Cat 8', 1
在代码中:
// Grab ordered data
$query = '';
$query .= 'SELECT node.title, (COUNT(parent.title) - 1) AS depth';
$query .= ' FROM t_categories AS node';
$query .= ' CROSS JOIN t_categories AS parent';
$query .= ' WHERE node.lft BETWEEN parent.lft AND parent.rght';
$query .= ' GROUP BY node.title';
$query .= ' ORDER BY node.lft';
$result = mysql_query($query);
// Build array
$tree = array();
while ($row = mysql_fetch_assoc($result)) {
$tree[] = $row;
}
结果数组如下所示:
Array
(
[0] => Array
(
[title] => Cat 1
[depth] => 0
)
[1] => Array
(
[title] => Cat 2
[depth] => 1
)
...
)
第2步 - 输出为HTML列表片段:
使用while循环:
// bootstrap loop
$result = '';
$currDepth = -1; // -1 to get the outer <ul>
while (!empty($tree)) {
$currNode = array_shift($tree);
// Level down?
if ($currNode['depth'] > $currDepth) {
// Yes, open <ul>
$result .= '<ul>';
}
// Level up?
if ($currNode['depth'] < $currDepth) {
// Yes, close n open <ul>
$result .= str_repeat('</ul>', $currDepth - $currNode['depth']);
}
// Always add node
$result .= '<li>' . $currNode['title'] . '</li>';
// Adjust current depth
$currDepth = $currNode['depth'];
// Are we finished?
if (empty($tree)) {
// Yes, close n open <ul>
$result .= str_repeat('</ul>', $currDepth + 1);
}
}
print $result;
与递归函数相同的逻辑:
function renderTree($tree, $currDepth = -1) {
$currNode = array_shift($tree);
$result = '';
// Going down?
if ($currNode['depth'] > $currDepth) {
// Yes, prepend <ul>
$result .= '<ul>';
}
// Going up?
if ($currNode['depth'] < $currDepth) {
// Yes, close n open <ul>
$result .= str_repeat('</ul>', $currDepth - $currNode['depth']);
}
// Always add the node
$result .= '<li>' . $currNode['title'] . '</li>';
// Anything left?
if (!empty($tree)) {
// Yes, recurse
$result .= renderTree($tree, $currNode['depth']);
}
else {
// No, close remaining <ul>
$result .= str_repeat('</ul>', $currNode['depth'] + 1);
}
return $result;
}
print renderTree($tree);
两者都将输出以下结构:
<ul>
<li>Cat 1</li>
<li>
<ul>
<li>Cat 2</li>
<li>Cat 3</li>
<li>
<ul>
<li>Cat 4</li>
</ul>
</li>
<li>Cat 5</li>
<li>
<ul>
<li>Cat 6</li>
<li>
<ul>
<li>Cat 7</li>
</ul>
</li>
</ul>
</li>
<li>Cat 8</li>
</ul>
</li>
</ul>
Nitpickers corner:发件人明确要求<ul>
,但订购了无序列表!?来吧......
的; - )强>
答案 1 :(得分:16)
更好的渲染树函数对我有用( php函数准备用于jsTree jQuery插件的html源代码)而不是Henrik Opel的那个:
function MyRenderTree ( $tree = array(array('name'=>'','depth'=>'')) ){
$current_depth = 0;
$counter = 0;
$result = '<ul>';
foreach($tree as $node){
$node_depth = $node['depth'];
$node_name = $node['name'];
$node_id = $node['category_id'];
if($node_depth == $current_depth){
if($counter > 0) $result .= '</li>';
}
elseif($node_depth > $current_depth){
$result .= '<ul>';
$current_depth = $current_depth + ($node_depth - $current_depth);
}
elseif($node_depth < $current_depth){
$result .= str_repeat('</li></ul>',$current_depth - $node_depth).'</li>';
$current_depth = $current_depth - ($current_depth - $node_depth);
}
$result .= '<li id="c'.$node_id.'"';
$result .= $node_depth < 2 ?' class="open"':'';
$result .= '><a href="#"><ins> </ins>'.$node_name.'</a>';
++$counter;
}
$result .= str_repeat('</li></ul>',$node_depth).'</li>';
$result .= '</ul>';
return $result;}
结果HTML:
<ul>
<li id="c1" class="open"><a href="#"><ins> </ins>ELECTRONICS</a>
<ul>
<li id="c2" class="open"><a href="#"><ins> </ins>TELEVISIONS</a>
<ul>
<li id="c3"><a href="#"><ins> </ins>TUBE</a></li>
<li id="c4"><a href="#"><ins> </ins>LCD</a></li>
<li id="c5"><a href="#"><ins> </ins>PLASMA</a>
<ul>
<li id="c14"><a href="#"><ins> </ins>PLASMA1</a></li>
<li id="c15"><a href="#"><ins> </ins>PLASMA2</a></li>
</ul>
</li>
</ul>
</li>
<li id="c6" class="open"><a href="#"><ins> </ins>PORTABLE ELECTRONICS</a>
<ul>
<li id="c7"><a href="#"><ins> </ins>MP3 PLAYERS</a>
<ul>
<li id="c8"><a href="#"><ins> </ins>FLASH</a></li>
</ul>
</li>
<li id="c9"><a href="#"><ins> </ins>CD PLAYERS</a></li>
<li id="c10"><a href="#"><ins> </ins>2 WAY RADIOS</a></li>
</ul>
</li>
</ul>
</li>
</ul>
答案 2 :(得分:4)
有一个用于处理嵌套集的PEAR包:DB_NestedSet 您可能还对文章Managing Hierarchical Data in MySQL感兴趣。
答案 3 :(得分:1)
这应该是你要找的东西:
function getCats($left = null, $right = null)
{
$sql = array();
$result = null;
if (isset($left) === true)
{
$sql[] = 'lft >= ' . intval($left);
}
if (isset($right) === true)
{
$sql[] = 'rght <= ' . intval($right);
}
if (empty($sql) === true)
{
$sql[] = 'lft = 1';
}
$sql = 'SELECT * FROM t_categories WHERE ' . implode(' AND ', $sql) . ';';
if ($rs = C_DB::fetchRecordset($sql))
{
// you need to make sure that the query returns
// something to correctly display the ULs
if (empty($rs) === false)
{
$result .= '<ul>' . "\n";
while ($row = C_DB::fetchRow($rs))
{
$result .= '<li>' . $row['title'] . '</li>' . "\n";
$result .= getCats($row['lft'], $row['rght']);
}
$result .= '</ul>' . "\n";
}
}
return $result;
}
要获取嵌套树的HTML,您应该这样做:
echo getCats();
请注意你的嵌套集样本看起来不对,你也应该确保我是否在调用你的C_DB类时没有犯错,我不知道,因为我不熟悉它。
答案 4 :(得分:1)
直接通过结果循环:
$sql = "SELECT node.name, (COUNT(parent.name) - 1) AS depth
FROM nested_category AS node,
nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.name
ORDER BY node.lft";
$query_result = mysql_query($sql)
$result = "<ul>";
$currDepth = 0;
while($row = mysql_fetch_assoc($query_result))
{
if($row['depth'] > $currDepth)
{
$result .= "<li><ul>"; // open sub tree if level up
}
if($row['depth'] < $currDepth)
{
$result .= str_repeat("</ul></li>", $currDepth - $row['depth']); // close sub tree if level down
}
$result .= "<li>$row['name']</li>"; // Always add node
$currDepth = $row['depth'];
}
$result .= "</ul>";
echo $result;
答案 5 :(得分:0)
$linaje='';
$lastnode='';
$sides['izq']=array();
$sides['der']=array();
$print = '<ul>';
foreach ($array as $key1 => $value1){ //Proyectos
if(strpos($info[$key1]['linaje'],'-') !== false)
$compare = strstr($info[$key1]['linaje'],'-',true);
else
$compare = $info[$key1]['linaje'];
if($linaje != ''){
if ($linaje != $compare){
$linaje= $compare;
$sides['izq']=array();
$sides['der']=array();
//for($i=1;$i <= substr_count($lastnode,'`')-substr_count($value1,'`');$i++)
//$print .= '</ul></li>';
}
}
if ($lastnode != '')
for($i=1;$i<= substr_count($lastnode,'`')-substr_count($value1,'`');$i++)
$print .= '</ul></li>';
if (count($sides['der'])>0)
if ($sides['der'][count($sides['der'])-1] > $info[$key1]['der'])
$print .= '<ul>';
$print .= '<li><a href="#'.$info[$key1]['id'].'#'.$info[$key1]['linaje'].'">'.substr($value1,substr_count($value1,'`')).'</a>';
if ($info[$key1]['der'] - $info[$key1]['izq'] == 1)
$print .= '</li>';
if ($key1 == count($info)-1)
for($i=1;$i <= substr_count($lastnode,'`')-1;$i++)
$print .= '</ul></li>';
$sides['der'][] = $info[$key1]['der'];
$sides['izq'][] = $info[$key1]['izq'];
if ($linaje =='')
$linaje = $info[$key1]['linaje'];
$lastnode = $value1;
}
$print .= '</ul>';
echo $print;
不同之处在于您可以渲染X个树,这适用于我的一个项目。当我从DB中获取行时,我使用char作为深度参考
答案 6 :(得分:0)
我正在使用CROSS JOIN查询显示jsTree jQuery菜单;一切都很棒! 现有表格我为该职位添加了一列。但是,当我定义位置并按位置排序时,相应的项目未正确分组。我想这是一个查询问题,尝试了很多组合,但没有成功。