我正在试图弄清楚如何使用php显示嵌套的MySQL数据。我设法将所有“叶子节点”放在一边,但后来我被困住了。我需要显示整棵树及其所有元素的关系。 这是表格
category_id, name, lft, rgt
1 Saws 1 12
2 Chainsaws 2 7
3 Red 3 4
4 Yellow 5 6
5 Circular saws 8 9
6 Other saws 10 11
这是代码:
$query = 'SELECT node.name, node.lft, node.rgt
FROM item_cats AS node,
item_cats AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt AND parent.name = "' . SAWS . '"
ORDER BY node.lft';
$result = mysql_query($query, $db) or die (mysql_error($db));
while ($row = mysql_fetch_assoc($result)) {
if ($row['rgt'] == $row['lft']+1) {
echo '==>';
}
echo $row['lft'];
echo $row['name'];
echo $row['rgt'];
echo '<br />';
echo '<br />';
}
这就是我得到的:
1Saws12
2Chainsaws7
==>3Red4
==>5Yellow6
==>8Circular saws9
==>10Other saws11
答案 0 :(得分:3)
根据Stu向我展示的链接,本教程显示了用于确定深度的查询:
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
所以这样的事情应该有效:
<?PHP
$query = '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';
$result = mysql_query($query, $db) or die (mysql_error($db));
while ($row = mysql_fetch_assoc($result)) {
for ($i = 0; $i < $row['depth']; $i++) {
echo '==>';
}
echo $row['name'];
echo '<br />';
echo '<br />';
}
?>
这应输出:
Saws
==>Chainsaws
==>==>Red
==>==>Yellow
==>Circular Saws
==>Other Saws
答案 1 :(得分:1)
<?PHP
$query = '
select if(
count(a.name) - 1 = 0,
a.name,
concat(repeat(' ', count(a.name) - 2), '+--', b.name)
)name
from nested_category b, nested_category a
where node.lft between a.lft and a.rgt
group by b.name
order by b.lft';
$result = mysql_query($query, $db) or die (mysql_error($db));
while ($row = mysql_fetch_assoc($result)) echo "{$row['name']}<br>";
?>
应该这样做:
Saws
+--Chainsaws
+--Red
+--Yellow
+--Circular Saws
+--Other Saws