嘿,伙计们,我想从mysql在我的html视图中创建一个基于树的数据。我有一个表,有两列pay_month和pay_year。
数据是这样的:
pay_month pay_year
February 2014
march 2014
January 2015
February 2015
April 2015
may 2015
June 2015
January 2016
march 2016
Decem
你好
对不起,法语,我会尝试英语,我有像这样的树形数组
$keys = array("elment1","elment1","elment2","elment1");
// this one can have duplicates values ,
$operator = array("=","<",">","=");
// operators for Mysql query
$queries = array("query1",
要在Oracle-Apex中创建树形图,我需要做什么?我已经尝试了所有的方法,但是我不能生成树形图。
我正在尝试使用siguienet查询生成图表:
select case when connect_by_isleaf = 1 then 0
when level = 1 then 1 else -1 end as status,
level,
ename as title,
'icon-tree-folder' as icon,
empno as value,
ename as tooltip,
我在mySQL表中存储了一个类别的树形结构,其中包含category_id和parent_id关系。Parent_id = Null对应于根节点。
Category (category_id, category_name, parent_id)
我要做的是获取所有的叶子节点,给出一个节点的category_id。我关注了的文章。它讨论了如何使用以下查询获取所有叶节点:
SELECT t1.category_name FROM
category AS t1 LEFT JOIN category as t2
ON t1.category_id = t2.parent_id
WHERE t2.cat
我有一个多层的父子关系,我想把它们列在一个树形结构中。我有一个表,里面有像id,parent_id.I这样的字段
$sql = 'SELECT * FROM list WHERE parent_id = 0';
$result = mysql_query($sql, $link);
while ($row = mysql_fetch_array($result)) {
echo $row['id'].'<br>';
$child1_sql = "SELECT * FROM list WHERE paren
我正在开发一个Qt应用程序,我想知道树形视图的项目是否在委托函数中展开。
这是我的树形视图的委托..
void roster_item_delegate::paint(QPainter *painter,
const QStyleOptionViewItem &option,
const QModelIndex &index) const
{
/* How can I know whether this item is expanded o