所以我有一个表,列是id,parent-id和name。
出现了什么。
1 Web Design
- 5 Templates
- 6 Finished Websites
2 Graphic Design
3 Photography
4 Image Manipulation
来自:
while($row = mysql_fetch_array( $result )) {
echo $row['id'] . " ";
echo $row['name'] . "<br/ >";
while($row = mysql_fetch_array( $result2 )) {
echo "- " . $row['id'] . " ";
echo $row['name'] . "<br/ >";
}
}
如何动态列出各自父母的所有孩子,例如......
1 Web Design
- 5 Templates
- 6 Finished Websites
2 Graphic Design
- 7 T-Shirts
- 8 Logos
3 Photography
- 9 Portraits
- 10 Nature
- 11 Animal
- 12 Architecture
4 Image Manipulation
答案 0 :(得分:1)
没关系,明白了......
// Get all the data from the "example" table
$result = mysql_query("SELECT * FROM `portfolio-categories` WHERE `parent-id` ='0'")
or die(mysql_error());
// keeps getting the next row until there are no more to get
$parent = 0;
while($row = mysql_fetch_array( $result )) {
echo $row['id'] . " ";
echo $row['name'];
$parent += 1;
echo "#" . $parent . "#";
echo "<br/ >";
$result2 = mysql_query("SELECT * FROM `portfolio-categories` WHERE `parent-id`=$parent")
or die(mysql_error());
while($row = mysql_fetch_array( $result2 )) {
echo "- " . $row['id'] . " ";
echo $row['name'] . "<br/ >";
}
}
答案 1 :(得分:0)
function items($parent_id) { $query = "SELECT * FROM mytable WHERE parent_id = $parent_id" $result = mysql_fetch_array(mysql_query($query)) foreach ($result as &$res) { echo $res['id'] . " "; echo $res['name'] . ""; echo items($parent_id) } }