目前我有2张桌子,只能使用PHP(没有javascript)。
table category:
id, name
1, mymother
2, hismother
和
table subcategory:
id, name, cat_id
1, cool, 2
2, uncool, 1
3, milf, 2
4, ugly, 1
首先,您可以将它们组合在一个下拉框中 所以结果将是:
mymother
-uncool
-ugly
hismother
-cool
-milf
只有子类别可选。
第二,可以将类别/子类别再次拆分为2个原始值,再次重新发布到mysql并插入表nmbr.3
table parents:
id, phonenmbr, cat_id, subcat_id
对于单个值,这很简单:
$result = $mysqli->query("SELECT id, name FROM category");
echo "<select name='category'>";
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['id'] . "'>" . $row['name']
"</option>";
为了再次组合AND拆分category_id和sub_category_id,我想我需要做两个单独的sql查询而不是使用JOIN ....但不确定
答案 0 :(得分:0)
可能在代码下方可以让您了解自己需要做什么:
<form method="get" action="example.php">
<select id="cd" name="cd">
<?php
$mysqlserver="localhost";
$mysqlusername="test";
$mysqlpassword="test";
$link=mysql_connect(localhost, $mysqlusername, $mysqlpassword) or die ("Error connecting to mysql server: ".mysql_error());
$dbname = 'test';
mysql_select_db($dbname, $link) or die ("Error selecting specified database on mysql server: ".mysql_error());
$cdquery="SELECT cdTitle, cdArtist, cdReference FROM firsttable";
$cdresult=mysql_query($cdquery) or die ("Query to get data from firsttable failed: ".mysql_error());
while ($cdrow=mysql_fetch_array($cdresult)) {
$cdTitle=$cdrow["cdTitle"];
$cdArtist=$cdrow["cdArtist"];
$cdReference=$cdrow["cdReference"];
echo "<option value=\"$cdReference\">
$cdTitle $cdArtist
</option>";
}
?>
</select>
</form>