我的查询不是像/search.php？cat=2，我希望它是/search.php？cat=SomeName。
我刚开始学习PHP和MYSQL，当然我觉得有很多我不知道的标准实践。我只是掌握了窍门（你可能不同意；），并构建了一个将名称替换为ID的函数，这很有效。
有更标准的方法吗？
工作代码：
function swap_name_for_id($name, $search) {
global $connection;
if ($search == 'category') {
$query = "SELECT categories.id, categories.name
FROM categories
WHERE categories.name = '{$name}'";
}
$swap_set = mysqli_query($connection, $query);
confirm_query($swap_set);
while ($swap = mysqli_fetch_array($swap_set)) {
$search_id = $swap["id"];
}
return $search_id;
}
function get_parts_for_category($category_name) {
$category_id = swap_name_for_id($category_name, 'category');
global $connection;
$query = "SELECT parts.id, parts.name, parts. part_category
FROM parts
WHERE part_category = {$category_id}
ORDER BY name ASC";
$part_set = mysqli_query($connection, $query);
confirm_query($part_set);
return $part_set;
}
据我所知，通常情况下，您只需像您一样执行第二个查询，尽管它们通常组合成一个查询，如下所示：
function get_parts_for_category($category_name) {
global $connection;
$query = "SELECT parts.id, parts.name, parts.part_category
FROM parts
WHERE part_category = (SELECT categories.id FROM categories WHERE categories.name = '{$category_name}')
ORDER BY name ASC";
$part_set = mysqli_query($connection, $query);
confirm_query($part_set);
return $part_set;
}