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我希望有人可以帮助我将这段代码从对象形式转换为过程形式。如何将PHP代码从对象转换为过程需要帮助?
因此,当我尝试转换代码片段的所有对象形式的代码片段时,我可以对此做些什么。
我是一个全新的初学者。
$sql = "SELECT id, label, link_url, parent_id FROM dyn_menu ORDER BY parent_id, id ASC";
$items = mysql_query($sql);
while ($obj = mysql_fetch_object($items)) {
if ($obj->parent_id == 0) {
$parent_menu[$obj->id]['label'] = $obj->label;
$parent_menu[$obj->id]['link'] = $obj->link_url;
} else {
$sub_menu[$obj->id]['parent'] = $obj->parent_id;
$sub_menu[$obj->id]['label'] = $obj->label;
$sub_menu[$obj->id]['link'] = $obj->link_url;
$parent_menu[$obj->parent_id]['count']++;
}
}
mysql_free_result($items);
我知道这段代码有问题吗?
$mysqli = new mysqli("localhost", "root", "", "sitename");
$dbc = mysqli_query($mysqli,"SELECT id, label, link_url, parent_id FROM dyn_menu ORDER BY parent_id, id ASC");
if (!$dbc) {
// There was an error...do something about it here...
print mysqli_error();
}
while ($obj = mysql_fetch_assoc($dbc)) {
if ($obj->parent_id == 0) {
$parent_menu $obj['id']['label'] = $obj['label'];
$parent_menu $obj['id']['link'] = $obj['link_url'];
} else {
$sub_menu $obj['id']['parent'] = $obj['parent_id'];
$sub_menu $obj['id']['label'] = $obj['label'];
$sub_menu $obj['id']['link'] = $obj['link_url'];
$parent_menu $obj['parent_id']['count']++;
}
}
mysql_free_result($dbc);
不要只是为了这样做。 – ceejayoz 2009-10-26 13:29:50
你有什么是一个干净的方法,为什么改变它? – 2009-10-26 13:31:14
为什么不会很有帮助。 – stack 2009-10-26 13:31:38