将 PDO::FETCH_ASSOC 转换为 SQLite
Convert PDO::FETCH_ASSOC to SQLite
我希望将我的一些基本设置(例如我的导航)、mysql 设置(主机、用户名、密码)转换为 SQLite,但从我今天早上读到的内容来看,似乎没有等效命令PDO::FETCH_ASSOC
。有没有其他方法可以让我执行与 SQLite 兼容的查询?
<?
$sql = "SELECT * FROM menu_items WHERE status = 'ACTIVE' ORDER BY menu_parent_id ASC, sortorder ASC, menu_item_name ASC";
$query = $db->query($sql);
$menu_items = array();
while($data = $query->fetch(PDO::FETCH_ASSOC)) {
if($data['menu_parent_id'] == 0) {
$menu_items[$data['menu_item_id']] = array();
$menu_items[$data['menu_item_id']]['menu_item_id'] = $data['menu_item_id'];
$menu_items[$data['menu_item_id']]['name'] = $data['menu_item_name'];
$menu_items[$data['menu_item_id']]['url'] = $data['menu_url'];
$menu_items[$data['menu_item_id']]['fontawesome'] = $data['fontawesome'];
$menu_items[$data['menu_item_id']]['children'] = array();
} else if($data['menu_parent_id'] != 0) {
$tmp = array();
$tmp['menu_item_id'] = $data['menu_item_id'];
$tmp['name'] = $data['menu_item_name'];
$tmp['url'] = $data['menu_url'];
$tmp['fontawesome'] = $data['fontawesome'];
array_push($menu_items[$data['menu_parent_id']]['children'],$tmp);
unset($tmp);
}
}
function create_list($arr)
{
$html = "";
foreach($arr as $key => $value) {
//Here the menu item has children
if(count($value['children']) > 0) {
$html .= '<!-- PARENT --> <li class="mm-dropdown"> <a href="'. $value['url'] .'"><i class="menu-icon fa '. $value['fontawesome']. '"></i><span class="mm-text">'.$value['name'].'</span></a>
<ul>';
// Here it is the child
foreach($value['children'] AS $child) {
$html .= '
<!-- child --><li><a href="'. $child['url'] .'" tabindex="-1" id="'.$child['menu_item_id'].'"><i class="menu-icon fa '. $child['fontawesome']. '"></i>'.$child['name'].'</a></li>';
}
$html .= ' </ul>
';
} else{
$html .= ' <a class="menu-icon fa '.$value['fontawesome'].'" id="'.$value['menu_item_id'].'" >'.$value['name'].'</a>';
}
}
return $html;
}
?>
所以你的 $query
是 SQLite3Result ??
那么试试这个方法:
while($data = $query->fetchArray(SQLITE3_ASSOC)) {
我希望将我的一些基本设置(例如我的导航)、mysql 设置(主机、用户名、密码)转换为 SQLite,但从我今天早上读到的内容来看,似乎没有等效命令PDO::FETCH_ASSOC
。有没有其他方法可以让我执行与 SQLite 兼容的查询?
<?
$sql = "SELECT * FROM menu_items WHERE status = 'ACTIVE' ORDER BY menu_parent_id ASC, sortorder ASC, menu_item_name ASC";
$query = $db->query($sql);
$menu_items = array();
while($data = $query->fetch(PDO::FETCH_ASSOC)) {
if($data['menu_parent_id'] == 0) {
$menu_items[$data['menu_item_id']] = array();
$menu_items[$data['menu_item_id']]['menu_item_id'] = $data['menu_item_id'];
$menu_items[$data['menu_item_id']]['name'] = $data['menu_item_name'];
$menu_items[$data['menu_item_id']]['url'] = $data['menu_url'];
$menu_items[$data['menu_item_id']]['fontawesome'] = $data['fontawesome'];
$menu_items[$data['menu_item_id']]['children'] = array();
} else if($data['menu_parent_id'] != 0) {
$tmp = array();
$tmp['menu_item_id'] = $data['menu_item_id'];
$tmp['name'] = $data['menu_item_name'];
$tmp['url'] = $data['menu_url'];
$tmp['fontawesome'] = $data['fontawesome'];
array_push($menu_items[$data['menu_parent_id']]['children'],$tmp);
unset($tmp);
}
}
function create_list($arr)
{
$html = "";
foreach($arr as $key => $value) {
//Here the menu item has children
if(count($value['children']) > 0) {
$html .= '<!-- PARENT --> <li class="mm-dropdown"> <a href="'. $value['url'] .'"><i class="menu-icon fa '. $value['fontawesome']. '"></i><span class="mm-text">'.$value['name'].'</span></a>
<ul>';
// Here it is the child
foreach($value['children'] AS $child) {
$html .= '
<!-- child --><li><a href="'. $child['url'] .'" tabindex="-1" id="'.$child['menu_item_id'].'"><i class="menu-icon fa '. $child['fontawesome']. '"></i>'.$child['name'].'</a></li>';
}
$html .= ' </ul>
';
} else{
$html .= ' <a class="menu-icon fa '.$value['fontawesome'].'" id="'.$value['menu_item_id'].'" >'.$value['name'].'</a>';
}
}
return $html;
}
?>
所以你的 $query
是 SQLite3Result ??
那么试试这个方法:
while($data = $query->fetchArray(SQLITE3_ASSOC)) {