function get_data()
{
$connect = mysqli_connect("localhost", "root", "root", "Database");
$query = "SELECT * FROM Furniture";
$result = mysqli_query($connect, $query);
$categories = array();
$row = mysqli_fetch_array($result);
{
$categories[] = array(
'mapwidth' => "2000",
'mapheight' => "2000",
);
}
这是我到目前为止所做的,但是我怎么能够创建一个“类别”数组,以便json文件看起来像这样,
{
"mapwidth": "2000",
"mapheight": "2000",
"categories": [
{
"id": "furniture",
"title": "furniture",
"color": "#4cd3b8",
"show": "false"
},
{
"id": "rooms",
"title": "Rooms",
"color": "#63aa9c",
"show": "true"
}
],
我尝试这样做,但我只能显示一行,但不知道如何遍历数据库,以便每一行都显示“id”,“title”,“color”和“显示“在json文件中看到的括号内。所以我只是想让json文件看起来与这里显示的完全一样。
答案 0 :(得分:0)
试试这样:
$employee_data = array();
$categories = array();
$employee_data["mapwidth"] ="2000";
$employee_data["mapheight"] ="2000";
while($row = $result->fetch_array(MYSQL_ASSOC)) {
$categories[] = $row;
}
$employee_data["categories"] =$categories;
echo json_encode($employee_data);
请参阅this answer以进一步改进您的代码。