php - 将来自 mysql 查询的 json 响应填充到另一个文件中的 html 列表

Question

您好我正在尝试填充从 MySQL 数据库获得的数据。我在这样的 json 中得到了所有查询结果：

function getUserInvoicesList($id){
    $table = "facturas";
    $query = "Select * from $table WHERE idcliente = ".$id;
    $resQuery = mysqli_query($this->conn, $query);
    $user = mysqli_fetch_all($resQuery);
    // print_r($user);
    if($user){
        return array('status'=>true, 'data' => $user);
    }
    return array('status'=>false);;
}

这显示了 json 中的结果，如：

{status: 200, data: [,…]}
data
:
[,…]
0
:
["00000431", "0", "000010", "2018-01-01", "2018-01-30", null, "0.37", "Contado", "No pagado", "1",…]
1
:
["00000432", "431", "000010", "2018-01-01", "2018-01-30", null, "1.39", "Contado", "No pagado", "2",…]
status
:
200

我将代码放在一个文件中，而 html 是一个名为 invoices.html 的单独文件

<div class="page-content">
    <div class="list no-hairlines media-list inv-list">
        <ul>
            <li class="swipeout">
                <a href="/invoice-view/" class="item-link item-content swipeout-content">
                    <div class="item-inner">
                        <div class="item-title-row">
                            <div class="item-title">#INV-0001</div>
                            <div class="item-after inv-amount">$500</div>
                        </div>
                        <div class="item-subtitle m-t-5"><span class="inv-client">Global Technologies</span> <span class="badge color-green inv-status">Paid</span></div>
                    </div>
                </a>
                <div class="swipeout-actions-right">
                    <a href="/edit-invoice/" class="color-green">Edit</a>
                    <a href="#" class="color-red swipeout-delete">Delete</a>
                </div>
            </li>

这就是我获取结果的方式：

if(isset($_POST)){
    $user = new User();
    $response = array();
    $data = json_decode(file_get_contents("php://input"));
    if($data->action == 'getuserinvoices'){
        $res = $user->getUserInvoices($data->id);
        if($res['status']){
            $response['status'] = 200;
            $response['data'] = $res['data'];
        }
        else {
            $response['status'] = 400;
            $response['message'] = 'Unable to get the Data';
        }

如何将 MySQL 查询的所有结果填充到该 html 列表中，它应该如下所示：

标签： phpmysqljsonhtml