我希望我的jquery的结果显示在我在另一个文件中的表中。
我已尝试更改表的ID，但我的表来自引导程序，因此
如果我更改 ID，css 会更改。我不知道还能做什么。
这些是我的文件
配置.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$db = "stock-stock-stock";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $db );
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
?>
read_students.php
include_once('../config.php');
$sql = "SELECT * FROM games WHERE status='available'";
$query = $conn->query($sql);
$result = array("data" => array());
while($data = $query->fetch_assoc()){
$checkbox = '<input type="checkbox" class="checkbox" id="checkbox_' . $data['id'] . '"
value="' . $data['id'] . '" name="groupCheckBox" onchange="enableDeleteAllButton(this)">';
$image = '<img width="100" height="100" class="" src="' . $data['game_pic'] . '">';
$buttons = '<a href="game-update.php?$id='. $data['id'] . '" class="btn btn-info btn-sm">
<i class="fa fa-edit" aria-hidden="true"></i></a><a onclick="removeGame(' . $data['id'] . ')" class="btn btn-danger btn-sm"><i class="fa fa-trash" aria-hidden="true"></i></a>';
// need "data" string
$result["data"][] = array(
$checkbox,
$data['id'],
$data['game_name'],
$data['players'],
$data['game_descrip'],
$image,
$data['status'],
$buttons,
);
}
echo json_encode($result);
?>
跃然一笑
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