我有点卡住了，我想把这个PHP列表。
<?php
include 'conexion.php';
$queri= mysqli_query($est, "SELECT * FROM objeto_gasto order by codigo" );
?>
<!DOCTYPE html>
<html lang="en">
<head>
<title>lista objetos</title>
</head>
<body>
<div>
<select name="objetos">
<?php
while($datos_obj= mysqli_fetch_array($queri))
{
?>
<option value="<?php echo $datos_obj['codigo'] ?>"><?php echo $datos_obj['detalle']?> </option>
<?php
}
?>
</select>
</div>
与数据库的连接工作正常，并且它还在所示代码的选择中显示了列表，但现在我想将其放入我的 HTML 页面中，而不将 HTML 文件更改为 PHP。
此外，学习和提高我的编程技能并将其付诸实践。
如果我理解正确的话，你需要这个（不完整，抱歉）
使用 JavaScript 和 Ajax
newphp.php
<?php
include_once 'config.php';
$select = $connection->prepare("SELECT username FROM table");
$select->execute();
$select = $select->fetchAll();
foreach ($select as $row) {
echo $row['username'] . ',';
}
?>
newhtml.html
<!DOCTYPE html>
<html lang="en">
<body>
<div id="demo"></div>
<script>
function receive() {
const http = new XMLHttpRequest();
http.onload = function () {
var splits = this.responseText.split(','), select = '';
for (let i = 0; i < splits.length - 1; i++) {
select += '<option>' + splits[i] + '</option>';
}
document.getElementById('demo').innerHTML = '<select>' + select + '</select>';
};
http.open("POST", "newphp.php");
http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
http.send();
}
receive();
</script>
</body>
</html>