接口提交示例 - 获取跳转链接

<?php
if ($_SERVER["REQUEST_METHOD"] == "POST") {
    $input = $_POST["input"];
    $input2 = $_POST["input2"];
    $url = "https://www.host99.cn/freehost/api.php?name=".$input."&pass=".$input2."&ip=69.165.74.89&key=psychzdl&id=4";

    $curl = curl_init($url);
    curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);
    $response = curl_exec($curl);

    if ($response === false) {
        echo "请求失败";
    } else {
        echo '<a href="http://qq.com"><button>跳转链接</button></a>';
    }

    curl_close($curl);
}
?>