接口提交示例 - 获取返回值并跳转
<?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 ($response === false) {
echo "请求失败";
} else {
echo "<a href='http://qq.com'>点击跳转</a>";
}
curl_close($curl);
}
?>
原文地址: https://www.cveoy.top/t/topic/RnT 著作权归作者所有。请勿转载和采集!