php - 如何使用 PHP 在 API 服务器的 json 结果响应中加载 URL?
问题描述
<?
$password = substr(md5(microtime()) , rand(0, 26) , 8);
$username = 'SAG' . $password;
$params = ['username' => $username, 'mobile' => 'yes', 'lang' => 'en', 'game_code' => 'lobby', 'page_site' => 'live'];
$curl = curl_init();
curl_setopt_array($curl, array(
CURLOPT_URL => "https://api.gmaster8.com/BBIN/game/open",
CURLOPT_CUSTOMREQUEST => "POST",
CURLOPT_POSTFIELDS => $params,
CURLOPT_RETURNTRANSFER => true,
CURLOPT_HTTPHEADER => array(
"Authorization: Basic U0FHQVBJOjEyMzRxd2Vy"
) ,
CURLOPT_USERPWD => "$username:$password",
));
$result = curl_exec($curl);
echo $result;
?>
我得到了这样的 $result:
{"url":"https:\/\/888.gsoftbb.com\/app\/WebService\/JSON\/display.php\/Login?website=gamingsoft&uppername=dgmaster8rmb&username=C71SAGabfbbcc9&lang=en-us&page_site=live&page_present=live&key=t7ksrtefa9e54610003dfb2b05986de7fcfa6btsmtpho&"}
如何发布 API,然后从 API 服务器加载 url 以响应?我应该用什么代码替换该行
echo $result;
从 API 服务器加载 json 响应中的 url?
解决方案
在你有 echo $result 的地方试试这个
$result = json_decode($result, true);
header("Location: {$result['url']}");