
https://developers.google.com/safe-browsing/v4/lookup-api
Google 提供的检测网址安全性的接口,不过按照文档 curl 返回的 json 是空的,又用 Stackoverflow 上看到的代码测试了,返回的也是空的。是我姿势不对嘛?
附代码:
<?php $url = 'https://v2ex.com'; $apiKey = 'YOUR API KEY'; $apiUrl = 'https://safebrowsing.googleapis.com/v4/threatMatches:find?key='.$apiKey; $params = [ 'client' => [ 'clientId' => 'foobar', 'clientVersion' => '1.2.3' ], 'threatInfo' => [ "threatTypes" =>["MALWARE", "SOCIAL_ENGINEERING"], "platformTypes" => ["WINDOWS"], 'threatEntryTypes' => ['URL'], 'threatEntries' => [ [ 'url' => $url ] ] ] ]; $ch = curl_init($apiUrl); curl_setopt_array($ch, [ CURLOPT_POST => 1, CURLOPT_RETURNTRANSFER => 1, CURLOPT_HEADER => 1, CURLOPT_POSTFIELDS => json_encode($params), CURLOPT_HTTPHEADER => [ 'Content-Type: text/json' ] ]); $res = curl_exec($ch); ?> <pre><?php echo print_r($res, true); ?></pre> <?php ?> 1 learningman 2020-03-08 20:54:40 +08:00 有没有想过是我国的独特网络环境 |
2 dot2017 2020-03-08 21:33:00 +08:00 @learningman 然而 safebrowsing.googleapis.com 是解析刀谷歌北京的 |
3 mashirozx OP |
4 o0 2020-03-09 16:01:50 +08:00 谷歌还有这么神奇的接口?以前找过,但不是说要停止使用了吗,或者每天限 100 次 |