我尝试使用google safe browsing API
来检查任何给定网址的分类。
根据Google安全浏览API页面(https://developers.google.com/safe-browsing/v4/lookup-api),我发出了cURL
POST
HTTP
个#{}
请求,并检查网站分类。
作为回报,我没有看到任何回应我的POST' ed。
收到回复:
#curl -H "Content-Type: application/json" -X POST -d ' { "client": { "clientId": "Test", "clientVersion": "1.0.0" }, "threatInfo": { "threatTypes": ["MALWARE", "SOCIAL_ENGINEERING"], "platformTypes": ["WINDOWS"], "threatEntryTypes": ["URL"], "threatEntries": [ {"url": "http://www.facebook.com"} ] } }' https://safebrowsing.googleapis.com/v4/threatMatches:find?key=AIzaSyD1IMgjaHEza6e9m_jwtjBgPmJX0IMKKIs HTTP/1.1
使用的命令:
<?php
ob_start();
if($_POST["nickname"] and $_POST["password"]) {
$username = $_POST["nickname"];
$password = $_POST["password"];
}
else
{
echo "Please fill the form";
exit();
}
$conn = new mysqli("localhost", $username, $password, "db");
if ($conn->connect_error) {
echo "Connection failed"; // $conn->connect_error
}
$conn->close();
session_start();
$_SESSION["nickname"] = $username;
$_SESSION["password"] = $password;
ob_end_clean();
header("location: ../interno/home/home.php");
exit();
?>
我检查了命令,但不确定我在哪里犯错误。人们有什么想法?
答案 0 :(得分:0)
好的,根据Google的说法,您测试的网址( www.facebook.com )可能是安全的 - 因为如果我发出您的命令,但 ianfette.org ,我得到这个回复:
{
"matches": [
{
"threatType": "MALWARE",
"platformType": "WINDOWS",
"threat": {
"url": "http://ianfette.org"
},
"cacheDuration": "300s",
"threatEntryType": "URL"
}
]
}