显示访客来源的问题
function getIp(){
$onlineip='';
if(getenv('HTTP_CLIENT_IP')&&strcasecmp(getenv('HTTP_CLIENT_IP'),'unknown')){
$onlineip=getenv('HTTP_CLIENT_IP');
} elseif(getenv('HTTP_X_FORWARDED_FOR')&&strcasecmp(getenv('HTTP_X_FORWARDED_FOR'),'unknown')){
$onlineip=getenv('HTTP_X_FORWARDED_FOR');
} elseif(getenv('REMOTE_ADDR')&&strcasecmp(getenv('REMOTE_ADDR'),'unknown')){
$onlineip=getenv('REMOTE_ADDR');
} elseif(isset($_SERVER['REMOTE_ADDR'])&&$_SERVER['REMOTE_ADDR']&&strcasecmp($_SERVER['REMOTE_ADDR'],'unknown')){
$onlineip=$_SERVER['REMOTE_ADDR'];
}
return $onlineip;
}
$onlineip = getIp();
$ipinfo=json_decode(file_get_contents("http://ip.taobao.com/service/getIpInfo.php?ip=".$onlineip),true);//将API的json格式解码保存到ipinfo变量
?>
IP: | |
国家: | |
省份: | |
城市: | |
县: | |
地区: | |
ISP: |
我这么写的,可以查到IP,但是只能查到一次。以后别的访客来查,显示的还是第一个访客的来源。
用CMS弄的,访客来源没有建立相应数据表。
问问是代码的问题呢,还是没建立相应的数据表。
回复讨论(解决方案)
你说的别的访客和你用的是不同的网络么?如果是同一个公网IP出去的两个人,访问到的看起来就是同一个IP啊。如果你的服务在外网,发给你的其它网络环境的朋友看一下,看是否也这样。
你这段代码单独拿出来也可以测试啊,与CMS没啥关系,而且你这是从
http://ip.taobao.com/service/getIpInfo.php
取的数据,就这个示例来说,不需要建什么表吧。
表现是不是所有的访问都是同意来源呢? 使用的web服务器是否中间经过了代理?
你说的别的访客和你用的是不同的网络么?如果是同一个公网IP出去的两个人,访问到的看起来就是同一个IP啊。如果你的服务在外网,发给你的其它网络环境的朋友看一下,看是否也这样。
你这段代码单独拿出来也可以测试啊,与CMS没啥关系,而且你这是从
http://ip.taobao.com/service/getIpInfo.php
取的数据,就这个示例来说,不需要建什么表吧。
别的访客不是同一个公网的,发给外地的朋友测试过了。显示的是我的IP
获取ip那个方法本来就不严谨,没想到一直流传到现在还有人用
获取ip那个方法本来就不严谨,没想到一直流传到现在还有人用
求个严谨点的
你先只取 $_SERVER['REMOTE_ADDR'] 试一下。
或者你自己应该找出来那段代码里最后生效的是哪个变量,就找到问题所在了。而且这和你自己的服务器网络结构也是有关系的。你应该在PHP手册里Predefined Variables一节搞懂那些环境变量的含义先。不应该直接把一段代码这么丢出来,就算别人帮你解决了,这块对你来说还是盲点。
说的好,解决了,是因为我网页是静态的,现在改成动态的,测试成功了。

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