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Home Backend Development PHP Tutorial 显示访客来源的问题

显示访客来源的问题

Jun 23, 2016 pm 01:37 PM


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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