如何获取当前访问的html页面并存储在php我的sql中

Question

I need the currently viewed page to be stored in php mysql. I have a single line of html page and i need to store the data like if users visit about us then the url followed by about. Since it is a single html page index.html and all the menus are provided within that index.html

I already have a db and PHP code but did not work as expected

 $page = "https://chennaifreelancers.com".$_SERVER['HTTPS_HOST']."".$_SERVER['PHP_SELF'];

I tried

$page="https://chennaifreelancers.com".$_SERVER['HTTPS_HOST']."".$_SERVER['index.html']; 

and

$page="https://chennaifreelancers.com".$_SERVER['HTTPS_HOST']."".$_SERVER['SERVER_NAME'];

Included this as well in my html page

<header class="bgimg-1 w3-display-container w3-grayscale-min" id="home" form method="post" action="get_details.php">

PHP Version: 5.5 and Maria DB: 10.2

#get_details.php

<?php
// require ReCaptcha class

  require_once("DBConfig.php");

 $ipaddress = $_SERVER['REMOTE_ADDR'];
 $page= https://chennaifreelancers.com".$_SERVER['HTTPS_HOST']."".$_SERVER['index.html'];
 $referrer = $_SERVER['HTTP_REFERER'];
 $datetime = date("F j, Y, g:i a");
 $useragent = $_SERVER['HTTP_USER_AGENT'];


$query = mysql_query("SELECT `ip` FROM `visitors_table` WHERE `ip` = '$ip' ");
if(mysql_num_rows($query)> 0) { // if return 1, email exist.

} else { // else not, insert to the table

  $query = mysql_query("INSERT INTO `visitors_table` (`ip` ,`current_page`,`referrer`,`time` ,`user_agent`)
                            VALUES ('$ipaddress','$page','$referrer','$datetime','$useragent')");


    $result=mysql_query($query);




     $result=mysql_query($query);




    } 



 close($query); 

Actual : https://chennaifreelancers.com/get_details.php this is recorded in the DB but

Expected: https://chennaifreelancers.com/#about etc i require this to be stored in DB

Answer 1

You could do with ajax call while hashchange() event

window.addEventListener("hashchange", store, false);

function store() {
  //call get_details.php via ajax call
  $.ajax({
    type: "post",
    url: "get_details.php",
    success: function(response) {
      $("#responsecontainer").html(response);
    }
  });
}