Number of online PHP websites

Just used PHP to write a website online number of procedures, please come in to guide!
I use Php+mysql to write, principle: Website online number of program code + backstage has MySQL database support, you can directly statistics the current number of online Web site.
First I create a MySQL database table.
CREATE TABLE TableName (
field type (max_length) DEFAULT ' Default_value ' (not) NULL
The SQL statement that you can use.
CREATE TABLE Useronline (
timestamp int () DEFAULT ' 0 ' not NULL,
IP varchar not NULL,
File varchar not NULL,
PRIMARY KEY (timestamp),
KEY IP (IP),
KEY file (file)
Below we are PHP scripts, first I define MySQL information.
$server = "localhost"; Your server
$db _user = "root"; The username of your MySQL
$db _pass = "password"; your MySQL password.
$database = "Users"; Table Name SET STATISTICS time (number of seconds online)
$timeoutseconds = 300, take the current time.
$timestamp = time (); The complete code above:
<?php
$server = "localhost"; Your server
$db _user = "root"; Your MySQL database username
$db _pass = "password"; Your MySQL database password if any
$database = "Users"; The DB name
$timeoutseconds = 300;//timeoutseconds Limit
Get the current time
$timestamp = time ();
Calculate the lowest timestamp allowed
$timeout = $timestamp-$timeoutseconds;
?> connection MySQL
mysql_connect (' localhost ', ' username ', ' password '); Variable forms are also allowed.
Mysql_connect ($server, $db _user, $db _pass); If the MySQL database does not have a password, you can use the following code to connect
Mysql_connect ($server, $db _user); Query the code for the database:
Mysql_db_query (' database ', ' query '); We need to add a record as long as we have visitors.
$insert = Mysql_db_query ($database, insert into Useronline VALUES
(' $timestamp ', ' ". $_server[' REMOTE_ADDR ']." ', ' ". $_server[' php_self ');
If the user uses the error message, this is done.
if (!) ( $insert)) {
Print "Useronline Insert Failed >";
Then implement to delete the user record when the setting is exceeded.
$delete = Mysql_db_query ($database, "delete from Useronline WHERE timestamp< $timeout");
if (!) ( $delete)) {
Print "Useronline Delete Failed >";
Below we solve the problem of different IP in the database
$result = Mysql_db_query ($database, "SELECT DISTINCT IP from useronline WHERE file= '". $_server[' Php_self ']. " "); use
Mysql_num_rows (query) to count the users, the code is as follows.
$user = mysql_num_rows ($result), and finally close the database.
Mysql_close () show the number of people online.
if ($user = = 1) {
Print ("1 user online\n");
} else {
Print ("$user users online\n");
Finally write the above code as a PHP file as follows.
<?php
Put your basic server info
$server = "localhost"; Normally localhost
$db _user = "root"; Your MySQL database username
$db _pass = "password"; Your MySQL database password
$database = "Users";
$timeoutseconds = 300; It'll delete all people which haven ' t refreshed (so probbably are
Offline or inactive) in $timieoutseconds time (so it actually checks the people this are active in the last
$timeoutseconds seconds)
This is where PHP gets the time
$timestamp = time ();
Counts the timeout, all people which have been seen last online in earlier than this timestamp, 'll get removed
$timeout = $timestamp-$timeoutseconds;
Connect to Database
Mysql_connect ($server, $db _user);
Add the timestamp from the user to the online list
$insert = Mysql_db_query ($database, insert into Useronline VALUES
(' $timestamp ', ' ". $_server[' REMOTE_ADDR ']." ', ' ". $_server[' php_self ');
if (!) ( $insert)) {
Print "Useronline Insert Failed >";
}
Delete the peoples which haven ' t been online/active in the last $timeoutseconds seconds.
$delete = Mysql_db_query ($database, "delete from Useronline WHERE timestamp< $timeout");
if (!) ( $delete)) {
Print "Useronline Delete Failed >";
}
Select the amount of people online, all uniques, which are online in this page
$result = Mysql_db_query ($database, "SELECT DISTINCT IP from useronline WHERE file= '". $_server[' Php_self ']. " ");
if (!) ( $result)) {
Print "Useronline Select Error >";
}
Count the number of rows = The number of people online
$user = mysql_num_rows ($result);
Spit out the results
Mysql_close ();
if ($user = = 1) {
Print ("1 user online\n");
} else {
Print ("$user users online\n");
}
?>