[SOLVED] Last Visited. Weigh in Please!

[Trium918]

What are the steps that I will need in order to keep

up with Users last visited and diplay it to their member

page.

 

Last Visited  5/6/2007

[skali]

1- You need to keep a field in your users table

2- when user logs into you website update this field with the timestamp of server at that moment

3- display this field where you want

[Trium918]

When you say the user table are you refering to the

database?

[Barand]

1- You need to keep a field in your users table

2- when user logs into you website update this field with the timestamp of server at that moment

3- display this field where you want

 

Step 1a, before you update, save the existing value as $lastVisit, otherwise you be displaying the time of this visit

[Trium918]

This is what I have so far, and what is wrong with it?

 

Question: When would the lastVisit column

update. During registeration or during login.

If its during login the field is lefted blank at

registeration.

 

CREATE TABLE users (username varchar(30),password varchar(32), 
lastVisit TIMESTAMP NULL DEFAULT NULL);

<?php
$currentVisit = date();
mysql_query( "update users set lastVisit= '$currentVisit'
where user_name = '$user_name'");
?>

[Trium918]

Step 1a, before you update, save the existing value as $lastVisit, otherwise you be displaying the time of this visit

 

When would this process take place?

[rcorlew]

What I do is to only track logins. You have to concate the the time of login into a mysql field using a common delimiter to separate time entries.

 

When the time of login is checked you can simply manipulate the $row['login_time'] by exploding it into an array.

 

$log_times = explode(",", $row['login_time']);//If you use a comma as the delimiter

 

You could then count the array using:

$log_check = count($log_times -1); Php starts arrays at position 0, this will get you current login

 

Now you can use the exploded array along with the array count to display the current login, if you want to you can use other variables to get various logins by manipulating the math formula on $log_check:

 

$last_log =  "$log_times[$log_check]";

 

echo "$last_log";

 

I hope that gets you going in the right direction.

[Trium918]

What I do is to only track logins. You have to concate the the time of login into a mysql field using a common delimiter to separate time entries.

 

When the time of login is checked you can simply manipulate the $row['login_time'] by exploding it into an array.

 

$log_times = explode(",", $row['login_time']);//If you use a comma as the delimiter

 

You could then count the array using:

$log_check = count($log_times -1); Php starts arrays at position 0, this will get you current login

 

Now you can use the exploded array along with the array count to display the current login, if you want to you can use other variables to get various logins by manipulating the math formula on $log_check:

 

$last_log =  "$log_times[$log_check]";

 

echo "$last_log";

 

I hope that gets you going in the right direction.

 

Where does $row['login_time']; get its value from?

[rcorlew]

That would be in the user table. You would have a field that would track their login times, I just used the name login_time to make it easier to follow.

[Trium918]

That would be in the user table. You would have a field that would track their login times, I just used the name login_time to make it easier to follow.

 

This would be the field in database correct?

login_time TIMESTAMP NULL DEFAULT NULL

 

[rcorlew]

That looks correct.

 

Here is what your query would look like:

 

<?php
$result = mysql_query("UPDATE users SET login_time = CONCAT(COALESCE(login_time, ''), ' , $time)");
?>

[Trium918]

Correct me if I am wrong, but wouldn't the syntax

be similar to the code below for saving the existing

value as $lastVisit.

 

Note: The database field is set as TIMESTAMP NULL DEFAULT NULL

Also, I noticed that when I enter data during registeration the TIMESTAMP

field auto updated time. Not sure if this is what I want.

<?php
$currentVisit = date();
mysql_query( "update users set lastVisit= '$currentVisit'
where user_name = '$user_name'");
?>

[tauchai83]

create a field for user-> Last Visit  (datetime) NOT NULL  default:0000-00-00 00:00:00

 

then:

<?php

$sql1="SELECT Last_visit FROM user WHERE user_id='$userid'";
$result1=mysql_query($sql1);
$row=mysql_fetch_assoc($result1);
$last=$row['Last_visit'];
$_SESSION['last']=$last;   //use the session variable to hold the previous last visit date.


$sql="UPDATE user SET Last_visit=NOW() WHERE user_id='$userid'";   //now only update it
$result=mysql_query($sql);

?>

 

it should work well for you.

 

Regards,

chai

 

[Trium918]

Thanks tauchai8! I got it working, but I would I

get the date to display 5/07/2007 format?

[Trium918]

Why doesn't DATE_FORMAT(date_column, '%m d% Y%')

work in the code below?

 

<?php
$sql1="SELECT DATE_FORMAT(Last_visit, '%m %d %Y') FROM members_info WHERE user_name='$user_name'";
?>

[benjaminbeazy]

depends, if you're field is unix timestamp, try

 

FROM_UNIXTIME(Last_vist, '%m %d %Y')

[Trium918]

The data field is set as Datetime not NULL default'0000-00-00'

[Trium918]

What am I doing wrong?

I am looking for this format MMDDYYYY

 

last_visit datetime not NULL default'0000-00-00' 
returns YYYYMMDDHHMMSS

last_visit TIMESTAMP NULL DEFAULT NULL
returns YYYYMMDDHHMMSS

 

<?php

$sql1="SELECT last_visit FROM users WHERE user_name='$user_name'";
$result1=mysql_query($sql1);
$row=mysql_fetch_assoc($result1);
$last=$row['last_visit'];
$_SESSION['last']=$last;   //use the session variable to hold the previous last visit date.


$sql="UPDATE users SET last_visit=NOW() WHERE user_name='$user_name'";   
//now only update it
$result2=mysql_query($sql);  
?>

[Trium918]

Were are them PHP Gurus?

[Trium918]

Can anyone explain to me what is it that

I am doing wrong. I an trying to get the

last visit to display as 5/8/2007. Currently,

I'm getting the output 20070508155335.

 

If there are any question that will help you

in guiding me, please ask!

[Barand]

<?php
$d = '20070508155335';

echo date ('n/d/Y', strtotime($d));             //--> 5/08/2007
?>

[Trium918]

<?php
$d = '20070508155335';

echo date ('n/d/Y', strtotime($d));             //--> 5/08/2007
?>

 

Shouldn't $d= what ever date is being pulled from the

database, just asking?

[per1os]

Wow, yes $d should be whatever date you want to be converted to a specific format.

 

strtotime is a wonderful function, glad the php coders included it. Makes date conversion a ton easier!

[Trium918]

<?php
$d = '20070508155335';

echo date ('n/d/Y', strtotime($d));             //--> 5/08/2007
?>

 

I just tried this as a test and I got a blank screen!

[Trium918]

<?php
$d = '20070508155335';

echo date ('n/d/Y', strtotime($d));             //--> 5/08/2007
?>

 

I just tried this as a test and I got a blank screen!

 

The problem was the time added at the end 155335. Is there

a way to do away with this?