members in x days

[brown2005]

$x = "30";

 

$sql = "SELECT COUNT(*) AS members_count FROM members WHERE ( code here )";

 

now where code here is.. i want to select all members who have logged in in $x days with members_login being the field with the date...

 

thanks in advance

 

 

[pocobueno1388]

<?php

$x = "30";
$sql = "SELECT COUNT(*) AS members_count FROM members WHERE members_login < (NOW() - INTERVAL $x DAY)";

?>

[brown2005]

hi thanks for the help, but that returns a 0 count

[brown2005]

1 2008-03-12 17:22:43

2 2008-03-12 17:22:43

3 2008-02-11 17:22:43

 

i have the above 3 results in my table.. and it comes up with 1.. which is number 3... which is out of 30 days. the code above looks like it gives all members that have logged in before $x amount of days.. but i want from $x days to now....

 

any ideas would be much appreciated..

 

thanks in advance.

 

[pocobueno1388]

Try changing the query to:

 

SELECT COUNT(*) AS members_count FROM members WHERE members_login <= (NOW() - INTERVAL $x DAY)

[brown2005]

nope still 1... wat does

 

<= (NOW() - INTERVAL $x DAY)

 

actually mean?

[pocobueno1388]

This part:

members_login <= (NOW() - INTERVAL $x DAY)

 

means If the date members_login is less than or equal to x days ago.

 

 

I think I got the symbol messed up and turned the wrong way So change it to this:

SELECT COUNT(*) AS members_count FROM members WHERE members_login >= (NOW() - INTERVAL $x DAY)

 

[brown2005]

yes that works.. why on earth did i not spot that..... tut tut..

 

anyway thanks very much