[SOLVED] Need Help!

dhossai · #1 Oct 24th, 2007, 19:06

Hi Friends!

Here I am again, looking for some help. In my page I have people registered from different country. Now I want to show that in my page at a glance how many registered people from which country. And it should change the number automatically and should add the country name automatically for any new entry. Is it too complicated? If anyone would kind enough to show me the code and explain I would be graeful to you. Thanks

c010depunkk · #2 Oct 24th, 2007, 20:27

What kind of a registration system do you have? A database? If so, which fields do you have in your users table?

dhossai · #3 Oct 24th, 2007, 20:42

Hi c010,
I am using MySQL and filed is userid(I user email address), password, Name , address, age, profession, tel, country, active. Hope that is you are asking.

c010depunkk · #4 Oct 25th, 2007, 06:14

OK, then this should be relatively easy....

If you could explain a bit more in detail what kind of help you need (PHP code, or the mySQL query....). Or, even better, post some code....

Lucleonhart · #5 Oct 25th, 2007, 06:34

pMaybe i can add a simple demonstration for you:

PHP: Select all


<?php
// At this point, the connection to the database is already made (mysql_connect...)

// get all countries of the database, order by countryname
$allc_query = mysql_query("SELECT DISTINCT country FROM registertable ORDER BY country");
while($allc_row = mysql_fetch_array($allc_query))
{
  // get the number of users from this country
  $thisc_query = mysql_query("SELECT id FROM registertable WHERE country LIKE '".$allc_row['country']."'");
  $thisc_num = mysql_num_rows($thisc_query);

  // Print that!
  echo("Users from country ".$allc_row['country'].": ".$thisc_num."<br />");
}
?>

Have fun!

dhossai · #6 Oct 25th, 2007, 13:47

Hi c010,
I need the full code for that as I am a novice in php. Your help woud be much appreciated. I want the result to be visible on my page in a text formant as
Country No. of User
Canada 10 Something like this. Thanks

Hi Lucleonhart,
Thanks for your code, but it is showing me only "Users from country: this nothing else....Pls let me know.

dhossai · #7 Oct 25th, 2007, 14:03

Hi Lucleonhart,

Sory it was my mistake. It give me the number of people but not the country name eg: it is like this

Users from country: 10 but no country name?

Lucleonhart · #8 Oct 25th, 2007, 14:08

Oh! there was an mistake from me. just delete one "$" of $$allc_row['country'] in the display line.
I have edited my last post.

dhossai · #9 Oct 25th, 2007, 14:52

Great! it works fine. Now how we can show the result of the scripts on the page without manually running it. and I want that in two colum. first country and 2nd No. of user. Will it automatically update the info everytime you refresh the page ? Hope I am not asking too much?

Lucleonhart · #10 Oct 25th, 2007, 15:08

Ehm.. it actually IS working automatically.. no?

So, you just place it at the position, where the results should appear:

PHP: Select all


<body>
~~here is the head~~
~~then some content~~
<div>~a div maybe~</div>
<table><tr><td>~now comes the stats:
*INSERTSCRIPTHERE*
</td></tr></table>
</body>

You see, it can be placed, whereever you want.

For a table style with two collums, you start the table before the script and end it afterwards. Also you have to edit the display a bit. Here comes the example:

PHP: Select all


<span style='font-weight:bold;'>The Country stats</span>
<table>
<?php
// At this point, the connection to the database is already made (mysql_connect...)

// get all countries of the database, order by countryname
$allc_query = mysql_query("SELECT DISTINCT country FROM registertable ORDER BY country");
while($allc_row = mysql_fetch_array($allc_query))
{
  // get the number of users from this country
  $thisc_query = mysql_query("SELECT id FROM registertable WHERE country LIKE '".$allc_row['country']."'");
  $thisc_num = mysql_num_rows($thisc_query);

  // Print that!
  echo("<tr><th>Users from country ".$allc_row['country'].":</th><td>".$thisc_num."</td></tr>");
}
?>
</table>

dhossai · #11 Oct 25th, 2007, 16:10

Hi,
I was just trying the basic one with following code and getting on page just "); } ?> any idea

PHP: Select all


<html>
<head>
<meta http-equiv="Content-Language" content="en-us">
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<title>New Page 1</title>
</head>
<body bgcolor="#60B773">
<?php
// At this point, the connection to the database is already made (mysql_connect...)
include "include/z_db.php";
// get all countries of the database, order by countryname
$allc_query = mysql_query("SELECT DISTINCT country FROM tbl_login ORDER BY country");
while($allc_row = mysql_fetch_array($allc_query))
{
  // get the number of users from this country
  $thisc_query = mysql_query("SELECT id FROM tbl_login WHERE country LIKE '".$allc_row['country']."'");
  $thisc_num = mysql_num_rows($thisc_query);
  // Print that!
  echo("country ".$allc_row['country'].": ".$thisc_num."<br />");
}
?></body>
</html>

Lucleonhart · #12 Oct 25th, 2007, 20:09

You have to consider, that this is a PHP script! You cannot open it like a normal document. Ithas to be parsed through a PHP server first.
To check it manually, you should install a package like xampp (http://www.apachefriends.org/de/xampp.html). It will install an Apache Webserver with PHP support and other additional featurs like MYSQL.

So after you installed it, put your file in the "htdocs" directory of xampp, start xampp using the "xampp-control.exe" and open your browser.
DO NOT simply open the file, instead type
http://localhost/yourfilename.php
localhost points to the "htdocs" folder of xampp. Now, the PHP script will be transmitted through the PHP server and will be executed.

dhossai · #13 Oct 25th, 2007, 20:54

Hi Lucleonhart,
Yes, I am using a php server where it works fine with your first code. It just not working after I put that in to a existing html file as you described. Thanks

c010depunkk · #14 Oct 25th, 2007, 21:26

How about posting the code you have. We can then have a look and see if we can spot the problem.

dhossai · #15 Oct 25th, 2007, 22:07

Hi,
Just see the three postings earlier , I have posted the code with error message, thanks

Rob · #16 Oct 25th, 2007, 23:42

can you upload the code in .txt format using an attachment?

dab42pat · #17 Oct 26th, 2007, 00:06

HTML: Select all

<html>
<head>
<meta http-equiv="Content-Language" content="en-us">
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<title>New Page 1</title>
</head>
<body bgcolor="#60B773">
<?php
// At this point, the connection to the database is already made (mysql_connect...)
include "include/z_db.php";
// get all countries of the database, order by countryname
$allc_query = mysql_query("SELECT DISTINCT country FROM tbl_login ORDER BY country");
while($allc_row = mysql_fetch_array($allc_query))
{
  // get the number of users from this country
  $thisc_query = mysql_query("SELECT id FROM tbl_login WHERE country LIKE '".$allc_row['country']."'");
  $thisc_num = mysql_num_rows($thisc_query);
  // Print that!
  echo("country ".$allc_row['country'].": ".$thisc_num."<br />");
}
?></body>
</html>

Sorry if I am pointing out the obvious, but have you opened your db connection on line one of the above page before the first <html> tag.

dhossai · #18 Oct 26th, 2007, 00:34

Hi dab42pat

PHP: Select all


include "include/z_db.php";

this is my db connection page

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[SOLVED] Need Help!