display a mysql cel in a page???

sis3970

Hi,
I have a site, with a few members, now i want to create a page that displays all of my members. i use MySQL and php.

I'm new to all this php stuff so forgive me for stupid questions

Greets
Sis3970

daynah

Hello there. I think this belongs in the database talk forum, so hopefully one of the admins will move it later. 🙂

But to answer your question, you may have to research more on mysql.com with the things I talk about in this post.

Now I'm assuming your members are stored in some sort of mysql database. So in your script, you want to extract the data out of this database. The steps to doing this are:

1) Connect to the database
2) Query the data you want to extract
3) Print out the data on a nice webpage

So the start of your code may look like this:

<?
// Now into the MySQL. Open the database
$db = mysql_connect( $host, $username, $dbpword );
$connection = mysql_select_db( $database, $db );

This connects to your database. You will have to fill in $host, $username, and $dbpword with the correct data. $host is 99% of the time "localhost."

// Query: Get all members and their info
$sql = "SELECT * FROM members";
$the_members = mysql_query($sql);

Now you want to run the query to get the members. the 'members' using in the $sql will be your database table with all the member information in it.

while($my_members = mysql_fetch_array($the_members))
{

printf("%s, %s <br />",
$my_members['username'], $my_members['birthday']);
}
?>

Now you want to print out the page with the data. In this example, I printed out the username and their birthday, but you can change it to your liking.

The while statement says, "while there is still data from the query, print 'username, birthday<br />"

I hope that's understandable and helps you. 🙂

sis3970

Thank you mate.

Just what i needed, i'll try to figure this out now

Greets
Sis

sis3970

This is what i figured out

<?
// Now into the MySQL. Open the database
$host = "localhost"
$db = "cscforum"
$username = "cscforum"
$dbpword = "bk146265"
$db = mysql_connect( $host, $username, $dbpword );
$connection = mysql_select_db( $database, $db );

$sql = "t_users";
$the_members = mysql_query($sql);

while($my_members = mysql_fetch_array($the_members))
{

printf("%s, %s <br />",
$my_members['username'], );

}
?>

And this is the error i get....

Parse error: parse error in /usr/local/psa/home/vhosts/sis3970.com/httpdocs/cscclan/viewmem/index.php on line 4

Can you see what's wrong?
Tnx
Sis

Shankspony

<?
// Now into the MySQL. Open the database
$host = "localhost"
$db = "cscforum"
$username = "cscforum"
$dbpword = "bk146265"
$db = mysql_connect( $host, $username, $dbpword );
$connection = mysql_select_db( $database, $db );

all this nooB notices is that you don't have semi-colons to terminate your first 4 lines... when parsed it will see this part as something like:

$host = "localhost" $db = "cscforum" $username = "cscforum" $dbpword = "bk146265" $db = mysql_connect( $host, $username, $dbpword );

ergo an error... most of my own errors are caused by lack of termination whether it's line code (😉 or entire blocks (?>).

daynah

<? 
// Now into the MySQL. Open the database 
$host = "localhost"; 
$db = "*********"; 
$username = "******"; 
$dbpword = "********"; 
$db = mysql_connect( $host, $username, $dbpword ); 
$connection = mysql_select_db( $database, $db ); 

$sql = "SELECT * FROM t_users"; 
$the_members = mysql_query($sql); 

while($my_members = mysql_fetch_array($the_members)) 
{ 

printf("%s <br />", 
$my_members['username'] ); 
} 
?>

This is how the code should be, except you replace the *** with your information.

PLEASE removed your password from your post. You don't want anyone to have access to your site.

P.S. I'm a girl. heh

siwis

Just a little advice. When posting your code, don't reveal your db usernames or passwords.
It doesn't take much to query a domain for the host , so anybody could theoretically then crack into your database remotely and edit the contents.

But as for your problem

; should be at the end of every command, as already correctly stated by shankspony :-) 😃

siwis

sorry daynah, still an error there:

$dbpword = "********"

should be

$dbpword = "********";

sis3970

Thanx for the reply,

Can you visit my site's forum, and tell me
or i should give up or keep trying :-)

Greets
Sis

www.cscclan.com

daynah

sis3970, where do I find it on your page?

siwis, thanks! I just fixed the script. 🙂

sis3970

http://www.cscclan.com

Greets
Sis

btw; sorry for the girl/guy thing...

daynah

I'm not too sure what I'm suppose to do after I click on your page. The members section takes me to a login page. And I don't see a member list on your forum. Did the script above work?

sis3970

I seem to not understand this line:
while($my_members = mysql_fetch_array($UserName))

I guess $UserName = the celname containing the loginnames??

I know, Iknow
I am a noob, but someday, I'll help other noobs like me. :-)

Greets
Sis

siwis

If you look carefully at what daynah did:

Create and perform a query.
$the_members = mysql_query($sql);

Cycle through the array to print the values to screen: Each row returned from the database query assigned to $the_members temparaily gets assigned as an array to $my_members:

while($my_members = mysql_fetch_array($the_members)) {

// Then (and this would be my method, 
// assign the current row value to a single
// variable and then echo it to the screen 
// (plus a standard html line break.
// Daynah example assumes that the database fieldname is called "username"

$member = $my_members['username']
echo "$member<br>";

// Then terminate the loop:

};

the script will keep running through the while loop until it runs out of rows to process at which point it then moves on.

sis3970

this is the answer:

<?
// Now into the MySQL. Open the database
$db = mysql_connect("*","","");
$connection = mysql_select_db("*");

$sql = "SELECT * FROM t_users";
$db = mysql_query($sql);

while($row = mysql_fetch_array($db)) {

printf("%s <br />", $row['UserName'] ); //
}
?>

Now my next question: in this table there is also a UserLevel.
I only whant a specific level to be printed.
Is this possible?

Tnx
Sis

Imitation

Something like this? I'm only a newbie at this so.. yeh.. I think this should work. Syntax is most likely wrong though.

$sql = "SELECT * FROM t_users WHERE userlevel == 100";

http://www.mysql.com/doc/en/SELECT.html

"You can in the WHERE clause use any of the functions that MySQL support."

siwis

Unfortunately, Imitation's syntax is incorrect.

== is a php comparator NOT a MySQL comparator, so the true sql would be:

$sql = "SELECT * FROM t_users WHERE userlevel = $condition";

Substitute the correct name for the database field containing your userlevels, and feed in the value you are defining against under $condition.

You may also specify more than one value for userlevel by using something like "WHERE userlevel > 5" or for more specific (lets say you want just userlevel 5 and userlevel 10:
WHERE ((userlevel = 5) OR (userlevel = 10))

siwis

sis3970, I meant to add in my last post:

Try to get away from using the %s method of inserting an array value. Declare the array value to a unique variable name and then echo that variable (or print or printf whichever you prefer)

The %s method, whilst perfectly valid as a method can get particularly complex when it comes to print statements with multiple array values, plus if you need to repeat the value more than once, you keep having to add it to the list appended to the end of your line of code. PLUS if you screw up the order in which you list your array values, %s doesn't care, it just enters array values in the order they are listed.

sis3970

Tnx siwis,

for your time and info.
I'l do that.

Greets
Sis