displaying info

nielsho

Hello people. I got a problem but i think it will be solved very fast. But ican't find the answer. PLZZZZZZZZZZ help me. Tnx

I got this on first page :

<html><head><script LANGUAGE="JavaScript">
document.write('<link rel="stylesheet" href="../dutchstyle.css" type="text/css">');
</script>head>
<BODY topmargin="0" bgproperties="fixed" background="../images/achtergrond4.jpg">
<?php
include("connect.php");

if ($do_this=="list_all"):

            $result = mysql_query("SELECT *  FROM $glossary_table   ORDER BY username");
            echo "<table style='$glossary_table_style'><tr style='$glossary_header_style'><td>$feild1name (",mysql_num_rows($result),")</td></tr>";
            while ($row = mysql_fetch_array($result,MYSQL_NUM)) {
            echo "<tr><td style='$glossary_word_style'><a href=\"check_players.php?id_$row[0]\">$row[1]</a></td></tr>";
            }
            echo "</table>";
            mysql_close($link);
            exit;

else :
echo "";
echo " ";

endif;
?>

</body></html>

$row[0] are id's $row[1] are all the usernames

The meaning is when i press a certain username it will display the info according to the id

2nd page:

<html>
<head><script LANGUAGE="JavaScript">
document.write('<link rel="stylesheet" href="../dutchstyle.css" type="text/css">');
</script><title>Welcome to UT2K3League.net</title></head>
<BODY topmargin="0" bgproperties="fixed" background="../images/achtergrond4.jpg">
<?
/ declare some relevant variables /
$DBhost = "localhost";
$DBuser = "user";
$DBpass = "pass";
$DBName = "db";
$table = "table";

mysql_connect($DBhost,$DBuser,$DBpass) or die("Unable to connect to database");
@mysql_select_db("$DBName") or die("Unable to select database $DBName"); $sqlquery = "SELECT * FROM $table limit 1";
$result = mysql_query($sqlquery);
$numrows = mysql_num_rows($result);

//set counter variable
$i=1;

while($myrow = mysql_fetch_array($result))
{
echo"<TABLE BORDER='0' CELLPADDING='2' CELLSPACING='2' class=\"formtext\">";
echo"<TR>";
echo"<tr>";
echo"<td class=\"formtext\">Player Name:$myrow[username] </td>";
echo"</tr>";
echo"<tr>";
echo"<td class=\"formtext\">Email: <a href=mailto:$myrow[email]>$myrow[email]</a></td>";
echo"</tr>";
echo"<tr>";
echo"<td class=\"formtext\"> </td>";
echo"</tr>";
echo"<tr>";
echo"<td class=\"formtext\">Clan(s):</td>";
echo"</tr>";

echo"</TABLE>";

//increment counter variable
$i++;
}
?>
</BODY></HTML>

When i use this it will only display the first record in the database when i press a link.

Thanks allready

hand

What's your problem?

leatherback

Hi,

Could you please tell us WHAT the problem is?
We know what you want to do, but not what the problem is: Where does it stop? (I do not feel like looking at all the code without knowing wether you don't get the link don't get the output ...

I have been looking at the code, and I see many things which coulf give you some errors...
J.

nielsho

the problem is that when i click on the records on the first page it will display all good id in statsubar. but when i press for example on lets say php noob it will display id 2 in status bar but on the new page( that will load) it will display id1 with all the toher info along with it. And that is for all the other records the same prob. It willdisplay each time the same record.

Hope the problem is more detailed now. If not plz tell me. Sorry for some errors in grammatic but english isn't my native language

example : here

leatherback

Aha,

Well, one of the things: Your code for querying the database:

$sqlquery = "SELECT * FROM $table limit 1";

Change this to something like:

$sqlquery = "SELECT * FROM $table where id=$id";

In the link it is probably wise to change the link to something to:
.php?id=row[0]

and I think you might have solved the problem.

Also: You do not need javascript to include a link to your stylesheet. Just <link rel="stylesheet" href="../dutchstyle.css" type="text/css"> should do.

nielsho

No i changed it like you said but this time it won't display any record............
It links to the right id but there isnt any record shown

PLz can you suggest something else?

leatherback

Hi,

Let's have a look at th ecode, shall we?

I'll assume you're connection is correct (Perhaps you should use an include: You set up the connection properly once, and it works..)

mysql_connect($DBhost,$DBuser,$DBpass) or die("Unable to connect to database");
@mysql_select_db("$DBName") or die("Unable to select database $DBName");

// Connected, now get data

// Did you call you table table? Then I will use the name instead of a variable

$sqlquery = "SELECT * FROM table where id = $id limit 1";
$result = mysql_query($sqlquery);
$numrows = mysql_num_rows($result);

// Unless you are interested you don't need a loop variable

while($myrow = mysql_fetch_array($result))
{
echo"<TABLE>";
echo"<TR>";
echo"<tr>";
echo"<td class=\"formtext\">Player Name: $myrow[username]</td>";
echo"</tr>";
echo"<tr>";
echo"<td class=\"formtext\">Email: <a href=mailto:$myrow[email]>$myrow[email]</a></td>";
echo"</tr>";
echo"<tr>";
echo"<td class=\"formtext\"> </td>";
echo"</tr>";
echo"<tr>";
echo"<td class=\"formtext\">Clan(s):</td>";
echo"</tr>";
echo"</TABLE>";

}
?>

</BODY></HTML>

// I personally prefer to place all the output in one variable, and print it later, but this should work. Does it? If not: make sure caps and names are consistent between script and database! Also check the msql_query in your MySQL front, to see wether the script is ok (Include in your script: echo $sqlquery and copy - paste into mysql querytester

leatherback

Hey,

while writing the last article, I gave myselve an idea. Went to your website. Checked the correct address:
http://www.ut2k3league.net/members/check_players.php?id=1
And behold... It seems to work. Change 1 for 12.. and the user changes.

??
'
'J.

nielsho

THnx very much i changed <a href=\"check_players.php?id_$row[0]\"> into <a href=\"check_players.php?id=$row[0]\"> and it worked tnx you very much. or to leatherback heel erg bedankt man nu kan ik tenminste weer verder en het blijkt 1 enkel klote dingetje te zijn. 🙁

leatherback

no problem man. That's what this is for'; A fresh set of eyes!
"Geen enkel probleem man. Daar is dit toch voor? Een stel 'verse'ogen!?"