printing table by user

DwarV

i have this script:

if(mysql_num_rows($result)<1){
echo "
</table><CENTER>No Entry's detected</CENTER>";
include('inc/footer.inc.php');
}

else
{
while($row = mysql_fetch_array($result))
{

            ?>
            <TR bgColor=#ffdddd>
                    <TD><? print $row[date]; ?></TD>
            <TD><a href="/pu/datadump/files/<? print $row[file]."\">".$row[file]; ?></a></TD>
            <TD><? print $row[type] ?></TD>
            <TD><? print $row[size] ?></TD>
            <TD><? print $row[user] ?></TD>
            <TD><? print $row[comment]?></TD>
    </TR>
    <?

    }
}

mysql_free_result($result);

my question:
for each user in the database i want to print one table, instead of one row. how do i manage this? reprhased: i want to print:

for each value of $row["user"]

cgraz

I'm not understanding your question. Do you want to pull out only one row? Do you want a seperate table for each row? What you want to print is exactly what you have in your while loop. Could you possibly rephrase your question?

Cgraz

DwarV

what i want to do is the following..

well.. compare these two:

Datadump v1.2

and
datadump v2.1

as you can see each user has its own table in datadump v1.2.
and not in v2.1.. that's my exact problem

DwarV

or do i just have to make separate tables for each of the users?

cgraz

Yeah, I think so. How does the database look? Each user seems to have a few records. Then what does your select query look like?

The69Pimp

may wanna try this

<?
$query = ("SELECT * FROM tblname");
$result = MYSQL_QUERY($query);
if (!$result) print ("failed\n");

WHILE ($row=MYSQL_FETCH_ROW($result))

//count the rows

$number = mysql_numrows($result);
if (!$number) print("failed 2 \n");
$i = 0;
//echo rows
         echo  "There are $number entries found";
WHILE ($i < $number):
    $id = mysql_result($result, $i,id);
    $type = mysql_result($result, $i,type);
    $file = mysql_result($result, $i,file);
    $size = mysql_result($result, $i,size);
    $user = mysql_result($result, $i,user);
    $comments = mysql_result($result, $i,comments);
        if ($i%2 == 0) {
//make the table

<TR bgColor=#ffdddd> 
<TD><? print $row[date]; ?></TD> 
<TD><a href="/pu/datadump/files/<? print $row[file]."\">".$row[file]; ?></a></TD> 
<TD><? print $row[type] ?></TD> 
<TD><? print $row[size] ?></TD> 
<TD><? print $row[user] ?></TD> 
<TD><? print $row[comment]?></TD> 
</TR> 


    } else {

<TR bgColor=#ffdddd> 
<TD><? print $row[date]; ?></TD> 
<TD><a href="/pu/datadump/files/<? print $row[file]."\">".$row[file]; ?></a></TD> 
<TD><? print $row[type] ?></TD> 
<TD><? print $row[size] ?></TD> 
<TD><? print $row[user] ?></TD> 
<TD><? print $row[comment]?></TD> 
</TR> 

}
$i++;
ENDWHILE;
echo "here"; 
?>

should work, and print all the users

DwarV

got this error:

Warning: comments not found in MySQL result index 2 in /var/www/html/pu/datadump/test.php on line 22

line 22 says:

$user = mysql_result($result, $i,user);

my tabe looks like this:
id (autoincrement, primary)
user
date
file
size
type
comment

all are text type, except for id, which is int(5)

my query is very simple:
$sql = "SELECT * FROM $table ORDER BY id";
$res = mysql_query($sql);

as i am quite new to both php and mysql