Re: [PHP-DB] Failed to query correctly.

[Dobromir Velev]

Hi,
The code looks fine to me - it will output all records in the selected table.

I assume that you want to display the field names as a first row. If this is the case 
you could use something like this

$query="SELECT * from $tbl";
$result=mysql_db_query($db,$query) or mysql_die();
echo "<table>";

while ($line = mysql_fetch_array($result,MYSQL_ASSOC)) {
if (!$first){
echo "<tr>";
while(list($col_name, $col_value) = each($line)) {
  print "\t\t<td><p align=center>$col_name</p></td>\n";
}
echo "</tr>";
$first=1;
}
echo "<tr>";
while(list($col_name, $col_value) = each($line)) {
  print "\t\t<td><p align=center>$col_value</p></td>\n";
}
echo "</tr>";
}
echo "</table>";

Hope this helps

Dobromir Velev
Web Developer
http://www.websitepulse.com/


-----Original Message-----
From: Denny <[EMAIL PROTECTED]>
To: [EMAIL PROTECTED] <[EMAIL PROTECTED]>
Date: Tuesday, December 04, 2001 09:14
Subject: [PHP-DB] Failed to query correctly.


Hi,

I got this problem when i tried to do a sql query and print the ouput. But the script 
below always failed to return the first row of
the records. But i do a manual query, i got the results i wanted.


Thanks,
Denny


    $query="SELECT * from $tbl ";

    $result=mysql_query($query);


    while ($line = mysql_fetch_row($result)) {
       while(list($col_name, $col_value) = each($line)) {
        print "\t\t<td><p align=center>$col_value</p></td>\n";
      }

     }





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