Hello! Please forgive newbie question

[jimmyoctane]

I have the following SQL statement

$result=mysql_query("SELECT long_name, count(*) 
FROM  `userlog` 
GROUP  BY long_name
ORDER  BY long_name");

I can access the browser name with no problem.

for($i=0;$i<$num;$i++)
{
$thebrowser = mysql_result($result,$i,"long_name");
echo $thebrowser;
echo "&nbsp;&nbsp;&nbsp;&nbsp;";
echo "<br>";
}

how do I define/reference the count(*) field to display????


Thanks
Jimmy

[jimmyoctane]

I ended up doing this within the loop:

$result2=mysql_query("SELECT * 
FROM  `userlog` where long_name =  '$thebrowser' ");
$num2=mysql_numrows($result2);

What should I have done?

thanks
Jimmy

[redhead]

PHP Code:

$result=mysql_query("SELECT long_name, count(*) 
                     FROM  `userlog` 
                     GROUP  BY long_name
                     ORDER  BY long_name");
$row = mysql_fetch_array($result);
echo "Long_name: " . $row[0] ."<br>";
echo "Count: " . $row[1] . "<br>";
mysql_free_result($result); 


But without any error checking.

[sde]

alternatively you can rename count(*) in your query:

PHP Code:

$result = mysql_query("select long_name,count(*) as count 
      from userlog
      group by long_name
      order by long_name"); 


now you can use mysql_result($result,$i,"count");

however, as redhead included in his example above, i prefer to use mysql_fetch_array() like so:

PHP Code:

while($row = mysql_fetch_array($result)){
  echo "browser: ".$row['long_name']." /  count: ".$row['count']."<br>\n";
} 


welcome back mr. octane, .. haven't seen ya in a while.

[jimmyoctane]

See what happend when you get an old green screen programmer hacking with a PC language.

Like a one legged man in a A$$ kicking contest...


SDE - I will be calling you soon bout that other thing



Great to be back with all my stuff working again



Take care
Jimmy