You are right about my query. I am only pulling down one column.
As you guys can probably tell I am pretty inexperienced with this. I tried
what you said and still got the same results.
Advertising
Here are snippets of my code and maybe this will depict what I am doing
wrong:
After creating the connection:
$query = "select server_name from servers where midtier = '$server'";
$dbResult = mysql_query($query,$dblink);
$c = 1;
print("<tr>");
while($row = mysql_fetch_array($dbResult))
{
print("<td>{$row['server_name']}</td>");
if(($c % 4) == 0)
{
print("</tr><tr>");
}
}
-----Original Message-----
From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]]
Sent: Thursday, December 05, 2002 3:26 PM
To: Ryan Jameson (USA); Art Chevalier; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] MySQL Array
I think he meant that he is only pulling one column, i.e. "field1" with each
mysql_fetch_array and he wants that in four columns.
$c = 1;
echo "<tr>";
while($ar = mysql_fetch_array($result))
{
echo "<td>{$ar['field1']}</td>";
if(($c % 4) == 0) { echo "</tr><tr>"; }
}
You'll have to account for incomplete rows and clean up the output, but
hopefully that gives you an idea.
---John Holmes...
----- Original Message -----
From: "Ryan Jameson (USA)" <[EMAIL PROTECTED]>
To: "Art Chevalier" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Thursday, December 05, 2002 3:05 PM
Subject: RE: [PHP-DB] MySQL Array
while ($ar = mysql_fetch_array($rs))
echo "<tr><td>". $ar['field1']."</td><td>".$ar['field2']."</td></tr>";
Make sense?
-----Original Message-----
From: Art Chevalier [mailto:[EMAIL PROTECTED]]
Sent: Thursday, December 05, 2002 12:44 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] MySQL Array
Hello,
I am pulling one column out of a MySQL table in an array. I want to place
each element into a HTML table 4 rows across. I am currently doing this
with the mysql_fetch_array() function. How can I pull out 4 array elements
in one pass through a while loop?
Thanks
Art Chevalier
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