very important

[user4fun]

$Letter= c ;i need a statment that would look in table Main column 'names' and the records that have the fifith letter to match the $Letter.or if there is any other ideasAll the records in the 'names' column start with www.i want to find all the names that i have listed ( obviously websites) to match the $Letter ( hence the fifth letter )how can i do that.examplewww.charleybrown.comwould come up if $Letter = c

[vijay]

Hi... You can check 5th letter='c' in column 'name' via following query:Query1: "select if (substring(name,5,1)='c',1,0) as cThere from tblName;" this query will return 1 if c present as 5th and 0 if not. and for starting name with 'www' for that Query2: select * from tblName where name='www%' suppose this will help you..Regards,Vijay

[user4fun]

this is what i have so far.not working thoughconfimred = the table that has the infromation

$result = mysql_query("SELECT Website,Email FROM confirmed"); echo "<table border='1'><tr><th>Website</th><th>Email</th></tr>";while($row = mysql_fetch_array($result))  {  $rest = substr($row['Website'], 5, 1);		if ($rest == c)		  {		  echo "<td>"  . $row['Website'] .  "</td>";   	 echo "<td>"  . $row['Email'] . "</td>";	  		   }		else 		  {		  echo " ";		   }echo "</tr>";  }echo "</table>";

[justsomeguy]

This query will return all records where the first 5 characters are "www.<letter>":

$sql = "SELECT Website, Email FROM confirmed WHERE Website LIKE 'www.{$letter}%'";

You don't need to do any additional checking with PHP.

[user4fun]

$letter= "o"; $sql = "SELECT Website, Email FROM confirmed WHERE Website LIKE 'www.{$letter}%'";echo "<table border='1'><tr><th>Website</th><th>Email</th></tr>";while($row = mysql_fetch_array($result)) { echo "<td>" . $row['Website'] . "</td>"; echo "<td>" . $row['Email'] . "</td>"; echo "</tr>"; }echo "</table>";I am not getting any results back, the table shows up but it is empty?i know there are entries starting with letter o

[justsomeguy]

Well you never sent the query to the database, that will probably help.

[user4fun]

it is a good idea, but unfortunatly i did. That part is skipped out because it has username, password information.

[justsomeguy]

Between this:$sql = "SELECT Website, Email FROM confirmed WHERE Website LIKE 'www.{$letter}%'";and this:while($row = mysql_fetch_array($result))you don't do anything with the database. If you are actually doing that and just decided not to include it, run the query on the database server and see what it is doing.

[user4fun]

What is it that need to be done, here is the full code

$link = mysql_connect("mysql", " ", " ") or die ("No connection");   mysql_select_db("business") or die ("no database");  $letter= "o";	   $result = "SELECT Website, Email FROM confirmed WHERE Website LIKE 'www.{$letter}%'";echo "<table border='1'><tr><th>Website</th><th>Email</th></tr>";while($row = mysql_fetch_array($result))  {  		  echo "<td>"  . $row['Website'] .  "</td>";   	 echo "<td>"  . $row['Email'] . "</td>";	  		echo "</tr>";  }echo "</table>";?>more stuff</td>	</tr>  </table>  </center></div><?include ("../main_btm1.php");?>

[justsomeguy]

You need to use the mysql_query function to send the query to the database and get the result back.