problem with OPTION

Discussion in 'PHP' started by abhiN, Feb 27, 2008.

0
  1. #1
    Hi

    i m trying to get the values from database and put it into the select options..
    i m using code-
    
<?php
$con = mysql_connect("localhost","cdccpl","d123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
					  mysql_select_db("cdccpl_aus", $con);
$result = mysql_query("SELECT * FROM bak_bill_cust");
 while( $row = mysql_fetch_array( $result ) ):
 $option = $row["cust_code"];

echo "<option value='$row[cust_code]'>$row[cust_code]</option>";
endwhile;
 mysql_close($con);?>
    Code (markup):
    but i m getting displayes with the values as--
    
Customer auswidestonemanpnicholsonllwmplnewuserguestanftasturnbullanglic
    Code (markup):
    means values getting displayed one after another and not in option..

    why so..
    plz help
     
    abhiN, Feb 27, 2008 IP
  2. ghannam

    ghannam Peon

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    #2
    try this
    <?php
$con = mysql_connect("localhost","cdccpl","d123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
					  mysql_select_db("cdccpl_aus", $con);
$result = mysql_query("SELECT * FROM bak_bill_cust");
echo"<select name='D1'>";
 while( $row = mysql_fetch_array( $result ) ):
 $option = $row["cust_code"];

echo "<option value='$row[cust_code]'>$row[cust_code]</option>";
endwhile;
echo "</select>";
 mysql_close($con);?>
    PHP:
     
    ghannam, Feb 27, 2008 IP