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i have been having the following problem .I am able to retrieve the data from database and display it in a drop down menu.But i dont know how to use the same drop down list as the input and embed it in a form.My code is as follows.Here project names are retrieved from the database and are displayed in the form of drop down list.Now i need to use the same drop down list in the form(HTML form) which will be used to input data into another table.

error_reporting(E_ALL ^ E_DEPRECATED);
include 'connect.php' ;

$sql="SELECT project_name FROM $tbl_name ";


if($result === FALSE) {

<form name="resources" action="hourssubmit.php"  method="post" >

echo "<select name='project_name'>";
while ($row = mysql_fetch_array($result)) {

echo "<option value='" . $row['project_name'] ."'>" . $row['project_name']."</option>";
echo "</select>";

Updated 6-Apr-16 0:36am
Mohibur Rashid 18-Sep-14 23:02pm    
Your question does not make any sense. I would ask you to explain what 'use same drop down list in html form'. You already have used it in a form. What else are you trying to do?

1 solution

mysql_connect("localhost","root") or die("not conected");
mysql_select_db("db_resolve") or die("not find the db");
$output = '';
//Logic Part
$searchq = $_POST['search'];
$searchq = preg_replace("#[^0-9a-z]#i","",$searchq);
$query = mysql_query("SELECT * FROM scrb WHERE district LIKE '%$searchq%' OR PS_SCRB LIKE '%$searchq%' OR Supplier LIKE '%$searchq%'") or die("could not search!");
$count = mysql_num_rows($query);
if($count == 0)
$output = 'there was no search results!';
while($row = mysql_fetch_array($query))
$dist = $row['District'];
$ps = $row['PS_SCRB'];
$supl = $row['Supplier'];
$cntc = $row['Contact'];
$mob = $row['Mobile'];

$output .= '
'.$dist.' - '.$ps.' - '.$supl.' - '.$cntc.' - '.$mob.'
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