retrieve data from database and display in form -php

rosy

Hi,
I am having 3 forms, one is client list form(Form 1), another one is adding records to each client selected from client list form"Add record"(Form 2), and another one is after entering the required fields in form 2, it should show some data and have another "Add Record" button in (3rd form).If i click add record button in 3rd form it redirects me to first form called client list form.
If i choose a client and add a record i add only some values based on item selected in combo box in form 2.
If i choose the same client , to add record , i should have the previously added values displayed in the form so that the user know what are the other values to be entered, so that i can avoid inserting values again and again in the table.

how can i do it,
please help me the code.
Here are my sample code.., in (form 2)
<form action="inc/form3.php" method="post" onsubmit="return check(this);">
<td align="right"><B>Client Name:</B></td>
<td><?php print $contact_company; ?><INPUT TYPE="hidden" NAME="clientname" VALUE="<?php print $contact_company; ?>" SIZE="25" CLASS="text"></td>
</TR>
<td ><b>Contact:</b></td>
<td><INPUT TYPE="text" NAME="contactname" VALUE="<?php print $contactname; ?>" SIZE="25" id="contactname" CLASS="text" tabindex="19"></td>
<td>Title:</td>
<td><INPUT TYPE="text" NAME="contacttitle" VALUE="<?php print $contacttitle; ?>" SIZE="25" CLASS="text" tabindex="20"></td>
</TR>
........
....
<INPUT TYPE="submit" NAME="submit" VALUE="Add Sales Record" >

How i can make this form both used as display and input form, based on the values already entered for the same client.
Thanks

NogDog

<?php // example using $_POST array and the ternary operator ?>
<INPUT TYPE="text" NAME="contactname" VALUE="<?php 
echo isset($_POST['contactname']) ? $_POST['contactname'] : '';
?>" SIZE="25" id="contactname" CLASS="text" tabindex="19">

rosy

I have to check when the form loads, in database if it already has data for the client.If so print the values in the form and the remaining values should be entered in the form.
So, i checked,

[code=php]
$salesrecordsql = mysql_query("SELECT * FROM sales_records WHERE clientname='$contact_company' AND rep_sid = '$rep_id'");
$salesrecordrow = mysql_fetch_array($salesrecordsql);
extract($salesrecordrow);

[/CODE]
how i can do it on the same page.

[code=html]
<td><INPUT TYPE="text" NAME="contactname" VALUE="<?php if($salesrecordsql) ?  print $contactname : '' ; ?>" SIZE="25" id="contactname" CLASS="text" tabindex="19"> </td>

[/CODE]

or is there any other method to do this? and also i want to insert once not duplicate items in table.
please help me out of that .
Thanks.