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Dynamically populated form for Insert?? It can't be this hard.

Thread began 10/24/2009 1:50 am by troyd | Last modified 10/27/2009 3:43 pm by troyd | 1138 views | 4 replies


I might be going in the wrong direction here, but I have tried this so far with minor results.

<form id="form1" name="form1" method="post" action="">
<select name="select_id" id="select_id">
<option value="">Select ID</option>
do {
<option value="<?php echo $row_rs['id']?>"><?php echo $row_rs['name']?></option>
} while ($row_rs = mysql_fetch_assoc($rs));
$rows = mysql_num_rows($rs);
if($rows > 0) {
mysql_data_seek($rs, 0);
$row_rs = mysql_fetch_assoc($rs);
<input name="name" type="text" value="<? echo $_POST['select_id']?>" />
<input type="submit" name="submit" id="submit" value="Submit" />


I do have the insert behavior and it inserts a record but does not insert anything into the field I have the hidden field assigned to.

What I ultimately want is for there to be 2 hidden fields. So the Select will provide the ID dynamically from table 1, and the two hidden fields will provide the Name and Image_Name.

When I tested it by just submitting to itself, the value of the hidden field was updated correctly. However not with the name associated with the id as I would like it to.

SO, how can I update the values in the hidden fields and at the same time post them to the new table? AND exchange the ID value from table 1 with the associated Name value from table 1 as well?

Any thoughts?


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