Posted by admin in This allow you to take form data and post it to a mysql database, poll it by the form id, and return the database identifier.
<form id=”form1″ name=”form1″ method=”post” action=”form.php?pid=1″>
<table width=”574″ height=”191″ border=”1″>
<tr>
<td width=”189″> </td>
<td width=”369″><span id=”sprytextfield1″>
<label>
<input type=”text” name=”name” id=”name” />
</label>
<span class=”textfieldRequiredMsg”><span class=”error_style1″>Minimum number of characters not met.</span></span></span></td>
</tr>
<tr>
<td> </td>
<td>
<span id=”sprytextfield2″>
<label>
<input type=”text” name=”phone” id=”phone” />
</label>
<span class=”textfieldRequiredMsg”><span class=”error_style1″>Minimum number of characters not met.</span></span>
<span class=”textfieldInvalidFormatMsg”><span class=”error_style1″>Must be numerical in value.</span></span>
</span></td>
</tr>
<tr>
<td> </td>
<td><label>
<input type=”submit” name=”submit” id=”submit” value=”Submit” />
</label></td>
</tr>
</table>
</form>
<?php
$pid = $_GET['pid'];
$name = $_POST['name'];
$phone = $_POST['phone'];
if ($pid == 1)
{
include ‘dbloc.php’;
srand(time());
$form_id = (rand()%4000);
$query = “INSERT INTO address (name, phone, form_id) VALUES (’$name’, ‘$phone’, $form_id)”;
mysql_query($query) or die(’FAILED: Insert into database failed’);
$query = “SELECT * FROM address WHERE form_id = $form_id” or die(mysql_error());
$result = mysql_query($query);
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
$db_id = $row['db_id'];
}
$query = “UPDATE address SET form_id = 0 WHERE db_id = $db_id”;
mysql_query($query) or die(’FAILED: Updating form identifier’);
echo “Datbase Identifier: $db_id <br>”;
$query = “SELECT * FROM address”;
$result = mysql_query($query);
echo “Form id: $form_id <br><br>”;
echo “<table>”;
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo ”<tr>”.
“<td>{$row['name']}</td>” .
“<td>{$row['phone']} </td>” .
“<td>{$row['form_id']} </td>” .
“</tr>”;
}
echo “</table>”;
}
?>