Sending id through select menu using PHP - php

Question

here is my problem, I want to send id number through select menu using PHP.
Here is the code:
<form name="update" method="post" action="ex_update.php?id=<?php echo ((int)$_POST['get_id']); ?>">
<p><strong>Enter Name:</strong>
<input type="text" name="name">
<br />
ID:
<label for="select"></label>
<select name="get_id">
<?php
$query = "SELECT * FROM test";
$run = mysql_query($query);
while($output = mysql_fetch_array($run)){
echo "<option value=\"{$output['id']}\">{$output['id']}</option>";}
?>
</select>
</p>
<p>
<input type="submit" name="submit" value="Update!">
</p>
</form>
I have tried but when I submit the id in the URL equals to zero. how can I send id to the URL??
here is the ex_update.php >>>
<?php
$connect = mysql_connect("localhost","root","");
$sel_database = mysql_select_db("test");
$id = (int)$_GET["get_id"];
$name = mysql_real_escape_string( $_POST["name"] );
$query = "UPDATE test SET name='{$name}' WHERE id=={$id}";
if($run = mysql_query($query)){
}else{mysql_error();}
?>
Thanks in advance

Answer 1

You can use the form GET method
<form name="update" method="GET" action="ex_update.php">
You can access that select box value using $_GET['get_id'] in ex_update.php

Answer 2

Here is your First Page
Note action of form...
<form name="update" method="post" action="ex_update.php">
<p><strong>Enter Name:</strong>
<input type="text" name="name">
<br />
ID:
<label for="select"></label>
<select name="get_id">
<?php
$query = "SELECT * FROM test";
$run = mysql_query($query);
while($output = mysql_fetch_array($run)){
echo "<option value=\"{$output['id']}\">{$output['id']}</option>";}
?>
</select>
</p>
<p>
<input type="submit" name="submit" value="Update!">
</p>
</form>
And here you can find ex_update.php. Note: $id = (int)$_POST["get_id"];
<?php
$connect = mysql_connect("localhost","root","");
$sel_database = mysql_select_db("test");
$id = (int)$_POST["get_id"];
$name = mysql_real_escape_string( $_POST["name"] );
$query = "UPDATE test SET name='{$name}' WHERE id={$id}";
if($run = mysql_query($query)){
}else{mysql_error();}
?>