Updating a table with PHP and MYSQL - php

Question

<?php require("inc_connect.php"); ?>
<h1 align="center">Farris Website</h1>
<hr width="1000">
<p align="center">
<table align="center" width="1000" border="3" bordercolor="#0066FF" >
<tr>
<td align="left" valign="top">
<form name="update" method="post" action="ex_update.php?id=<?php echo urlencode($_POST['id']); ?>">
<p><strong>Enter Name:</strong>
<input type="text" name="name">
<br />
ID:
<label for="select"></label>
<select name="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></td>
<td width="300" align="left" valign="top"><?php include("inc_output.php"); ?></td>
</tr>
</table>
</p>
The above is the index page ...
<?php
$connect = mysql_connect("localhost","root","");
$sel_database = mysql_select_db("test");
$name = mysql_real_escape_string( $_POST["name"] );
$id = (int) $_GET['id'];
$query = "UPDATE test SET name='{$name}'";
if($run = mysql_query($query)){
header("location: index.php");
exit;
}else{mysql_error();}
?>
And this is the page that processes the form.
The problem is that the record won't update if i set the id={$_GET['id']}
and if I remove that part it updates all the rows.
So updating according to id ...
Thanks in Advance
FarrisFahad

Answer 1

Try changing your form action to
<form name="update" method="post" action="ex_update.php?id=<?php echo urlencode($_GET['id']); ?>">
Also, doing an echo of $query might help debug your problem.

Answer 2

First, just be aware of SQL Injection - your code is wide open to it. See http://bobby-tables.com/
PHP Code
<?php
$connect = mysql_connect("localhost","root","");
$sel_database = mysql_select_db("test");
$name = mysql_real_escape_string( $_POST["name"] );
$id = (int) $_GET['id'];
$query = "UPDATE test SET name='{$name}' WHERE id = {$id}";
if($run = mysql_query($query)){
header("location: index.php");
exit;
}else{
# In production, don't show raw errors to users - log them to a file and
# present the user with a generic "There was a problem with the database"
# error. Or people can start sniffing for vulnerabilities in your site.
echo mysql_error();
}
?>
Page
<?php
require("inc_connect.php");
?>
<h1 align="center">Farris Website</h1>
<hr width="1000">
<p align="center">
<table align="center" width="1000" border="3" bordercolor="#0066FF" >
<tr>
<td><form name="update" method="post" action="ex_update.php?id=<?php echo urlencode($_GET['id']); ?>">
<p><strong>Enter Name:</strong>
<input type="text" name="name"><br />
<label for="select">ID:</label>
<select name="id" id="select">
<?php
$query = "SELECT * FROM test";
$run = mysql_query($query);
while( $r = mysql_fetch_array($run) ){
# I always use short, single character, variables when in loops.
# Saves alot of characters and potential confusion.
echo " <option value='{$r['id']}'>{$r['id']}</option>\n";
}
?>
</select>
</p>
<p>
<input type="submit" name="submit" value="Update!">
</p>
</form></td>
<td><?php include("inc_output.php"); ?></td>
</tr>
</table>
</p>

Answer 3

As you want to update that record which is selected from your dropdown. Moreover u have set your form method to POST. So you should try following:
<form name="update" method="post" action="ex_update.php?id=<?php echo urlencode($_POST['id']); ?>">