Submitting a PHP form with multiple submit buttons - php

Question

EDIT AGAIN: NEW REVISED VERSION
So I have corrected the lack of tag in my table, I have added id's onto my forms, I have added a hidden field to each form with the same value as the form id also (as suggested)
I was trying to modify my original if isset to include form id's but thats where I have came a little unstuck.
As mentioned, here is the HTML that the page creates:
<br><table border=1><tr><th>customer</th><th>product</th><th>employee</th><th>return_date</th><th>complete</th></tr><br><tr><td>marne1</td><td>basketball</td><td>user</td><td>01-10-12</td><td> <FORM id="1" method ="post" action ="g.php">
<input type='hidden' name='x' value='1'>
<Input type = 'Radio' Name ='ir' value= '47'> 47 <input type="submit" name="return"/>
</FORM>
</td></tr><br><tr><td>marne1</td><td>basketball</td><td>user</td><td>01-10-12</td><td> <FORM id="2" method ="post" action ="g.php">
<input type='hidden' name='x' value='2'>
<Input type = 'Radio' Name ='ir' value= '48'> 48 <input type="submit" name="return"/>
</FORM>
</td></tr><br><tr><td>marne1</td><td>basketball</td><td>user</td><td>02-10-12</td><td> <FORM id="3" method ="post" action ="g.php">
<input type='hidden' name='x' value='3'>
<Input type = 'Radio' Name ='ir' value= '49'> 49 <input type="submit" name="return"/>
</FORM>
</td></tr><br><tr><td>marne1</td><td>basketball</td><td>user</td><td>29-09-12</td><td> <FORM id="4" method ="post" action ="g.php">
<input type='hidden' name='x' value='4'>
<Input type = 'Radio' Name ='ir' value= '50'> 50 <input type="submit" name="return"/>
</FORM>
</td></tr><br><tr><td>marne1</td><td>basketball</td><td>user</td><td>01-10-12</td><td> <FORM id="5" method ="post" action ="g.php">
<input type='hidden' name='x' value='5'>
<Input type = 'Radio' Name ='ir' value= '51'> 51 <input type="submit" name="return"/>
</FORM>
</td></tr><br><tr><td>marne1</td><td>basketball</td><td>user</td><td>01-10-12</td><td> <FORM id="6" method ="post" action ="g.php">
<input type='hidden' name='x' value='6'>
<Input type = 'Radio' Name ='ir' value= '52'> 52 <input type="submit" name="return"/>
</FORM>
</td></tr><br><tr><td>marne1</td><td>basketball</td><td>user</td><td>02-10-12</td><td> <FORM id="7" method ="post" action ="g.php">
<input type='hidden' name='x' value='7'>
<Input type = 'Radio' Name ='ir' value= '60'> 60 <input type="submit" name="return"/>
</FORM>
</td></tr><br><tr><td>marne1</td><td>basketball</td><td>user</td><td>02-10-12</td><td> <FORM id="8" method ="post" action ="g.php">
<input type='hidden' name='x' value='8'>
<Input type = 'Radio' Name ='ir' value= '61'> 61 <input type="submit" name="return"/>
</FORM>
</td></tr><br></table>
<fieldset>
<legend>???</legend>
<form action="z.php" method="post">
Customer Name: <input type="text" name="ir" /> <br><br>
<input type="submit" name="return"/>
</form>
This is the PHP page that is creating this HTML (with my porly attempted form id test):
<?php
$db_host = "localhost";
$db_username = "root";
$db_pass = "usbw";
$db_name = "test";
mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql");
mysql_select_db("$db_name") or die ("no database");
$result = mysql_query("SELECT * FROM rental WHERE customer = '$_POST[customer]' AND complete = '0' ") or die (mysql_error());
?>
<?php
echo '<br>';
echo '<table border=1>';
echo '<tr><th>customer</th><th>product</th><th>employee</th><th>return_date</th><th>complete</th></tr>';
echo "<br>";
$x=1;
while($row=mysql_fetch_array($result))
{
echo '<tr>';
echo '<td>';
echo $row ['customer'];
echo '</td><td>' ;
echo $row ['product'];
echo '</td><td>';
echo $row ['employee'];
echo '</td><td>';
echo $row ['return_date'];
echo '</td><td>';
$cur_return = $row ['rental_id'];
?>
<FORM id="<?php echo $x; ?>" method ="post" action ="g.php">
<input type='hidden' name='x' value='<?php echo $x; ?>'>
<Input type = 'Radio' Name ='ir' value= '<?php echo $cur_return ?>'> <?php echo $cur_return ?>
<input type="submit" name="return"/>
</FORM>
<?php
echo '</td></tr>';
echo '<br>';
$x++;
}
echo '</table>';
?>
<fieldset>
<legend>???</legend>
<form action="z.php" method="post">
Customer Name: <input type="text" name="ir" /> <br><br>
<input type="submit" name="return"/>
</form>
<?php
if (isset($_POST['return']))
{
if ($_POST['x']==1)
{
mysql_query("UPDATE rental SET complete = '1' WHERE rental_id = '$_POST[ir]'") or die (mysql_error());
}
}
?>
I really do appreciate all the help you guys are suggesting, and apologies for the time it's taken me to get this sorted :)
Rich

Answer 1

Wait wait. You have multiple rows of data here and you want to submit one specific row of data, but have it all in the same form? I'm a little confused. It looks like what you're really looking for here is to generate a different form element for each row. But you cannot have multiple submit buttons submit different parts of the same form, that's not how forms work. A submit button will submit all data belonging to the form.
In that case, just move the opening form tag to the very beginning of the inside of your while loop, and move your submit button and closing form tag into the bottom of your while loop. It will generate each form and still maintain the structure of your table. But you cant have one form that submits only certain parts of itself, that goes against basic HTML principles.
I'd strongly suggest slapping an ID on each form, for future reference and update purposes.
Good luck.

Answer 2

You can do this with javascript, but not with php. But it's overcomplicating for no reason. Make different forms and assign them to different IDs. You can use the same method="post" and the same action="g.php". To differ on g.php, just put this in each form:
<input type='hidden' name='x' value='1'>
and then
<input type='hidden' name='x' value='2'>
Then it's a simple if $_POST['x']==1
Also, make sure you protect g.php, that's a pretty insecure name for a file that's handling forms.