Solved

php form with two buttons

Posted on 2009-03-29
6
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Last Modified: 2013-12-12
Hi,

I am trying to make a form that has two submit buttons that will send the form input one of  two different location depending on which submit button the user hits. My code (that doesn't work) is below.


<?php
 
if (isset($_POST['submit']))
{   
   $submit_value="products2.php";
}
else if (isset($_POST['creditcard']))
{
   $submit_value="products2.php";
}
?> 
        
<form  action="<?php $submit_value ?>" method="post" class="smalltxt1">
              <input type="hidden" name="cmd" value="_s-xclick" />
              <input type="hidden" name="hosted_button_id" value="2060293" />
              <table>
                <tr>
                  <td>
                    Gift Card Message:</td>
                </tr>
                <tr>
                  <td><input type="text" name="on0" maxlength="200" value="" /></td>
                </tr>
                <tr>
                  <td>Quantity:&nbsp;
                      <input type="text" name="quantity" value="" size="3" /></td>
                </tr>
              </table>
              <input type="image" src="images/button1.gif" border="0" name="submit" alt="" />
              <br />
              <input type="image" src="images/addtocart.gif" border="0" name="creditcard" alt="" />
            </form>

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Question by:jackie777
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6 Comments
 
LVL 28

Expert Comment

by:gamebits
ID: 24013217
What is not working? What I usually do is I set my form action to one page and do the split (one script or another) on that page.
0
 
LVL 28

Expert Comment

by:gamebits
ID: 24013255
I think there is a flaw in your logic (code logic)

1- the first time you open the page there is nothing in the form action, so the page will submit to herself.

2- when you click on one of the buttons to submit the form your code check which one has been clicked and set the form action appropriately but at this point it is to late the form has already been submited.
0
 

Expert Comment

by:toodfa
ID: 24013605
Aslo, there is one litle mistake in that code.
You actually dont print $submit_value variable, and because of that the action parameter is always empty.

Use something like this and try to initialize varaibles before using/printing.

<form  action="<?php echo $submit_value; ?>" method="post" class="smalltxt1">

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Author Comment

by:jackie777
ID: 24013611
Thanks gamebits.

Could you please show me an example of the code you are talking about.
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Accepted Solution

by:
gamebits earned 250 total points
ID: 24013656
Roughly it look like this

<form action="process.php" method="POST">
   Your form input fields here as usual
  <input type="image" src="images/button1.gif" border="0" name="submit" alt="" />
              <br />
              <input type="image" src="images/addtocart.gif" border="0" name="creditcard" alt="" />
            </form>

your process.php page will look like this

<?php

   if(isset($_POST['submit'])){

                           here you put your code to handle the form when the button submit has been clicked

                                            }else{

                           here you put your code to handle the form when the button creditcard has been clicked

                                                     }
?>

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Author Closing Comment

by:jackie777
ID: 31564053
thanks again gamebits
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