Why My Form Is Not Working?

[Alicia]

Hi,

 

I am trying to create a simple form with image buttons but why my php page can't capture the value of the image button clicked?

 

can somebody give me a hint about this.. thankyou

 

<form id="form1" name="form1" method="post" action="process.php">
  <tr>
  <td colspan="2"><input type="image" name="Submit" id="Submit" src="images/01.jpg" value="Mobile" /></td>
  <td>
    <input type="image" name="Submit" src="images/03.png" id="Submit" value="Hardware" /></td>
  <td> </td>
  <td> </td>
   </tr>
</form>

 

in my php file i put use this code.. it is very simple and straight forward but wonderng why nothing is printed.. Other elements in the form can be printed except the value of the submit image. =(

 

echo $_POST['Submit'];

[floridaflatlander]

Try

 

<form id="form1" name="form1" method="post" action="process.php" enctype="multipart/form-data" >

 

added

 

sorry I think I got what you want wrong. Are you trying to upload an image?

Edited September 28, 2012 by floridaflatlander

[Barand]

Clicking on images sends x,y coordinates.

 

Try this code so you can see what is in the POST array

echo '<pre>'.print_r($_POST, 1).'</pre>';

Edited September 28, 2012 by Barand

[jazzman1]

<input type="image" name="Submit" src="images/03.png" id="Submit" value="Hardware" />

Not Equal To

<input type="submit" name="Submit" id="Submit" value="Hardware" />

 

If you want to check whether the form has been submitted, you could use a hidden input field.

[Alicia]

i am not uploading any image so the form type is not what i need.. actually i have 2 image buttons for them to click..in my php file, i want to do conditional statement $_POST['Submit'] == a do this or b do that so hidden value wont work in this case since we need to perform an action based on the image button clicked\

 

any guru can advise

[ManiacDan]

Someone asked you to print_r post.

[PFMaBiSmAd]

The value attribute of an image being used as a submit button won't be submitted (well it will by some browsers, but those are not following the w3.org specification.)

 

To distinguish which image was clicked, that will work in all browsers, you must use different names for each one.

[Alicia]

print_r($_POST, 1) shows nothing...

 

no.. different name doesn't work either.. it still print nothing.. any guru can assist.. this looks simple but why not able to solve the issue =(

[ManiacDan]

print_r($_POST, 1) would show nothing...unless you put it inside the rest of the statement like you were asked to do.

 

Why not able to solve the issue is because you're not giving us any information. Copy and paste the command you were given. Show the output. It absolutely 100% will always print something. Show us that something.

[Alicia]

I managed to solve this one by using css button..

 

another new question if any guru can assist.

 

now I put this

<input type="hidden" name="checkboxes[]" value="<? echo $row['memo'] ?>" /><input type="hidden" name="checkboxes[]" value="<? echo $row['memo'] ?>" />

inisde a for loop

 

when I posted to the php file, it has error : Warning: Invalid argument supplied for foreach() in

when I use this :

foreach($_POST['checkbox'] as $copid){

 

The code works when I use it as checkbox but why it doesn't work anymore after I changed it to hidden value ? any idea??

[darkfreaks]

This will not work as stated sending images outputs X and Y cordinates

 

if(isset($_POST['submit'])) { //** Do Stuff **//
}

 

however this will because it includes the X & Y cordinates in the submit.

 

if(isset($_POST['submit_x') && isset($_POST['submit_y'])) { //**DO Stuff**//
}

[ManiacDan]

Checkboxes need to be checked in order to pass information.

 

You shouldn't use checkboxes for this anyway though, the session is more correct for passing data between page loads.

[Christian F.]

Also, "checkboxes" != "checkbox".

Edited September 28, 2012 by Christian F.