Button Element with Type Submit & PHP Challenges

Question

Recently I change from <input type="button"> to <button> in my forms however the form being processed by PHP wouldn't then submit to the database. Am I missing something in my code?

Basically all I have done is changed this:

<input type="submit" name="submitAdd" value="Ask Question! " />

To this:

<button type="submit" class="btn" name="submitAdd"><span><span>Ask Question!</span></span></button>

Here is the basic PHP processing Code:

//Extract question from submission
$question = (isset($_POST["question"]))?$_POST["question"]:"";
$question_date = (isset($_POST["question_date"]))?$_POST["question_date"]:"";
$submitAdd = (isset($_POST["submitAdd"]))?$_POST["submitAdd"]:"";

//Open connect to database 
include("include/session.php");

//Prepare data for submission
$db_question = addslashes($question);
$db_question_date = addslashes($question_date);

//If form has been submitted, insert question into database
if ($submitAdd) {
    $sql ="INSERT INTO questions
    (question,question_date)
    VALUES ('$db_question', '$db_question_date')";
    $result = mysql_query($sql);
    if (!$result) {
        $message = "Failed to add question. MySQL said " . mysql_error();
    } else {
        header("Location:http://localhost/grill/register.php"); 
    }
}

You want mysql_real_escape_string. addslashes isn't secure in all circumstances. — bobince
– bobince, Commented Nov 26, 2009 at 0:45

cletus · Accepted Answer · 2009-11-25 23:44:41Z

3

It doesn't work because the button version has no value. Your code says:

$submitAdd = (isset($_POST["submitAdd"]))?$_POST["submitAdd"]:"";

but you have:

<button type="submit" class="btn" name="submitAdd"><span><span>Ask Question!</span></span></button>

Compare this to:

<input type="submit" name="submitAdd" value="Ask Question! " />

which has a value attribute. This value is passed to the PHP script and is what you're testing. Your <button> doesn't have one.

With no value $submitAdd, even when clicked, will have a value of ''. An empty string evaluates to false when you do this:

if ($submitAdd) {

So, a couple of changes I would recommend. Firstly, change this:

$submitAdd = (isset($_POST["submitAdd"]))?$_POST["submitAdd"]:"";

to

$submitAdd = isset($_POST['submitAdd']);

since you don't really care about the value.

Secondly, unrelated to this but still worth mentioning, I would do this:

$db_question = mysql_real_escape_string($question);
$db_question_date = mysql_real_escape_string($question_date);
$sql = <<<END
INSERT INTO QUESTIONS (question, question_date)
VALUES ('$db_question', '$db_question_date')
END;

answered Nov 25, 2009 at 23:44

cletus

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3 Comments

Katundu Over a year ago

Thanks, I understand, this worked. So I understand and learn, why change the other section of code, what is the advantage?

cletus Over a year ago

The advantage is twofold: 1) your code exactly matches your intent (in that you only care if the button was pressed, not its value) and 2) brevity (the less code you write to achieve the same thing, the better it is generally speaking).

bobince Over a year ago

Even better, include a hidden field to signal that the form is being submitted. You don't generally want to rely on the submit-control giving you a value, because of quirks about when that control is submitted, and what the value is when it's a button.

Malitta N · Accepted Answer · 2016-06-13 17:32:41Z

1

Make the submitAdd input field hidden
Call a Javascript function on click of the button

Markup:

<input type="hidden" name="submitAdd" value="Ask Question! " />

<button type="submit" onclick="submitForm();" class="btn" name="submitAdd">
<span><span>Ask Question!</span></span></button>

Javascript:

function submitForm(){
    document.forms["form_name"].submit();
}

I wouldn't recommend it though as you would depend on javascript for form submission. But even gmail does it that way :P

edited Jun 13, 2016 at 17:32

Malitta N

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answered Nov 26, 2009 at 1:35

vsr

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Comments

Langdon · Accepted Answer · 2009-11-25 23:43:00Z

0

I'm guessing <button> tags don't get submitted, and thus don't end up in $_POST. Just do this at the top of your script to verify:

print $_POST["submitAdd"];

Why did you change from input to button anyway? I would recommend adding a hidden field called form_action or something similar and base your if logic on that... if ($_POST['form_action']) instead of if ($submitAdd).

answered Nov 25, 2009 at 23:43

Langdon

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1 Comment

Katundu Over a year ago

Yeah nothing gets printed. What should I be doing to remedy this? (PHP newbie) To answer your question button provides a better way of consistent styling (apart from IE6). Check out the buttons on Google or twitter. I'm doing a similar thing. stopdesign.com/eg/buttons/3.0/code.html

Rick Mogstad · Accepted Answer · 2009-11-25 23:43:34Z

0

Per w3schools.com

Important: If you use the button element in an HTML form, different browsers will submit different values. Internet Explorer will submit the text between the and tags, while other browsers will submit the content of the value attribute. Use the input element to create buttons in an HTML form.

http://www.w3schools.com/tags/tag_button.asp

answered Nov 25, 2009 at 23:43

Rick Mogstad

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Comments

Sergiu · Accepted Answer · 2014-01-07 13:12:02Z

0

Had the same problem, and my solution uses no javascript. It's mainly the same as user vsr previously answered.

Add a hidden input with the data you want to sumbit:

<input name="submitAdd" type="hidden" value=" Ask Question! ">

Name the button somehow else:

<button type="submit" class="btn" name="submitsForm"> Ask Question! </button>

The button will submit the form, and the data you want will be in the hidden input, and nicely submitted further. Hope it helps!

answered Jan 7, 2014 at 13:12

Sergiu

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