display a table when a button is pressed [duplicate] - php

Question

I'm having trouble adding the value of type="button" form elements to a mySql database, and I'm wondering if I'm missing something.
Edit - It doesn't look like the information for that element is being passed from the html to the php because it's not echoing a value. My only problem is with this one element and the rest of the form is being submitted properly.
I'm using this for an online quiz which builds a user profile based upon images they've selected, and am setting the images as background images for the button elements, and I'm trying to do this in straight html (as opposed to using javascript together with radio buttons or check boxes).
<input type="button" name="quiz_start" value="jeans" style="background: url(files/start1.jpg) no-repeat; width:54px;height:140px; cursor:pointer; border:none; color: transparent; font-size : 0">
I've simplified the php code for purposes of asking the question (including specifying the user id and limiting it to only one field). I've also included the full code below.
<?php
//Start session & connect to database
$user_id = 3;
$qry = "INSERT INTO style(user_id, quiz_start) VALUES('$user_id','$_POST[quiz_start]')";
$result = #mysql_query($qry);
header("location: page2.html");
exit();
?>
The full query is:
$fieldlist=$vallist='';
foreach ($_POST as $key => $value) {
$fieldlist.=$key.',';
$vallist.='\''.($value).'\',';
}
$fieldlist=substr($fieldlist, 0, -1);
$vallist=substr($vallist, 0, -1);
$fieldlist.=', user_id';
$vallist.=','.$user_id;
$setlist='';
foreach ($_POST as $key=>$value){
$setlist.=$key .'=\''.$value.'\',';
}
$setlist=substr($setlist, 0, -1);
$result = mysql_query('UPDATE style SET '.$setlist.' WHERE user_id='.$user_id);
if (mysql_affected_rows()==0) {
$result = mysql_query('INSERT INTO style ('.$fieldlist.') VALUES ('.$vallist.')');
}
header("location: page2.html");
exit();
?>

Answer 1

Seeing that you are unable to echo $_POST['quiz_start'] that means your value is not actually set. This is because when you use a class button as in <input type='button'> your form is not actually submitted like <input type='submit'>
One solution would be to change your button to an actual submit and format that... or you need to call a javascript function with an onClick from your button as in:
<input type="button" onClick="myfunction()">
For reference to what I am talking about look at this post.
If as you say the rest of the form values are submitting fine but just the button value is not working you have a few different possible solutions depending on your preference.
Use a select field or checkbox for people to select a type in which you can pass your data.
Submit your form in javascript with <input type="button" onClick="myfunction()"> then running your update query in javascript.
Finally if you still want to run your query in PHP you can run a javascript function to make an AJAX call to return JSON information in which you can define a php variable after the page has loaded in which you can then plug into your update query.

Answer 2

Try this:
<?php
//Start session & connect to database
$user_id = 3;
$qry = "INSERT INTO style(user_id, quiz_start) VALUES('".$user_id."','".$_POST['quiz_start']."')";
$result = #mysql_query($qry);
header("location: page2.html");
exit();
?>

Answer 3

Since it seems that input type="button" can't capture the user's selection, I wanted to share a really simple way I figured out for doing this with radio buttons or check boxes using only html.
All you need to do is set the input element to style="display:none", and surround both the image and input element with a label tag so that users can click anywhere on the image to select the element :-)
<label for="quiz_start">
<img src="files/start1.jpg" />
<input style="display:none" type="radio" id="quiz_start" name="quiz_start" value="jeans">
</label>