[Resolved] update record problem with dyanmic checkboxes

[vinny199]

Hi,

here is the issue.

I need people to vote for a picture with two choices:

hot
not hot

and the MySql database table: picturetbl

has 1 row for hot: pic_bvotes
and a row for not hot: pic_mvotes

I'm using checkboxes for choosing between the 2 options.

so, when i updtae the table, the value of the checked box ticked should be: the existing value+1, so the vote is added.

No problem there.

I did the following and it works:

<td width="68%"> <input type="radio" name="pic_bvotes" value="<?php echo $row_rslargepic['pic_bvotes']+1; ?>"></td>
</tr>
<tr>
<td>not hot</td>
<td> <input type="radio" name="pic_mvotes" value="<?php echo $row_rslargepic['pic_mvotes']+1; ?>"></td>

the problem is the checked value is updated correctly, but the unchecked box takes a value of NULL and overwrite the existing vote amount with a 0 value... not good.

My question would be:

Either:

1/ How could I say: If checkbox value is unckecked, then echo current row value (so the record would be updated with the same value for unchecked box)

2/ (this seems more logical to me)
if unchecked, do not update this field.

I'm really stuck and would really apreciate your help.

thanks,

vinny

[devinemke]

i think you are overcomplicating this. simply do:

hot: <input type="radio" name="vote" value="hot"><br>
not hot: <input type="radio" name="vote" value="not hot">

then check to see the value of $_POST['vote'] and update the appropriate database record.

[vinny199]

well, how is that going to show how many people voted for a pic as "hot" or a pic as "not hot" when displaying the poll results?

correct me if i'm wrong but you just seem to be suggesting that I write the value "hot" or "not hot" into the database rather than adding a vote.

This is not the case at all.

i'm trying to get say if 10 people voted for hot, the hot value is 10, and the next person to vote will make it 11

which i got to do fine.

the part i'm struggling with is the UNCHECKED box taking a NULL value and overwriting the existing amount of votes with 0

for instance, if 5 votes were "not hot" and the next person votes for "hot", "not hot" being unchecked will change the 5 to 0

that's the problem

[ahundiak]

Read devinemke's suggestion more carefully. All he is saying is to use a different approach to determine which way the user voted.

Do not store the vote count in a form field and then rely on it being posted. What if five people access the site at more or less the same time, get the same vote count and then vote? Only one of their votes would be counted. Even worse, what if some jolly joker edited the form and set the count to some outrageuos value?

Your code should check the vote and then do the incrementing at the server.

[vinny199]

Thanks for the reply,

I'm sorry to be a pain, but I seem to understand the theory of what you are saying, but not the implementation of it.

how do I do that?

How do you go about:

check the vote and then do the incrementing at the server

?

Is it something to to with chanching my field definition in MySql?

I apreciate both your reply and devinemke's, but i still don't really get it.

Would you have an example I could look at for something similar please?

[devinemke]

let's back up. i'm assuming you have multiple pictures that people are voting on. i'll further assume that the info about the photos is stored in a database table. why not simply add 2 more fields to your table: votes_hot and votes_not (these could be simple INT fields). then display a form like this (i'm using pic_id #123 as an example):

$pic_id = 123; // for example

echo '
<form action="" method="POST">
please vote for picture #' . $pic_id . ':<br>
hot: <input type="radio" name="vote" value="' . $pic_id . '|vote_hot"><br>
not hot: <input type="radio" name="vote" value="' . $pic_id . '|vote_not"><br>
<input type="submit" name="submit" value="submit">
</form>
';

then after the form is submitted your query might look like this:

$explode = explode('|', $_POST['vote']);
$sql = "UPDATE picture SET $explode[1] = $explode[1] + 1 WHERE pic_id = $explode[0]";

[vinny199]

Thanks devinemke,

err..just not quite there yet.
I've tried a few things.

here is where i am:

The form:

<?php echo '
<form name="form1" method="POST" form action="">

                      <table width="100%" border="0" cellspacing="1" cellpadding="1">
                        <tr> 
                          <td colspan="2">Hot or not Hot?please vote for picture #' . $pic_id . ':<br> 

</td>
</tr>
<tr>
<td width="32%">hot</td>
<td width="68%"> <input type="radio" name="vote" value="' . $pic_id . '|pic_bvotes"></td>
</tr>
<tr>
<td>not hot</td>
<td> <input type="radio" name="vote" value="' . $pic_id . '|pic_mvotes"></td>
</tr>
<tr>
<td> </td>
<td align="right"> <input type="submit" name="Submit" value="Vote">

                          </td>
                        </tr>
                      </table>

                    </form>'; ?>

Then I have tried to place the following both before the <head> tags and just after the form:

<?php $explode = explode('|', $_GET['vote']); 

$sql = "UPDATE picturetbl SET $explode[1] = $explode[1] + 1 WHERE pic_id = $explode[0]"; ?>

picturetbl is my table, pic_bvotes is the equivalent of the field you suggest: votes_hot and pic_mvotes equivalent of vote_not

At the moment, The page reloads but does not update the database table when i vote.

Can you see what I'm still doing wrong?

(note that the pic_id displays correctly on the page for the selected picture)

You can see a basic example of what i'm trying to do at:

http://www.purple-e3.com/purple/section_01/01_whatson/03_past_gallery1.php?pageNum_rsrspictures=0&totalRows_rsrspictures=17&eve_id=119&pic_id=2234&pic_id=2234

clicking on thumbnails loads the "big" pic on the right that you can vote for.

Thanks again for your help.

[devinemke]

2 obvious problems:

1. your form is POST method and your form handling code is looking for GET variables.

2. you are creating the query but you are never running it. you need to feed your $sql variable to [man]mysql_query[/man].

i would suggest you place the form handling code at the very top of the script like this:

// connect to db first

if (isset($_POST['Submit']) && $_POST['Submit'] == 'Vote')
{
	$explode = explode('|', $_POST['vote']);
	$result = mysql_query("UPDATE picturetbl SET $explode[1] = $explode[1] + 1 WHERE pic_id = $explode[0]");
	if (!$result) {exit(mysql_error());}	
}

[vinny199]

ok, thanks a lot, this is great.

I use dreamweaver MX and i'm not good when a problem occurs that requieres handcoding.

but i'm learning...

I'm not out of the woods yet as I have now to make the poll results appear when the page reloads instead of the "voting box" (current form).

but that's another stuggle to get started.

cheers,

Vinny

[vinny199]

Hi Devinemke,

Thanks a lot for your help.
don't want to abuse your time, but since you understand what I'm trying to do, i thought i'd ask you one more thing:

I have done my poll results, and the % and decimal formating.

all good.

But wht I'd like is for that new table (the results table) to only be showing on page reload AFTER a vote has been made and INSTEAD of the poll table you help me do yesterday.

So when someone votes, I'd like to get this code below SHOWING INSTEAD of the poll table, so when a person votes, the page reloads with the current results, but not the "voting" box again, so they don't keep voting.

I have no idea how to go about that and would really apreciate your help again,

                    <table width="100%" border="0" cellspacing="1" cellpadding="1">
                      <tr> 
                        <td colspan="2">current vote result for this pic</td>
                      </tr>
                      <tr> 
                        <td>hot</td>
                        <td><?php echo DoFormatNumber($number2, 2, '.', ''); ?> 
                          %</td>
                      </tr>
                      <tr> 
                        <td>not hot</td>
                        <td><?php echo DoFormatNumber($numbera2, 2, '.', ''); ?> 
                          &nbsp;%</td>
                      </tr>
                    </table>

[devinemke]

if (isset($_POST['Submit']) && $_POST['Submit'] == 'Vote')
{
	// echo the results of the poll
}
else
{
	// echo the form that lets people vote
}

[vinny199]

OK, great,

It's almost working,

well it is if Im not showing the function DoFormatNumber in my poll results.

I think it may just be that i write it the wrong way now it is included in a echo'

If i just replace it with a simple value it works.

is there a way around that?

{
// echo the results of the poll
echo ' <table width="100%" border="0" cellspacing="1" cellpadding="1">
<tr>
<td colspan="2">current vote result for this pic</td>
</tr>
<tr>
<td>hot</td>
<td> DoFormatNumber($number2, 2, '.', '');
%</td>
</tr>
<tr>
<td>not hot</td>
<td> DoFormatNumber($numbera2, 2, '.', '');

&nbsp;%</td>
</tr>
</table>

                  </td>
                </tr>
                <tr><td>';

}