Using PHP MySQL to grab data for a form, but having issues in $_POST

Question

Okay, I'm sure I'm missing something simple, but I'm populating a hidden field on a form with data from a MySQL While statement. the data shows up properly when on the page, but when submitted, the data submitted is always from the last loaded id.

while ($row = mysql_fetch_assoc($result)) {
    $id = $row['id'];
    echo '<form name="voting" action="thanks_for_voting.php" method="POST">
    <align="right"><input type="hidden" name="id" value="'.$id.'"
    <br><input type="submit" value="VOTE FOR '.$id.'"><br>';
}

What am I doing wrong? I'm trying to keep a single call to my DB to prevent overloading, I assume there must be a simple way to achieve my goal? But what?

Machineadict solved my problem below by noticing I hadn't been closing my form. I closed it, and the problem was solved. Thanks!

xpda · Accepted Answer · 2013-01-21 19:49:55Z

2

The form tag is not closed. Lose the html tags, you don't need them. And it's alright to use forms in loops, you don't have to change the names of inputs.

edited Jan 21, 2013 at 19:49

xpda

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answered Jan 28, 2012 at 10:29

machineaddict

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1 Comment

Tony Over a year ago

You got it! I forgot to close my form so it was left open!

Adam Zalcman · Accepted Answer · 2012-01-28 11:53:21Z

1

As machineaddict has mentioned, if you want to use multiple forms you must close the tag of the preceding one before you open the next.

You can also do this with a single form with a single hidden field assigning it a specific value depending on which button is clicked:

echo '<html><form name="voting" action="thanks_for_voting.php" method="POST"><align="right">';
echo '<input type="hidden" name="id" value="-1">';
while ($row = mysql_fetch_assoc($result)) {
  $id = $row['id'];
  echo '<input type="button" value="VOTE FOR '.$id.'" onclick="javascript:document.forms[\'voting\'].elements[\'id\'].value='.$id.'; document.forms[\'voting\'].submit();">';
}
echo '</form></html>';

Here you have a button for each of the IDs from the database. Each button executes the following JavaScript code in its onclick handler:

document.forms['voting'].elements['id'].value = YOUR_ID;
document.forms['voting'].submit();

This sets the ID to the desired value and then submits the form. You may want to put this code aside in a function and put the name of the function into the onclick handler.

Also, you probably don't want to generate multiple <html> tags.

edited Jan 28, 2012 at 11:53

answered Jan 28, 2012 at 10:12

Adam Zalcman

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1 Comment

Tony Over a year ago

What do you think the easiest way would be to accomplish that in the loop? I'm outputting 3 results with every loop by design.

SimonMayer · Accepted Answer · 2012-01-28 10:13:13Z

0

Your <form> is inside the while loop. And as Adam Zalcman says, use a unique name.

Try

echo '<html><form name="myform" action="thanks_for_voting.php" method="POST"><align="right">';
while ($row = mysql_fetch_assoc($result)) {
  $id = $row['id'];
  echo '<input type="hidden" name="id[]" value="'.$id.'"><br><input type="submit" value="VOTE FOR '.$id.'">';
}
echo '<br></html>';}

This will post an array of values, which you can retrieve as $_POST['id'][0], $_POST['id'][1] etc

answered Jan 28, 2012 at 10:13

SimonMayer

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2 Comments

SimonMayer Over a year ago

actually. This is wrong. I've missed the point of what you're trying to do.

Tony Over a year ago

my goal is to have three submit button on the page, each one for a different item, containing the unique id. that way when a user clicks the "vote for ___" button, it submits to the next page the id data from that item.

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