how to pass value of checked checkbox in php - php

Question

Hi i am generating checkbox list by using a for each loop, I want to pass the value of checked checkbox into database.
foreach($this->lis as $lst)
{?>
<tr>
<td>
<input type="checkbox" name="list" value="1" />
<label for="list_32"><?php echo $list->nList ?></label>
</td>
</tr>
<?php } ?>
when i checked the checkbox i want to pass the label name into the database can someone help.

Answer 1

Check this code. You need to take array as checkbox name i.e. list[]. Once form is submitted you can use it's value to insert in to database.
<?
if(isset($_POST['submit'])) // once form is submitted build your logic
{
$list = $_POST['list'];
foreach($list as $value)
{
echo "<br />Checked: $value";
// use $value to insert into database
$sql = "insert into...";
mysql_query($sql);
}
}
?>
<form name="frm" method="post">
<table>
<? foreach($this->lis as $lst)
{?>
<tr>
<td>
<input type="checkbox" name="list[]" value="<?php echo $list->nList ?>" />
<label for="list_32"><?php echo $list->nList ?></label>
</td>
</tr>
<?php } ?>
<tr>
<td><input name="submit" value="Submit" type="submit" /></td></tr></table>
</form>

Answer 2

Ok, you've got a few problems there..
Firstly, please avoid opening php tags and closing them constantly. Just echo back the HTML. Going in and out of php so often will cause performance issues.
So, first off, you need to set up a form on here, your form needs to point to a second page (or self-process using an isset function, but let's go with the former solution for clarity). The second page is what's going to dump things into the DB for you based on what the user ticked.
So, here's the HTML for your form;
<form action="database.php" method="POST">
<div>
<label for="checkbox1">This is the first option:</label>
<input type="checkbox" id="checkbox1" name="checkbox1" />
</div>
<div>
<label for="checkbox2">This is the second option:</label>
<input type="checkbox" id="checkbox2" name="checkbox2" />
</div>
<div>
<input type="submit" />
</div>
</form>
Again, I'd recommend echo'ing this back in your foreach loop - a short function could accomplish this for you.
So, once someone hits the "Submit" button, they'll be taken to your second page (database.php as we've called it), and their options will be stored in the POST array. For good practice, always use POST rather than GET when dealing with DB entries.
So on this page, we'd check what checkboxes were selected, and then update the db as required;
if (isset($_POST['checkbox1']))
{
// user ticked checkbox1
$sql = "UPDATE table
SET 'checkbox1' = 1
WHERE 'user' = 'username';"
mysqli_query($sql);
}
Note that's just a rough solution - you haven't told me exactly how you're doing this, so I'm assuming your DB is tied to user's votes etc. You can also update checkbox1 by incrementing the value.. etc etc.
Hope this helps.
Eoghan

Answer 3

In the name you should use [] (because it's an array). So in this case it'd be name="list[]". Then when you submit it (either post or get) you just get the checked data on the list[] array. I'm not sure if I understood your question. If I didn't; please share a little more information.

Answer 4

I think this is what you're looking for:
foreach($this->lis as $lst)
{?>
<tr>
<td>
<input type="checkbox" name="list[<?php echo $list->nList ?>]" value="1" />
<label for="list_32"><?php echo $list->nList ?></label>
</td>
</tr>
<?php } ?>
Then in PHP, you would do foreach($_POST['list'] as $listval) to loop over it again. Substitute $_POST for $_GET if you are using GET in your form.
Edit: I just noticed where you did echo $list->nList that this variable is either not going to change in your loop or it's a typo. Assuming it's a typo, it should be echo $lst->nList instead. If it's not a typo, then you want to just remove it and have name="list[]" in your HTML