Arrays Help

[thebutler]

Hi this is my first post, and yes it is on Arrays. 🙂

This is a simple example of what I am trying to do.

I have a form page with X number of check boxes on, all the check boxes have the same name but a different value
<input type="checkbox" name="stylecheckbox[]" value="Test">

When X number of the check boxes are sent
<form name="db1" method="post" action="dbtest1.php">

to the receiving page, I get all the information that was passed.

<?php
foreach($_POST['stylecheckbox'] as $value) {
echo "<BR>You clicked " . $value , "\n";
}
?>

Now to my question all this information is in $value and if I am correct this is an Array, how do I split this Array so it will put the information into a MySQL database as a separate column for each entery.

At this time I can upload the information to the database but it puts in into just one column.
Col1
Test1
Test2
Test3

But I need it in columns like this

Col1 Col2 Col3
Test1 Test2 Test3

Any help on this would be appreciated and a snippet of code to go with it would help me a lot.
TheButler

[devinemke]

<?php
if (!isset($_POST['submit']))
{
	echo '
	<form action="" method="POST">
	<input type="checkbox" name="stylecheckbox[Col1]" value="test1"> #1<br>
	<input type="checkbox" name="stylecheckbox[Col2]" value="test2"> #2<br>
	<input type="checkbox" name="stylecheckbox[Col3]" value="test3"> #3<br>
	<input type="submit" name="submit" value="submit">
	</form>
	';
}
else
{
	if (isset($_POST['stylecheckbox']))
	{
		$sql = 'INSERT INTO table SET ';
		foreach ($_POST['stylecheckbox'] as $key => $value)
		{
			$sql_array[] = $key . " = '" . $value . "'";
		}
		$sql .= implode(', ', $sql_array);
		echo $sql; // just for testing
	}
	else
	{
		echo 'you did not check anything';
	}
}
?>

[thebutler]

Thank you for such a quick reply, I do not post normally I try to work things out for my self but this had me stumped. I will now try it out and let you know how I got on.

Again thank you John Clarke

[e-mail]

[thebutler]

Well I am dumb, I tried your code and it looked as if it was working but when I put my login details to up load the information to the server not a thing happened.

Can you please put into the code a dummy login so I can look at where I am missing the post.

At this time I have

$hostname = "localhost";
$databaseusername = "any";
$databasepassword = "any";
$database = "any";
$userstable = "testtest";
@mysql_connect("localhost", "$databaseusername", "$databasepassword");
mysql_select_db("$database");

and $sql = 'INSERT INTO testtest SET ';

<?php
if (!isset($POST['submit']))
{
echo '
<form action="" method="POST">
<input type="checkbox" name="stylecheckbox[Col1]" value="test1"> #1<br>
<input type="checkbox" name="stylecheckbox[Col2]" value="test2"> #2<br>
<input type="checkbox" name="stylecheckbox[Col3]" value="test3"> #3<br>
<input type="submit" name="submit" value="submit">
</form>
';
}
else
{
if (isset($POST['stylecheckbox']))
{
$sql = 'INSERT INTO table SET ';
foreach ($_POST['stylecheckbox'] as $key => $value)
{
$sql_array[] = $key . " = '" . $value . "'";
}
$sql .= implode(', ', $sql_array);
echo $sql; // just for testing
}
else
{
echo 'you did not check anything';
}
}
?>

If you do not have the time thank you any way.

John….

[leatherback]

hey John,

You do not seem to run the queries anywhere? You specify the sql query, but you need a command to actually go to the database and run it, e.g.,
$ result = mysql_query($sql);

[thebutler]

Hi leatherback,
What can I say but thank you I have uploaded to my database a trillion times and used the $ result = mysql_query($sql); but this was the first time to try this new coding and just missed the end, for a week so I thank you big time on this I could not see it at all.

It is now working :p

John....