Insert labels of checked checkboxes in database

RLJ · December 29, 2010

I use the following code to generate a scrollable checkbox list of options:

<html>
<form action="SCRIPT.php">

<?php
	$expList = array('Engineering ','Science', 'Art', 'IT', 'Electronics', 'Communications');
        sort($expList);

                $tmp = array();
	$i = 1;
	$tmp[] = '<ul style="height: 95px; overflow: auto; width: 200px; border: 1px solid #480091; list-style-type: none; margin: 0; padding: 0;">';
	foreach ($expList as $option) {
     			$tmp[] = '<li id="li'  . $i . 'b"><label for="chk' . $i . 'b"><input name="chk' . $i . 'b" id="chk' . $i . 'b" type="checkbox" onchange="Enable(\'chk' . $i . 'b\',\'li' . $i . 'b\')">' . $option . '</label></li>';
     			$i++;
	}
	$tmp[] = '</ul>';
	echo implode("\n",$tmp) . "\n";
?>

<input type="submit">
</form>
</html>

But there are in fact over 50 options (not just 6) and they are in fact stored in an array $expList in an external php file called LISTS.php.

As you can see, the checkbox list consists of checkboxes chk1b, chk2b, chk3b, etc. with associated labels (Art, Communications, Electronics, etc.).

What I need in SCIPT.php is code that will insert each of the labels where the associated checkbox has been checked in my MySQL database. E.g. in this particular case: if chk1b & chk3b have been checked, 'Art' & 'Electronics' will be inserted. Also, I want the label of the first checked checkbox to be inserted in the (database) table column 'Exp1', the 2nd one in 'Exp2'.........25th one in 'Exp25' (I already have a script that allows a maximum of 25 checkboxes to be checked).

I'm guessing some sort of 'foreach' loop is required, but I can't quite work it out. Pls help!

Many thanks!

sasa · December 29, 2010

change chkbox name and add value attribute

$tmp[] = '<li id="li'  . $i . 'b"><label for="chk' . $i . 'b"><input name="chk_b[' . $i . ']" id="chk' . $i . 'b" type="checkbox" value ="'.$option.'"onchange="Enable(\'chk' . $i . 'b\',\'li' . $i . 'b\')">' . $option . '</label></li>';

and on submit page do

foreach $_POST['chk_b'] as $index => $option) echo $index, ' -> ', $option, '<br />\n';

RLJ · December 30, 2010

Thanks for your help so far!

I've modified your code slightly to the following:

$i = 1;
foreach ($_POST['chk_b'] as $index => $option) 
{
echo $option, '<br>';
"$"."EXP".$i = '$option';
$i++;
}

Because I want to do the following:

1) print each option on-screen

2) assign $EXP1=first selected option, $EXP2=2nd selected option......... $EXP6=6th selected option, etc. (N.B. not neccesarily $EXP6=option with index 6)

1) is working

2) is not, can you see why? (it's probably a very elementary error, but I'm new to this stuff)

3) Also, the final thing I want to do is, once $EXP1, $EXP2, etc. have been assigned, to then insert $EXP1 in the MySQL table column #n, $EXP2 in column #n+1, $EXP3 in column #n+2, etc.

I'm familiar with how to connect to MySQL and insert data when all the variable names are known, but in this case I don't know how many 'EXP's the user will select and so, again, I'm guessing a foreach loop is required and I'm stuck...

Greatly appreciated if you could help me out!

Thanks.

sasa · December 31, 2010

<?php
$i = 1;
foreach ($_POST['chk_b'] as $index => $option) 
{
echo $option, '<br>';
        $fields[$i] = "`EXP$i`";
$value[$i] = "'$option'";
$i++;
}
//echo
$sql = "INSERT INTO table_name (". implode(', ', $fields). ") VALUES (". implode(', ', $value). ")";
?>

RLJ · December 31, 2010

That's great! Thanks a lot!

Final question:

The part where I connect to database & insert is actually in a different PHP script from the part where I generate $fields and $value, so I've tried to pass on $fields and $value as follows:

first script:

$i = 1;
foreach ($chk_b as $index => $option) 
{
   echo $option, '<br>';
        $EXPfields[$i] = "`EXP$i`";
   $EXPvalue[$i] = "'$option'";
   $i++;
}

print "
<html><body><form action='2ndscript.php' method='POST'>
<input type='hidden' name='EXPfields' value='{$EXPfields}'>
<input type='hidden' name='EXPvalue' value='{$EXPvalue}'>
</form></body></html>";

2nd script:

$EXPfields = $_POST['EXPfields'];
$EXPvalue = $_POST['EXPvalue'];

$link = mysql_connect  ('localhost', 'root');
if (!$link) {die('Could not connect: ' . mysql_error());}

$selectDB = mysql_select_db ('database_name', $link);
if (!$selectDB) {die('Could not select database: ' . mysql_error());}

$insert= mysql_query   ("INSERT INTO table_name (". implode(', ', $EXPfields). ") VALUES (". implode(', ', $EXPvalue). ")"); 
if (!$insert) {die('Could not insert into database: ' . mysql_error());}

However, I get the following error message:

Could not insert into database: Column count doesn't match value count at row 1

If I put the "connect to database & insert" part in the first script it works fine.

What have I done wrong? Is it not possible to pass on arrays like this? (So far I've only passed on single variables in this manner).

Thanks again.

BlueSkyIS · December 31, 2010

i suggest that you store the arrays in session variables and pick them up in the second script.

http://www.google.com/search?client=safari&rls=en&q=php+session&ie=UTF-8&oe=UTF-8

RLJ · December 31, 2010

Thanks for your help, but I'd rather not use sessions, since I am a beginner at this stuff. Is there a way to simply POST through the arrays?

Thanks.

BlueSkyIS · December 31, 2010

Do you mean a way that does not require you to click a button to POST the values to the next page? If that's what you mean, I guess you could use header("location: nextpage.php?"); and append all values on the end of nextpage.php? thereby passing the values in the URL which is obviously undesirable.

FYI: POSTed information can be easily spoofed, much more easily than sessions. One of the first things I learned in PHP was to pass variables behind-the-scenes via sessions. Passing values via sessions takes fewer lines of code than what you've already got.

If you don't want to learn about sessions, then I would put all logic in the one file and then use header() to redirect to whatever page you need to go to after the logic is complete.

RLJ · January 1, 2011

OK, so to clarify:

I now use the following code to generate a scrollable list of options: (call this SCRIPT1.php)

<html>

<form action="SCRIPT2.php" method=POST>

<?php
	$expList = array('Engineering','Science', 'Art', 'IT', 'Electronics', 'Communications');
        sort($expList);

                $tmp = array();
	$i = 1;
	$tmp[] = '<ul style="height: 95px; overflow: auto; width: 200px; border: 1px solid #480091; background-color:#E7E7E7; list-style-type: none; margin: 0; padding: 0;">';
	foreach ($expList as $option) 
            {
     			$tmp[] = '<li id="li'  . $i . 'b"><label for="chk' . $i . 'b"><input name="chk_b[' . $i . ']" id="chk' . $i . 'b" type="checkbox" value ="'.$option.'">' . $option . '</label></li>';
		$i++;	            
}
	$tmp[] = '</ul>';
	echo implode("\n",$tmp) . "\n";
?>

<input type="submit">
</form>
</body>
</html>

SCRIPT2.php then retrieves the selected options as follows:

<?php

$chk_b = $_POST['chk_b'];
$i = 1;
foreach ($chk_b as $index => $option) 
{
   echo $option, '<br>';
        $fields[$i] = "`EXP$i`";
   $value[$i] = "'$option'";
   $i++;
}
$EXPfields = implode(', ', $fields);
$EXPvalues = implode(', ', $value);
//echo $EXPfields;
//echo $EXPvalues;

print "
<html><body><form action='SCRIPT3.php' method='POST'>
<input type='hidden' name='EXPfields' value='{$EXPfields}'>
<input type='hidden' name='EXPvalues' value='{$EXPvalues}'>
<input type='submit''>
</form></body></html>
";

?>

SCRIPT3.php then retrieves $EXPfields and $EXPvalues as follows:

<?php

$EXPfields = $_POST['EXPfields']; 
$EXPvalues = $_POST['EXPvalues'];

//echo $EXPfields;
//echo $EXPvalues;

?>

So as you can see, SCRIPT1.php creates a scrollable list of options. SCRIPT2.php creates an array from all selected options ($value) and an array of column names where the selected options should eventually be inserted in the database ($fields).

Then SCRIPT2 creates a variable $EXPfields which is an implosion of $fields and a variable $EXPvalues which is an implosion of $value.

These 2 variables are then passed on to SCRIPT3.php via hidden form elements.

If I 'un-comment-out' "echo $EXPfields;" and "echo $EXPvalues;" in SCRIPT2.php the 2 imploded arrays are printed on-screen as expected. But if I do the same in SCRIPT3.php, only $EXPfields; is printed: it seems like $EXPvalues; isn't passed through properly.

So, hopefully that is slightly clearer now. Any ideas? Thanks!

BlueSkyIS · January 1, 2011

I think I understand now. This is still far more trouble than sessions, especially if the user clicks the back button. but anyway...

You can't pass an array in a form field, per se. You will need to loop over the array values, setting each value in a hidden field.

$EXPfields = $_POST['EXPfields'];

foreach ($EXPfields AS $field_value) {
     echo "<input type='hidden' name='$EXPfields[]' value='$field_value'>";
}

... and then grab the array on the next page.

You should really just do

session_start();
$_SESSION['EXPfields'] = $_POST['EXPfields'];

and be done with it.

RLJ · January 2, 2011

Hmmmm OK I think I'll take your advice and go with sessions...

Thanks for your help you two.

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Insert labels of checked checkboxes in database

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