display data from database after chekcbox selection - php

Question

I want to have a form that allows the user to choose what data to display from a table through checking the checkboxes. If the user wants only 2 columns to be shown, should only 2 columns be shown. I have my codes, but after I submit, it displays nothing.Here's my code:
<form name="form1" method="post" action="view_emp.php">
<p>Select Option
<input type="checkbox" name="number[]" value="name" />Name
<input type="checkbox" name="number[]" value="hired" />Date Hired
<input type="checkbox" name="number[]" value="basic" />Basic Pay
<input type="checkbox" name="number[]" value="incentives">Incentives
</p>
<input type="submit" name="Submit" value="Submit">
</form>
here's my php:
<?php
$db = mysql_connect('localhost', 'root', '');
mysql_select_db('eis', $db) or die (mysql_error());
$employee = array();
foreach ($_POST['number'] as $employee) {
$number = mysql_real_escape_string($number);
$employee[] = "'{$number}'";
}
$sql = "select * from employees where type in (" .implode(", ", $number). ")";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
print $row['name'];
}
?>
i am a beginner in php and i need help from gurus and experts. thank you...

Answer 1

PHP's implode() and explode() functions might come in handy. You can easily turn your POST or GET attribute, 'number', into a comma-separated list by using implode($_POST['number']). This would be easy to store in one MySQL field, maybe a column in your user table.
If you want users to edit the form later, render the checkboxes with a loop and add a "checked" attribute to each checkbox element whose name exists in 'exploded' list (array) retrieved from your database.
This is basically serialization/deserialization. There are other ways to do it including serialize(), unserialize() or json_encode(), json_decode(). But since your data seems to be easily modeled as a basic list, you can keep it even simpler.