[RESOLVED] Array from multiple select into variables

edude

Hi,
I am not a programmer, work with php from time to time and mostly to do easy stuff, so my knowledge is very basic.
I am trying to get the values of an array, coming from a select with multiple options, to use them in the "where" part of mysql queries. After looking around I don't know how to do this. Here i paste my code:

<HTML>
<select name="enquesta[]" size="8" multiple>
<option value="All" SELECTED>All</option>
<option value="EB1">Eurobarometer 1</option>
<option value="EB2">Eurobarometer 2</option>
<option value="EB3">Eurobarometer 3</option>
<option value="EB4">Eurobarometer 4</option>
<option value="EB5">Eurobarometer 5</option>
<option value="EB6">Eurobarometer 6</option>
<option value="EB7">Eurobarometer 7</option>
<option value="EB8">Eurobarometer 8</option>
<option value="EB9">Eurobarometer 9</option>
<option value="EB10">Eurobarometer 10</option>
...
<option value="EB64">Eurobarometer 64</option>

</select>
</HTML>

and the php one:

<PHP>
$enquesta=$_POST['enquesta'];

$enq_explo = explode(', ', $_POST['enquesta']);
echo count($enquesta)."<br />";
echo $enq_explo[0];

if (count($enquesta)==1 & $enq_explo[0]=="All")
{$str_enquesta=="";
}
if (count($enquesta)==1 & $enq_explo[0]!="All")
{$str_enquesta=="AND q.questionnaireid LIKE '$enq_explo[0]'";
}

echo "<br>".$str_enquesta;
</PHP>

I have to continue the php code to recover the options selected by the user when they are more than one. But even with only option I cannot succed; with this code I only get the output "Array".
Any help will be welcomed. Thanks in advance.

devinemke

<?php
if (!isset($_POST['submit']))
{
	echo '
	<form action="" method="POST">
	<select name="enquesta[]" size="8" multiple>
	<option value="All">All</option>
	';

for ($i = 1; $i <= 64; $i++)
{
	echo '<option value="EB' . $i . '">Eurobarometer ' . $i . '</option>';
}

echo '
</select><br>
<input type="submit" name="submit" value="submit">
</form>
';
}
else
{
	if (isset($_POST['enquesta']))
	{
		$sql = 'SELECT * FROM table';
		if (!in_array('All', $_POST['enquesta']))
		{
			foreach ($_POST['enquesta'] as $key => $value) {$_POST['enquesta'][$key] = "'" . $value . "'";}
			$sql .= ' WHERE questionnaireid IN(' . implode(',', $_POST['enquesta']) . ')';
		}
		echo $sql;
	}
	else
	{
		echo 'you did not select anything';
	}
}
?>

in this example selecting "All" overides any other selection. this may or may not be what you want.

edude

Hi Devinemke,

Thank you very much for your answer; it worked, and additionally i learnt the use of mysql syntax "...IN..." to use in the "where".

Eduard