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dropdown menu items


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foreach($users as $u){print('<input type="radio" name="username" value="'.$u.'">'.$u.'<br>');}

Drop down is CSS finish off.

Edited by es131245
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  • 3 weeks later...

i created loop for selecting from all the rows of a table

like this---

<?$rows = mysql_num_rows($result);for ($n = 0 ; $n < $rows ; ++$n)	{		echo '' . mysql_result($result,$n,'item') . '<br />';	}?>

but couldn't insert it into a dropdown list. please guide...

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tried to put in dropdown menu like this ----


echo '<form action="#" id="test" name="test">';echo '<select name="item">';echo "<option></option>";echo "<option>";{echo '' . mysql_result($result,$n,'item') . '<br />';}echo "</option>";echo '</select>';echo '</form>';

but returned error


"Warning: mysql_result(): Unable to jump to row 2 on MySQL result index 5 in .................................."

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This is what I did:

		<label>Client:</label>		<?php			$result = mysqli_query($con,"SELECT * FROM clients");			echo "<select name='clientid'>";			while ($row = mysqli_fetch_array($result)) {				echo "<option value='" . $row['id'] . "'>" . $row['name'] . "</option>";			}			echo "</select>";    ?>
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