drop down menu/PHP/MySQL

ashz

Hi,

As a newbie in PHP I am looking for a simple and easy way to take a data from one of the field in MySQL and use them to create a drop down menu. How can I transfer the data dynamically?

Thanks,
Ashz.

rIkLuNd

This is how I would do it:

<?php

// Connecting to host
$link = mysql_connect($host,$user,$pass);
if(!$link)
die(mysql_error());

// Selecting database
$select = mysql_select_db($db);
if(!$select)
die(mysql_error());

// Getting the content of the table
$query = mysql_query("SELECT * FROM table");
if(!$query)
die(mysql_error());

// Print a little form
print "<form action=\"$PHP_SELF\" method=\"post\">\n";
print "<select name=\"value\">\n";

// Then we use a loop to create the menu
while($row = mysql_fetch_array($query))
{
print "<option name=\"$row[name]\" value=\"$row[value]\">$row[name]</option>\n";
}

print "</select>\n";
print "<input type=\"submit\" value=\"send\">\n";
print "</form>\n";

// Closing connection
$close = mysql_close($link);

?>

liteon

Hi,

The script worked fine but I tried to split it and failed.
Due to that is it possible to split the script to 2 files (HTML and PHP) and without $PHP_SELF.

Thx.

rIkLuNd

No, I don't think that it's possible to split this script, but I'll try it. I'll write the result here!

rIkLuNd

Nope, it's not possible, since you can't use the [man]while[/man] statement in a html file... Sorry!

liteon

Is their another way then.

Thx.

rIkLuNd

maybe... I can investigate it!

liteon

The reason that I need to split it, is since I have more then 4 fields that I need to put on diffrent drop-down menu and build from them a query.

daveyboy

Here is a function I use to return a pulldown box. Run your query and build an array from the results, then pass this as $items to the function.

For example:
$items = array( "dave"=>1, "kev"=>2, "sally"=>3 );
$form = get_dropdown( "fm_mate", $items, 2 );

$form contains the HTML code for the form, with "kev" selected. You would the use $POST['fm_mate'] to find the selected value. If the user chooses "dave", $POST['fm_mate'] will be 1.

/**
* Generates and returns a pulldown menu for a form.
*
* @access public
* @param string $name
* @param array $items Array of options. Title => value.
* @param string $selected Value of selected item in array $items.
* @return string
*/
function get_dropdown( $name, $items, $selected )
{
	$content = 	"<select name=\"$name\" id=\"$name\">";
	foreach ( $items as $title=>$value ) {
		$content .= "<option value=\"$value\" ";
			if ( $selected == $value ) {
				$content .= 'selected="selected"';
			}
		$content .= ">$title</option>";
	}
	$content .= "</select>";
	return $content;
}

TheReverend

You can split this up... (if by splitting you mean by having: form action="anotherpage.php")

Sure, watch this :-)

Page 1: form.php


<?php 

// all your database connect info here

?>
<!-- HTML header info, etc -->

<FORM ACTION="submit_info.php" METHOD="POST">
<SELECT NAME="dropdown_name[]"> 
<!-- IMPORTANT TO HAVE THE ARRAY brackets [] AFTER THE NAME OF THE DROPDOWNLIST -->

<?php
$query = "SELECT * FROM table";
$result = mysql_query($query);

while($row = mysql_fetch_array($result))
{
echo("<OPTION value=\"".$row[value]."\">".$row[name]."\n");
}
?>

</SELECT>
<INPUT TYPE="SUBMIT" VALUE="Submit">
</FORM>

Now for Page 2: submit_info.php

<?php

// Pull the info from the drop down list...
while(list($key,$val) = each($dropdown_name)) 
{
$dropdown_value = $dropdown_name[$key];
}

// $dropdown_value is what you put into your datasbase
//example:

$query = "INSERT INTO table (id) VALUES ('$dropdown_value')";
$result = mysql_query($query);
if($result)
echo("Info entered into DB");
else
echo("Something went wrong. ERROR: ".mysql_error());

?>

Hope this helps :-)