dynamic dropdown not working

sarmenhb · 02-17-2008, 04:07 AM

heres the function im using

when i load the page the drop down doesnt show anything. so im wondering if there is a syntax error u can find because my ide isnt finding any.

this is how im typing the html out

<select><?php dyn_dropdown(tbl_status,status); ?></select>

Code:

function dyn_dropdown($tablename,$columnname) {


	$query = mysql_query("SELECT $columnname FROM $tablename");
	
	if(mysql_num_rows($query))
	{
	while($row = mysql_fetch_assoc($query))
	{
	$block = "<option>";
	$block .=  $row['$columnname'];
	$block .= "</option>";

	}
	
	}
	else {
	echo "<option>no data</option>";
	}

}

here is the table structure

tbl_status

Code:

CREATE TABLE `tbl_status` (
  `id`      int AUTO_INCREMENT NOT NULL,
  `status`  varchar(20) NOT NULL,
  /* Keys */
  PRIMARY KEY (`id`)
) ENGINE = InnoDB;

heres the data it contains

Code:

        id status                                                       
        20 open                                                         
        21 op

DeMo · 02-17-2008, 05:58 AM

Try this:

PHP Code:


			
$block .=  $row[$columnname];

$columnname is a variable, so I guess you can use it directly inside the $row[] array. If it doesn't work you can also try $row["columnname"] (double quotes instead of single quotes). As far as I remember variables enclosed in double quotes will evaluate to their value... while single quotes won't.

Eg:

PHP Code:


			
$x = "test";



echo "this is a $x";

//output: this is a test



echo 'this is a $x';

//output: this is a $x

SOCK · 02-17-2008, 07:17 AM

Let's not post this in every forum we run across, shall we?

sarmenhb · 02-17-2008, 06:27 PM

i figured out the problem, thanks