repopulating drop down box problems

gabrielkolbe

Hi i would be grateful if someone can help me with this, I think i know what is wrong but i don't know how to fix it. I am trying to repopulate this drop down menu from the mysql db so that the new values can be selected again...

i get this error...
Notice: Undefined variable: menuexample in C:\Program Files\Apache Group\Apache2\htdocs\prop_manager\prop_search_form.php on line 117

IN the drop down box this error occurs..
Undefined offset:1 in C:\ ......bla bla bla ...on line 137

here is the code:

<?php 

     $query = "SELECT city FROM right_prop_1";    
$city="menuexample"; 
$selected_value=$menuexample; 
$topoption = 'Select an option'; 
selectmenu($selected_value,$query,$city,$topoption); 

function selectmenu($selected_value,$query,$city,$topoption) 
{ 
        $result = mysql_query($query); 
        echo '<select name="'.$city.'">'; 

    echo '<option> ---'.$topoption.' --- </option>'; 
   while($row = mysql_fetch_array($result)) 
            { 
            echo  '<option value="';                 
            echo  $row[0]; 
            echo  '" '; 
           if($row[0]==$selected_value) 
                    { 
                    echo  'selected'; 
                    } 
            echo  '>'; 
            echo  $row[1]; 
            echo  '</option>'; 
            } 
    echo '</select>'; 
} 
?>

devinemke

you are never setting the $menuexample variable.

gabrielkolbe

Thanks i new that, i can see it too unfortunately i am resonable new to php and have pirated this code form sowhere but i am struggling to get it to work..

here are the original code:

$query = "SELECT city FROM right_prop_1";    

$selectname="menuexample"; 
$selected_value=$menuexample; 
$topoption = 'Select an option'; 
selectmenu($selected_value,$query,$city,$topoption); 

function selectmenu($selected_value,$query,$selectname,$topoption) 
{ 
        $result = mysql_query($query); 
        echo '<select name="'.$selectname.'">'; 

    echo '<option> ---'.$topoption.' --- </option>'; 
   while($row = mysql_fetch_array($result)) 
            { 
            echo  '<option value="';                 
			echo  $row[0]; 
            echo  '" '; 
           if($row[0]==$selected_value) 
                    { 
                    echo  'selected'; 
                    } 
            echo  '>'; 
            echo  $row[1]; 
            echo  '</option>'; 
            } 
    echo '</select>'; 
}

_RL_Ian

Try something like this. I didn't test it so their may be errors or something, but I wasn't sure if you were trying to make a menu dropdown or a city dropdown, so I chose city. Anyway, let me know if you any more help:

<?php
$query = "SELECT city FROM right_prop_1";
$city = "Atlanta";
$topoption = 'Select an option';

function selectmenu($query, $city, $topoption) 
{ 
	$result = mysql_query($query); 
    echo '<select name="menuexample">'; 
    echo '<option> --- ' . $topoption . ' --- </option>'; 
    while($row = mysql_fetch_array($result)) 
    { 
    	echo ('<option value="' . $row['city'] . '" ' . ( ($row['city'] == $city) ? echo("selected";
        if ($row['city'] == $city)
        {
        	echo 'selected'; 
        }
		echo '>' . $row['city'] . '</option>';
	} 
	echo '</select>';
}

selectmenu($selected_value, $query, $city, $topoption);
?>

gabrielkolbe

Thanks for your help... but yes i do get an error with that script i have tried various options to solve it but to no avail.

here is the error:

Parse error: parse error in C:\Program Files\Apache Group\Apache2\htdocs\prop_manager\prop_search_form.php on line 133

line 133 is:

  echo ('<option value="' . $row['city'] . '" ' . ( ($row['city'] == $city) ? echo("selected";

I think maybe the brackets aren't right, i tried to put new ones in and move them but can't get the error solved..

_RL_Ian

gabrielkolbe wrote:
Thanks for your help... but yes i do get an error with that script i have tried various options to solve it but to no avail.

here is the error:
Parse error: parse error in C:\Program Files\Apache Group\Apache2\htdocs\prop_manager\prop_search_form.php on line 133
line 133 is:
  echo ('<option value="' . $row['city'] . '" ' . ( ($row['city'] == $city) ? echo("selected"; 
I think maybe the brackets aren't right, i tried to put new ones in and move them but can't get the error solved..

Not sure how that paste got in there. Try this one:

<?php
$query = "SELECT city FROM right_prop_1";
$city = "Atlanta";
$topoption = 'Select an option';
$selected_value = 'menuexample';

function selectmenu($selected_value, $query, $city, $topoption) 
{ 
    $result = mysql_query($query); 
    echo '<select name="' . $selected_value . '">'; 
    echo '<option> --- ' . $topoption . ' --- </option>'; 
    while($row = mysql_fetch_array($result)) 
    { 
        echo '<option value="' . $row['city'] . '" ';
        if ($row['city'] == $city)
        {
            echo 'selected'; 
        }
        echo '>' . $row['city'] . '</option>';
    } 
    echo '</select>';
}

selectmenu($selected_value, $query, $city, $topoption);
?>

gabrielkolbe

Thank you VERRRRY much it works like a dream! While we are on the subject would you mind if i pickle your brain a bit futher... how would you use the above script and when it prints the info in the drop down box make it so, so that only two entries of one city will not be in the drop down box i.o. w. if Atlanta is in the db twice it will desplay it twice, how do you get it to desplay Atlanta only once?

_RL_Ian

<?php
$query = "SELECT DISTINCT city FROM right_prop_1"; 
?>

Try that and let me know.