[RESOLVED] dropdown menu question

[trendMM]

hi,

i'm working on a website for a client and on their events page it is broken down into national, regional, and then every state in the us. well in the cms aspect of it i figured i would have a dropdown so they could select where the event falls into national, regional, or statewise. then on the public end people can sort it by that as well to find events that work best for them. anyways i know how to do the whole drop down thing but my question is, when they go to edit an event, i want it so it displays the dropdown again but automatically selects the one they had chosen when they created it. make sense? any help would be greatly appreciated!

evan

[leatherback]

In your dropdown-creation function, add something like:

$selected = "";
if(isset($selected_country))
  {
  if($row[id] == $selected_country)
    {
    $selected = " selected ";
    }
  }
$drop .= "<option $selected value=\"$row[id]\">$row['description']</option>

[trendMM]

Thanks, does this mean I need to store all of my regions in their own table in the db or what? Usually I just have the dropdown set to static within the code.

[leatherback]

ah, oh, sorry. I thought to remember that you were pulling them from a database.

nope, you do not need to get them from a database.

Two options:

• Put the fieldoptions in an array, and do a foreach loop, where you create the dropdown, like I showed you in my prev. post.
• Take the optionlist, modify each option:
  <option <?php echo $selected['OPTION_VALUE_HERE'] ?> value="OPTION_VALUE_HERE">>OPTION_DESCRIPTION</option>

And when you get your data from the post array:

$value = $_POST['value'];
$selected[$value] = " selected ";

The latter will give you warnings if you have set error reporting to all, but not if you have it to failures only (Which is the default, I think). Just remember it is not fully correct, but it is pretty much the default people are using.

[trendMM]

Awesome, thanks for your help, it's starting to make some sense to me now. But how exactly do I tie the dropdown into the value already stored in the DB for that field? Do I give the dropdown select area the name of the field in the DB for that or what? Thanks again!

[trendMM]

After looking at it I think I got a better idea of how it all works. So basically I either pull all of the stuff for the drop down from a db or put it all in an array. Including the description and value options within it.

Then I run the foreach and loop it through til it matchs up with the current $field that has been stored in the database until they match up and it gives that <option> tag the selected tag within it.

Now that I know how it works how would I go about putting the values all into an array and having it run foreach? A snippled of code would be great. For instance. If I had a yes and no option on my drop down with the value of yes and no of course. How would I put those in an array to check them up against the current selection stored in the db? I'm pretty much just stuck on the array function and how I plug it into your code pasted above.

Thanks in advance!

[leatherback]

How about a generic function?

// an array

$myarray = array(
1 => "Testje",
2 => "Surely",
3 => "this works",
4 => "if I say it does");

$myarray[5] = "Unless I forget one";

echo "You cannot echo an array".$myarray." see?<br><pre>";
echo "You need to print_r it:";
print_r($myarray);
echo "<br /> or print each value seperately: "
foreach($myarray as $key=>$value)
  {
  echo $value."<br />";
  }
echo "</pre>";


<?php
// generic function to create a dropdown body from an array
// $array: input data, each array element represents information about one dropdown enrtry, in an sub-array
// $selected_id: Identifier of the selected drop-down element
// $value_label: The name of the array-element that holds the value for the dropdown
// $name_label: The name of the array-element that holds the name for the dropdown element


function create_dropdown($myarray, $selected_id)
  {
  $dropdown = "";

  foreach($array as $key=>$value)
    {
    $selected = "";
    if($key == $selected_id)
      {
      $selected = "selected";
      }

$dropdown .="<option $selected value=\"".$key."\">".$value."</option>
            ";
}

  return $dropdown;
  }

$dropdown = "<select>";
$dropdown .= create_dropdown($myarray, 4);
$dropdown .= " </select>";

echo $dropdown;

[trendMM]

kind of got lost on this one so i tried posting your code to disect it and get a better idea of how to make it work for me and i keep getting an error...

Parse error: syntax error, unexpected T_FOREACH, expecting ',' or ';' in /home/evanbart/public_html/test.php on line 17

[leatherback]

Oh, shoop,

I thought I changed sometihng..

I modified one of my functions for you, but couldn't test it, since I am not behind my dev. machine (But at work 🙂)

this line:
function create_dropdown($myarray, $selected_id)

should read:
function create_dropdown($array, $selected_id)

and this: <?php
should not be there..

But it should work, and yes, I added some examples in there, with echoes so you can see how it works.

[trendMM]

still no worky. this whole drop down idea seems to be a little harder for me to figure out then expected. thanks for helping me out so far though!

[dmacman1962]

Leatherback was missing a few things,

1) a ; at the end of this line

echo "<br /> or print each value seperately: "

2) a space after

"You cannot echo an array".$myarray.

Here is the code working. Try it and see if it makes more sense now. In leatherbacks defense, I understand he is not on his testing computer.

// an array

$myarray = array(
1 => "Testje",
2 => "Surely",
3 => "this works",
4 => "if I say it does");

$myarray[5] = "Unless I forget one";

echo "You cannot echo an array ".$myarray." see?<br><pre>";
echo "You need to print_r it:";
print_r($myarray);
echo "<br /> or print each value seperately: ";
foreach($myarray as $key=>$value)
  {
  echo $value."<br />";
  }
echo "</pre>";



// generic function to create a dropdown body from an array
// $array: input data, each array element represents information about one dropdown enrtry, in an sub-array
// $selected_id: Identifier of the selected drop-down element
// $value_label: The name of the array-element that holds the value for the dropdown
// $name_label: The name of the array-element that holds the name for the dropdown element


function create_dropdown($array, $selected_id)
  {
  $dropdown = "";

  foreach($array as $key=>$value)
    {
    $selected = "";
    if($key == $selected_id)
      {
      $selected = "selected";
      }

$dropdown .="<option $selected value=\"".$key."\">".$value."</option>
            ";
}

  return $dropdown;
  }

$dropdown = "<select>";
$dropdown .= create_dropdown($myarray, 4);
$dropdown .= " </select>";

echo $dropdown;

enjoy,

Don

[leatherback]

Hey Don,

Thanks,

Nope, was writing a few replies in my coffee break at work. Home now, after seeing a horrible movie (davinci code). Bit slow.

[trendMM]

awesome, thanks a lot to both of you for your help. i really appreciate it. what i ended up doing before hooking it up to my db was run a variable with a value at the top and replacing the "4" in the create area to my variable and it worked great!

One thing I had a question about though. When you made the array part, why is #5 setup differently than the rest? Were you just trying to show me the ways I could format it or was there a reason for that? Thanks!

[dmacman1962]

Leatherback was just showing you how to add another item to the array.

Glad to hear you are good to go.

enjoy!

Don

[trendMM]

Oh cool, I was thinking that might be all it was so yeah. Thanks again!