selected in drop down list

raffael3d

I have an admin center where I can modify content. Now one line is a "select" drop down list. I want to make that depending what data is in this colum it shows as "selected"
It's not necessary that it fetchs all the possible categorys from the database, rather that I can write them in the script
<option>category1</option>
<option>category2</option> selected
<option>category3</option>

how can I achieve that the right one is "selected" ?

the_Igel

for ($i = 1; $i < 4; $i++)
{
$tmp = 'category'.$i;
echo '<option';
if ($selected_cat == $tmp) echo ' selected';
echo '>'.$tmp.'</option>';
}

GBahle

Add the word "selected" inside the option tag that you want to be selected (like this: <option selected>whatever</option>)

raffael3d

thanks a lot
i have a problem as I don't understand it competely, how can I make that e.g
the categories "car, home, base" appear in the "options" but only the appropriate one is selected (it looks in the database in the colum "catg" which one it should display as selected, thanks.

the_Igel

$catg = array('car', 'home', 'base');
$selected = 'car';

foreach ($catg as $cat)
{
echo '<option';
if ($selected == $cat) echo ' selected';
echo '>'.$cat.'</option>';
}

raffael3d

thanks a lot!! it's working, now I have one further question:

'$catg = array('1', '2', '3');'

how can I modify the "appearance" of it?
I made that the value that is filled in the "catg" is a number. Now I would like to display those numbers as a name
like 1= car
2= home etc.

I tried it several times to get it right, can you help me again? thanks

the_Igel

$catg = array(1 => 'car', 2 => 'home', 3 => 'base');
$selected = 1;

foreach ($catg as $key => $cat)
{
echo '<option value='.$key;
if ($selected == $key) echo ' selected';
echo '>'.$cat.'</option>';

it will produce

<option value=1 selected>car</option>
<option value=2>home</option>
<option value=3>base</option>

raffael3d

thank you very much!! many thanks!

only one thing I don't understand. Shouldn't this line be variable?

$selected = 1;

the_Igel

$selected should be set to whatever option is selected. Isn't it cler enough?

raffael3d

allright thanks a lot. I have it now this way and it works[. thanks again

raffael3d

ah unfortunately something is not working properly, I don't know if we have a misunderstanding there. the scripts works fine, except that it's not checking which data is already in the database.
depedning on which value is set in the database (1, 2 or 3) it should display the selected.
I tried something like this

echo "echo "<tr><td><select name='catg'> ";

$catg = array(1 => 'car', 2 => 'home', 3 => 'base');

$selected = 1;
$selected2 = 2;
$selected2 = 3;



foreach ($catg as $key => $cat)

{

echo '<option value='.$key;

if ($selected == $key) { 
	echo ' selected';
	}
if ($selected2 == $key) { 
echo ' selected2';
}
if ($selected3 == $key) {
echo ' selected3';
}

echo '>'.$cat.'</option>';
}

echo "</select></td></tr>";

the_Igel

Look,

you have a dbtable and there's a field which contains the index of category selected.
You retrieve it through mysql query or whatever:

$r = mysql_query("select cat_num from dbtable where id = $id);

then you assign that value to $selected :

$selected = mysql_result($r, 0);

and then everything in the way i described before.

Be aware that 'selected' in echo ' selected' isn't a variable and hasn't much to do with $selected. It's just html keyword that scrolls the dropdown list to the needed position.

Clear enough?

raffael3d

rigth! its working thanks a lot for your help!