PHP Dropdown List variable

Question

I am trying to have a drop down menu where you can choose from 3 options. I've got the following code, but I can't seem to get my third option to work. Any ideas?

        $_POST['faction'] = ($_POST['faction'] == "s")? "S" : "K";

I want to add a third option which is "R" and I can't get it to work.

Ideas please?

What do you mean dropdown? PHP is executed server side. Why not just repeatedly use if-else? — Harry
– Harry, Commented Dec 25, 2013 at 18:43

vee · Accepted Answer · 2013-12-25 18:49:17Z

1

Although adding another ternary statement to your statement is possible I think it will affect the code readability.

Using array will make this simpler.

$option_array = array('r' => 'R', 's' => 'S', 'k' => 'K');

if (array_key_exists($_POST['faction'])) { 
    $_POST['faction'] = $option_array[$_POST['faction']]
}

You could choose to add the array_key_exists to make sure you are not updating $_POST['faction'] if $_POST['faction'] contains other values, but this is your choice.

Another approach similar to yours with multiple ternary operations:

$_POST['faction'] = ($_POST['faction'] == "s") ? ($_POST['faction'] == "r" ? "R" : "S") : "K";

answered Dec 25, 2013 at 18:49

vee

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2 Comments

Richard Evans Over a year ago

What it is that in my table, I've got the 3 options and from a drop down menu in a html code, it's selecting that option and then displaying it as information. I tried your code with the multiple ternary operations but that didn't work.

vee Over a year ago

Please post output of var_dump($_POST)

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