get value from select inside $post request

Question

Could someone help my with an issue I have? I have made an select list with different options, based on the option the user did choice I want to run a certain function so i can later show the data which i have already saved in my database.

However how do I do this? right now I have been working with: if(!empty($_POST["option1"])) to check which option is selected but the code will never get inside those if statements. the code stops after: if(!empty($_POST["confirmOption"]))

<form method="post" action="index.php"> 
            <select id="list" name="list">
              <option selected><--select option--></option>
              <option value="option1">test1</option>
              <option value="option2">test2</option>
              <option value="option3">test3</option>
              <option value="option4">test4</option>
            </select>

            <button name="confirmOption">confirm option</button
          </form>

if($_SERVER["REQUEST_METHOD"] == "POST"){
if(!empty($_POST["confirmOption"])){
      if(!empty($_POST["option1"])){
        $conn = openDatabase();
        $query = $conn->prepare("SELECT * FROM choices WHERE status = 'option1'");
        $query->execute();
        return $query->fetchAll();
      }elseif(!empty($_POST["option2"])){
        $conn = openDatabase();
        $query = $conn->prepare("SELECT * FROM choices WHERE status = 'option2'");
        $query->execute();
        return $query->fetchAll();
      }elseif(!empty($_POST["option3"])){
        $conn = openDatabase();
        $query = $conn->prepare("SELECT * FROM choices WHERE status = 'option3'");
        $query->execute();
        return $query->fetchAll();
      }elseif(!empty($_POST["option4"])){
        $conn = openDatabase();
        $query = $conn->prepare("SELECT * FROM choices WHERE status = 'option4'");
        $query->execute();
        return $query->fetchAll();
      }
    }
  }

In the code select name is `list`, you needs get value of `$_POST['list']` — Lessmore, Mar 08 '22 at 18:31
The name of the select is `list`. `$_POST['list']` will have the value of option1/2/3/4. — aynber, Mar 08 '22 at 18:31
`confirmOption` is a button, which will not be an element in POST. — devlin carnate, Mar 08 '22 at 18:34
so i need to work with '$_POST['list']' instead of '$_POST['confirmOption']'. but how can i check the selected value? I assume '!empty($_POST["option1"]' should not be a $post again then? — , Mar 08 '22 at 18:39
[See here](https://stackoverflow.com/questions/17139501/using-post-to-get-select-option-value-from-html) for ways to get the selected value. — devlin carnate, Mar 08 '22 at 18:42
Also, side note...you might consider using some debugging techniques in your php. Like, in this particular case, using `var_dump()` on your POST would show you exactly what is being POSTed and its structure. You can use that info to help you figure out what you need to access in POST and how you can access it. — devlin carnate, Mar 08 '22 at 18:48
also consider refactoring right away. The options are all the same apart from the value of the select. That isn't duplicate code that is quadruple code ;) write a method, which accepts the value as a parameter. — Calamity Jane, Mar 09 '22 at 18:56

Christian Wyrwas · Accepted Answer · 2022-03-23T20:31:11.603

You need to access it via $_POST['list']. For instance, if I were using your form and I chose the second list item (not the first one without a value), then the value of $_POST['list'] would be set to 'option1'.

You'll need to modify your if condition. It should look something like this:

if ( !empty($_POST["confirmOption"]) && isset($_POST['list'] ){

    $selectedItem = $_POST['list'];
    $conn = openDatabase();

    if ($selectedItem == 'option1'){
      $query = $conn->prepare("SELECT * FROM choices WHERE status = 'option1'");
    } elseif ($selectedItem = 'option2'){)){
      $query = $conn->prepare("SELECT * FROM choices WHERE status = 'option2'");
    } elseif ($selectedItem == 'option3'){
      $query = $conn->prepare("SELECT * FROM choices WHERE status = 'option3'");
    } elseif ($selectedItem == 'option4'){
      $query = $conn->prepare("SELECT * FROM choices WHERE status = 'option4'");
    } else {
        return false
    }

    $query->execute();
    return $query->fetchAll();
}

I removed some of the duplication, because if an item is selected we know that we'll have to open the database, and ultimately will need to execute and return the query. You could even probably remove the duplicated lines where you're setting the query statement multiple times.

To clean it up even further - if the only thing that will be changing is the select value and you know the input is safe you could reliably do something like $conn->prepare("SELECT * FROM choices WHERE status = {$selectedItem}");

get value from select inside $post request

1 Answers1