Dynamically creating array elements

Question

The form below on a page a.php submits to a page b.php.

<form method="post" action="b.php">
   <select name="zzz" id="zzz">
    <option value="id1" label="aaa">aaa</option>
    ..................
    ..................

On the page b.php I have the following code created manually that returns the label of the option selected from the page a.php, i said the label and not the value.

<?php

  $options = array();

  $options["id1"] = "aaa";
  $options["id2"] = "bbb";
  $options["id3"] = "ccc";

  $label = $options[$_POST['zzz']];

  echo $label;

You can create these lines

 $options["id1"] = "aaa";
 $options["id2"] = "bbb";
 $options["id3"] = "ccc";

dynamically from a select being id1, id2 .... and aaa, bbb ... the values in the database table.

Is there any way to create these lines and work as manually created? That is, instead of typing in the code id1 = "aaa" do something like $row["id"] = "$row["xxx"]

Answer 1

Despite the downvotos, I made a point of reopening this post and show a solution.

On the page a.php create the file optionsArray.php

$servidor = "localhost";
$usuario = "Usuario";
$senha = "SENHA";
$dbname = "NOME_DB";

$mysqli = mysqli_connect($servidor, $usuario, $senha, $dbname);


$query = "SELECT id_regional, regional FROM t_regional ORDER BY regional";
$resultado=$mysqli->query($query);


$filename="optionsArray.php";

if (!file_exists($filename)) {

    $linhas="";

    while ($row = mysqli_fetch_assoc($resultado)){

    $linhas .= "\n".'$options["'.$row['id_regional'].'"] = "'.$row['regional'].'";';

    }

    $linhas ="<?php \n".$linhas."\n ?>";

    file_put_contents($filename, $linhas);

}

On the page b.php include the archive optionsArray.php created on the page a.php

$options = array();

include_once("optionsArray.php");

$label = $options[$_POST['cbx_regional']];

echo $label; 

The two pages above could merge into one without problems.

Question: why you want to create this file dynamically?

Answer: Anderson Carlos Woss' answer is certainly the most obvious.

However I do not know if I understood well the comment in the accepted reply of this post target HTML option label Instead of its value in PHP? what it says (google translation) _{"Okay, this works in a way. BUT, about whether there are 100 options on the page? Do I have to add $options ["music / three"] = "somename"; $ Options ["music / four"] = "somename"; For each one? Or is there another way to do this?"}.

That is the reason for the question and that is the reason for my reply.

Summary of the opera: The usefulness of this only the American could explain :)

NOTE: It is not clear where this information would come from, but I believe they are from a database.

Answer 2

Supposing we have the table browsers in the database:

id | name
---+------------------
1  + Google Chrome
---+------------------
2  + Firefox
---+------------------
3  + Opera
---+------------------
4  + Safari
---+------------------
5  + Internet Explorer
---+------------------

We can generate the following HTML with PHP:

<?php

//...

$result = mysqli_query($mysqli, "SELECT `id`, `name` FROM `browsers`");

echo "<select name='browser'>";

while ($row = mysqli_fetch_assoc($result)) {
    echo "<option value='{$row['id']}' label='{$row['name']}'>{$row['name']}</option>";
}

echo "</select>";

The HTML output will be:

<select name='browser'>
    <option value='1' label='Google Chrome'>Google Chrome</option>
    <option value='2' label='Firefox'>Firefox</option>
    <option value='3' label='Opera'>Opera</option>
    <option value='4' label='Safari'>Safari</option>
    <option value='5' label='Internet Explorer'>Internet Explorer</option>
</select>

When the form is submitted, in PHP you can recover the attribute value value selected. This value is the id of the selected table record browsers, then, to get the name of the selected browser, just select the respective database record:

<?php

// ...

if ($stmt = mysqli_prepare($mysqli, "SELECT `id`, `name` FROM `browsers` WHERE `id` = ?")) {

    mysqli_stmt_bind_param($stmt, "i", $_POST["browser"]);
    mysqli_stmt_execute($stmt);
    mysqli_stmt_bind_result($stmt, $id, $name);
    mysqli_stmt_fetch($stmt);

    echo "O navegador selecionado foi: ", $name;

}

The complete code tested was:

<form action="" method="post">
    <select name="browser">
        <?php

            $mysqli = mysqli_connect("localhost", "root", "", "sopt");

            $result = mysqli_query($mysqli, "SELECT `id`, `name` FROM `browsers`");

            while ($row = mysqli_fetch_assoc($result)) {
                echo "<option value='{$row['id']}' label='{$row['name']}'>{$row['name']}</option>";
            }

        ?>
    </select>
    <button>Enviar</button>
</form>

<?php

if ($_SERVER["REQUEST_METHOD"] == "POST") {

    if ($stmt = mysqli_prepare($mysqli, "SELECT `id`, `name` FROM `browsers` WHERE `id` = ?")) {

        mysqli_stmt_bind_param($stmt, "i", $_POST["browser"]);

        mysqli_stmt_execute($stmt);

        mysqli_stmt_bind_result($stmt, $id, $name);

        mysqli_stmt_fetch($stmt);

        echo "O navegador selecionado foi: ", $name;

    }

}

?>