PHP function only works on the first call

Question

I am calling the same php function, in which I pass as parameter a database result, twice in the script but it only works in the first call.

I realized that if I call the function that returns the query result and pass to another variable that in turn is passed as parameter to the function that previously did not work, it goes to work.

I know I’m missing some point of php working, I’d like a hand in this.

Remembering that I am calling the same function because I need a select exactly like the previous one, does anyone suggest any more optimized way to do this? Because the function carries approximately 13,000 bank records which makes the page take a long time to load.

<!--TIMES-->
<div class="page-header">
    <h4>Times</h4>
        <hr>
</div>
<?php
    $registro = lista_times();
    //$registro2 = lista_times();
?>
<div style="padding: 20px; align-content: center" class="row" id="times-row">
    <div class="col-sm-6">
        <div class="form-group">
            <label for="time_casa" class="col-sm-2 control-label">Casa</label>
            <div class="col-sm-10">
                <?php //var_dump($registro); ?>
                <select style="width: 300px" class="form-control chosen-select" id="time_casa" name="time_casa" data-error="Por favor, selecione o time de casa." required>
                    <?php echo preenche_time_combo($registro); ?>
                </select>
            <div class="help-block with-errors"></div>
        </div>
    </div>
</div>

<div class="col-sm-6">
    <div class="form-group">
        <?php //var_dump($registro); ?>
        <label for="time_fora" class="col-sm-2 control-label">Fora</label>
        <div class="col-sm-10">
            <select style="width: 300px" class="form-control chosen-select" id="time_fora" name="time_fora" data-error="Por favor, selecione o time de fora." required>
                <?php echo preenche_time_combo($registro); ?>
            </select>
        <div class="help-block with-errors"></div>
    </div>
</div>
</div>
</div>
<!--TIMES-->

lista_times() executes the query and returns the result and preenche_time_combo() returns a string with html code with an option for each element recovered from the database.

//Lista todos os jogos ativos
function lista_times()
{
    $link = conectar();
    $query = "SELECT tb_time.id as id_time, tb_time.nome_time, tb_campeonato.nome_camp
              FROM tb_campeonato, tb_time
              WHERE tb_time.tb_campeonato_id = tb_campeonato.id";
    $result = mysqli_query($link, $query) or die(print_r(mysqli_error()));

    return $result;

}

function preenche_time_combo($result)
{
    $header_atual="";

    $html="";

    while ($registro = mysqli_fetch_assoc($result)) {
        if($registro['nome_camp'] != $header_atual){
            if($header_atual != ""){
                $html .= "</optgroup>";
            }

            $html .= "<optgroup label='".$registro['nome_camp']."'>";
            $header_atual = $registro['nome_camp'];
        }

        $html .= "<option value='" . $registro['id_time'] . "'>" . $registro['nome_time'] . "</option>";
    }

    $html .= "</optgroup>";

    return $html;

    exit();
}

Answer 1

In preenche_time_combo() every call from mysqli_fetch_assoc($result) is altering/emptying the variable $registro out of function (change by reference), so the second call does not work.

The simplest way to solve this is to store the options already generated in a variable and then print it inside the <select>

Change the call:

<?php
   $registro = lista_times();

To:

<?php
   $options = preenche_time_combo(lista_times());

And finally (I removed the style to simplify the example):

<select>
   <?php echo $options;?>
</select>