PHP function only works on the first call

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PHP function only works on the first call

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#1 by (3 votes)

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

I noticed that if I call the function that returns the result of the query and move it to another variable that is passed as a parameter to the function that did not work before, it starts to work.

I know I'm letting some point of php work, I'd like a hand in it.

Remembering that I'm calling the same function because I need a select exactly the same as the above, does anyone suggest any more optimized way to do this? Because the function loads 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 the html code with an option for each element retrieved 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();
}

php mysql mysqli

asked by anonymous 12.08.2016 / 22:37

1 answer

C # - Reference a variable from another method [closed] Calculation in a DataGrid. W#

score 3 · Accepted Answer

In% with% each call of preenche_time_combo() is changing / emptying the%% variable outside the function (change by reference), so the second call does not work.

The simplest way to solve this is to store the already generated options in a variable and then print it within mysqli_fetch_assoc($result)

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>