how do I fix this error "Notice: Trying to get property of non-object"

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how do I fix this error "Notice: Trying to get property of non-object"

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#1 by (4 votes)

1

Notice: Trying to get property of non-object in C: \ xampp \ htdocs \ fairs-master \ fairs.php on line 84

    <?php
                        $rsFeriasMarcada = 
                        mysqli_query($link, "SELECT * FROM tab_ferias WHERE num_id_colaborador = $ferias->id_colaborador");

                        $feriasMarcada = mysqli_fetch_object($rsFeriasMarcada);

   linea 84---->        if ($feriasMarcada->num_id_colaborador == 0) {
                        ?>
                        <a href="marcar-ferias.php?id=<?php echo $ferias->id_colaborador; ?>">Marcar</a>
                        <?php
                        } else {                        
                            $dataFeriasMarcada = strtotime($feriasMarcada->dt_fim_ferias);
                            $dataAtual = strtotime(date('Y-m-d'));                      
                            if ($dataFeriasMarcada < $dataAtual) {

                            mysqli_query($link, "UPDATE tab_colaborador SET dt_ultferias_inicio = '$feriasMarcada->dt_inicio_ferias', dt_ultferias_fim = '$feriasMarcada->dt_fim_ferias' WHERE id_colaborador = $feriasMarcada->num_id_colaborador");
                            mysqli_query($link, "DELETE FROM tab_ferias WHERE num_id_colaborador = $ferias->id_colaborador");
                            }

                            echo DataBr($feriasMarcada->dt_inicio_ferias) . " &agrave; " . DataBr($feriasMarcada->dt_fim_ferias) . " <a href='ferias.php?deletarFerias=" . $ferias->id_colaborador . "'><img src='img/ico_deletar.png' width='9' height='9' /></a>";
                        }
                        ?>

mysql php mysqli xampp

asked by Eric Brandenberg 28.08.2017 в 01:19

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1 answer

Change the opacity and styles of different divs when the mouse passed over Search records by criterion shows nothing

score 4 · Accepted Answer

If the query does not get results, return a null:

Return values mysqli_fetch_object () :

Returns an object with the string properties that correspond to the row obtained or NULL if there are no more rows in the result set .

Before entering the if, you could do a check like:

if(!is_null($feriasMarcada )){...}