Problems with mysqli_result as array

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Problems with mysqli_result as array

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#1 by (1 votes)
#2 by (1 votes)

1

I'm trying to make the page display the documents sent by the user, but it always returns me the same error

Fatal error: Can not use object of type mysqli_result as array in> C: \ wamp64 \ www \ Project-EscolWeb \ student-query.php on line 158

The code on the line is this

<h1>Documentos:</h1>
<?php $arrayDadosDocumentos = ObtemDocumentos($rg);
$idTipo = $arrayDadosDocumentos[2];?>

I've already researched and always find the same solution, speaking to use '', but I do not know how to apply it in my code.

At that moment, thank you for your attention.

Q: If you need more information, please let me know.

Code of ObtemDocumento()

function ObtemDocumentos($rg){
    include("conexao.php");

    $SQL ="SELECT D.* FROM documento D JOIN alunodocumento AD WHERE AD.RG = ".$rg." AND D.IdCaixa = AD.IdCaixa AND D.IdLote = AD.IdLote AND D.idTipoDoc = AD.idTipoDoc ORDER BY D.idTipoDoc;" ;

    $resultado = mysqli_query($conexao, $SQL);

    if (!$resultado) {
        $mensagem_erro  = 'Erro de sintaxe na query: ' . mysqli_error() . "<br>";
        $mensagem_erro .= 'SQL executado: ' . $SQL;
        $destino = 'http://localhost/Projeto-escolweb/php/erro.php?msg='.$mensagem_erro;
        header("Location: $destino");
        exit;
    }

   return $resultado;

php mysqli

asked by anonymous 13.10.2016 / 15:47

2 answers

1

$query = ObtemDocumentos($rg);
while($dados = mysqli_fetch_array($query)){
    $idTipo = dados["idTipo"];
}
//se eu ajudei por favor vote na resposta

13.10.2016 / 15:53

How to do when I click button, open link in another window How to enter several data received from the same field?

score 1 · Accepted Answer

It may be more practical to make the function return the array with the data, then add the mysqli_fetch_assoc() call in the while and at the end of it returns the variable ( $arr ) containing the query data:

$resultado = mysqli_query($conexao, $SQL);
if (!$resultado) {
    $mensagem_erro  = 'Erro de sintaxe na query: ' . mysqli_error() . "<br>";
    $mensagem_erro .= 'SQL executado: ' . $SQL;
    $destino = 'http://localhost/Projeto-escolweb/php/erro.php?msg='.$mensagem_erro;
    header("Location: $destino");
    exit;
}

$arr = array();
while($row = mysqli_fetch_assoc($resultado)){
    $arr[] = $row;
}

return $arr;