调用表时sql为空

Question

我在SQL + PHP中有一个错误，我在generos表中带了一个带有特定ID的行，当我抛出一个带有信息错误的var_dump时，我得到了以下错误：

array(3) { [0]=> string(5) "00000" [1]=> NULL [2]=> NULL }

function getGEN($Genero_ID)
{
    $query = $this->db->prepare("SELECT * FROM generos WHERE Genero_ID= ?");

    $query->execute(array($Genero_ID));

    $genero = $query->fetch(PDO::FETCH_OBJ);

    var_dump($query->errorInfo());

    return $genero;
}

这是我的表单，我要在这里显示数据，然后编辑它。

<form class="form-alta" action="editadogen" method="post">
   <input type="hidden" name="Genero_ID" value="{$genero->Genero_ID}">
   <div class="form-group row margin-15px">
      <label for="autor" class="col-sm-2 col-form-label">Genero:</label>
      <div class="col-sm-10">
         <input type="text" class="form-control" name="Genero" value="{$genero->Genero}">
      </div>
   </div>
   <div class="form-group row margin-15px">
      <label for="autor" class="col-sm-2 col-form-label">Edad:</label>
      <div class="col-sm-10">
         <input type="text" class="form-control" name="Edad" value="{$genero->Edad}" >
      </div>
   </div>
   <div class="form-group row margin-15px">
      <div class="col-sm-10  btn-sub-center">
         <button type="submit" class="btn btn-primary"  id="submit-create-libro">Editar Genero</button>
      </div>
   </div>
</form>

Answer 1

试着：

function getGEN($Genero_ID)
{
    $query = $this->db->prepare("SELECT * FROM generos WHERE Genero_ID = :Genero_ID;");

    $query->execute(array('Genero_ID' => $Genero_ID), array(PDO::ATTR_CURSOR => PDO::CURSOR_FWDONLY));
    $data = $query->fethAll();

    var_dump($data);

    return $data;
}