Question

当我要显示流派表中的名称时出现问题...如何调用此功能以及如何从类型中获取名称...当我使用echo时，我得到的是未定义的$ data ...

public function getGenreName(){
    $query = mysqli_query($this->con, "SELECT * FROM genres WHERE id='$this->genre'");
    $data = mysqli_fetch_array($query);
    return $data['name'];
  }

Answer 1

您不是要检查错误，我认为您在查询行中有一个

public function getGenreName(){
    $query = mysqli_query($this->con, "SELECT * FROM genres WHERE id='{$this->genre}'");

    if ( ! $query ) {
        echo $this->con->error;
        return false;
    }
    $data = mysqli_fetch_array($query);
    return $data['name'];
}

您可能会更有效率，只需选择name，这就是您似乎感兴趣的全部内容。

public function getGenreName(){
    $query = mysqli_query($this->con, "SELECT name FROM genres WHERE id='{$this->genre}'");

    if ( ! $query ) {
        echo $this->con->error;
        return false;
    }
    $data = mysqli_fetch_array($query);
    return $data['name'];
}

虽然这仍然包含SQL Injection Attack甚至if you are escaping inputs, its not safe!的可能性在MYSQLI_或PDO API中使用prepared parameterized statements

所以您应该确实在做

public function getGenreName(){
    $stmt = $this->con->prepare("SELECT name 
                                FROM genres 
                                WHERE id=?");
    $stmt->bind_param('s', $this->genre );
    $query = $stmt->execute();

    if ( ! $query ) {
        echo $this->con->error;
        return false;
    }
    $result = $stmt->get_result();
    $result->fetch_array(MYSQLI_NUM);
    return $result[0];
}

如何从MySQL获取特定列的数据？

1 个答案: