如何将PHP变量传递给SQL Select查询

Question

main.php文件中的函数

public function get_email_details($email, $table_name)
{
    $query_verify_email = "SELECT * FROM $table_name where u_email='$email'";
    $result = mysqli_query($this->connection,$query_verify_email) or die(mysql_error());
    return $result;
}

我用它来称呼它

$result_get_email_details = $Object->get_email_details($email, $table_name);
$rows=mysql_fetch_array($result_get_email_details); // It is showing me error on this line

错误:mysql_fetch_assoc()期望参数1是资源,在我评论之后在上面的行给出的对象.

请帮我解决这个SQL查询中的错误

Answer 1

那么你正在使用mysql_fetch_query......

不应该mysqli_fetch_query.注意imysql中缺少的.