PHP选择登录错误

Question

我只是 PHP 的新手，我需要创建一个 php 登录表单，但它有一个语法错误，提示 Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\login.php 第 18 行
也许你可以在这个 tnx 上帮助我!

<?php

$con=mysqli_connect("localhost","root","","reservation");
if (mysqli_connect_errno())
  {
  echo "Failed to connect to Database: " . mysqli_connect_error();
  }
  else
  {
  }

if ($_POST["SUBMIT"]="SUBMIT")
{
$username = $_POST['username'];
$password =$_POST['password'];
$sql ="SELECT * FROM usersaccount where='$username' and password ='$password'";
$result=mysqli_query($con,$sql);
$count=mysqli_num_rows($result);
if($count==1)
{
echo "success";
}
//mysqli_close($con);
}
?>

Answer 1

你的SELECT语句是错误的……应该是

SELECT * FROM `usersaccount` WHERE `username` ='$username' and `password` ='$password'";
                                   ---------- // Here.. Added the column name

您没有为 username 字段提供列名。那是你的问题。由于您要将结果与条件匹配，因此您也需要 WHERE 子句!

^{免责声明:停止使用 mysql_* 函数，因为它们已被弃用。改用 MySQLi 或 PDO。}