未定义变量：第14行C:\xampp\htdocs\CRUD\login.php中的con

您可以将其设置为函数中的全局值，也可以将其传递到参数中以访问它

<?php

    $con = mysqli_connect('localhost', 'root', '', 'crud_db');
    if (mysqli_connect_errno()) 
    {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    function signin() 
    {
      $session_start;
      if (!empty($_POST['email'])) 
      {
        echo "not empty";
        $query = "SELECT * FROM 'employee' where email= '$_POST ['email']' AND password=   '$_POST ['pwd']'";
        $result = mysqli_query($con, $query);
        $row = mysqli_fetch_array($result);

        if (!empty($row['email']) AND !empty($row['password'])) {
            echo "successfully login";
        } else {
            echo "login fail";
      }
    }

    }

    if (isset($_POST['submit'])) 
    {
      signin();
    }
?> 

---

与问题无关，但：

$con = mysqli_connect('localhost', 'root', '', 'crud_db');
if (mysqli_connect_errno()) 
{
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

function signin($con) 
{
  session_start();
  if (!empty($_POST['email'])) 
  {
    echo "not empty";
    $query = "SELECT * FROM employee where email= '$_POST ['email']' AND password=   '$_POST ['pwd']'";
    $result = mysqli_query($con, $query);
    $row = mysqli_fetch_array($result);

    if (!empty($row['email']) AND !empty($row['password'])) {
        echo "successfully login";
    } else {
        echo "login fail";
  }
}

}

if (isset($_POST['submit'])) 
{
  signin($con);
}
?> 

这是行不通的

 $query = "SELECT * FROM 'employee' where email= '$_POST ['email']' AND password=   '$_POST ['pwd']'";

将工作，但它仍然是sql注入易受攻击的

---

读

用不同的php脚本编写数据库端和函数端，就像这样，它可能会工作

$query = "SELECT * FROM 'employee' where email= '{$_POST ['email']}' AND password=   '{$_POST['pwd']}'";

function.php
请提供代码请清理代码，太可怕了+1查找错误，但您应该建议将$con作为参数传递，而不是全局-1：在函数中将其声明为全局变量可能会解决问题，但这是一个混乱的解决方案。更好的方法是将其作为参数传递。有时，全局变量是正确的解决方案，但在这种情况下，不应依赖于在全局范围内初始化变量的假设-应将其显式声明为全局变量，即使预期范围是全局的，多亏了veerendra，它仍然有效。你能帮我解决这个错误吗？mysqli_fetch_array（）希望参数1是mysqli_result，C:\xampp\htdocs\CRUD\login.php中的第行中给出了null15@AshuChauhan我正在编辑答案，如果答案对您有帮助，请接受：）我认为错误应该现在解决，因为您现在正在获得连接，所以查询将返回result@symcbean是 啊我也不喜欢环球队，但是-1有点苛刻，尤其是当他解决了问题的时候
<?php 
$con = mysqli_connect('localhost', 'root', '', 'crud_db');
?>

function.php
<?php
include('database.php');
function signin() 
{
global $con;
  session_start();
  if (!empty($_POST['email'])) 
  {
    echo "not empty";
    $query = "SELECT * FROM employee where email= '$_POST ['email']' AND password=   '$_POST ['pwd']'";
    $result = mysqli_query($con, $query);
    $row = mysqli_fetch_array($result);

    if (!empty($row['email']) AND !empty($row['password'])) {
        echo "successfully login";
    } else {
        echo "login fail";
  }
}

}


?>