php mysql中的数据更新问题

Question

我试图创建一个更新系统，但我不知道为什么我的代码不能工作。我的代码是：

<?php 
 include "dbconnection.php";

  if(isset($_POST['submit'])){
        $residency_status = $_POST['residency_status'];
  $query="UPDATE profile SET 
           residential_status='$residency_status'
  WHERE id = '".user_id."' 

  ";
    }
    

    header('location:profile.php');
    ?>

我从profile.php页面添加了一些代码：

<?php 
session_start();
include_once ('dbconnection.php');


 if(!isset($_SESSION['logged_in'])){
    header("Location: login.php");
    die();
} 
    $usrname=$_SESSION['username']; 
    $pasword=$_SESSION['password'];
    $user_id= get_user_id($usrname, $pasword);   
    
   while($us_id= $user_id->fetch_assoc()) :
    $collected_id=$us_id['id'];
    $resi_status=$us_id['residential_status'];
 endwhile;

  ?>

表单的代码是:我尽量保持代码简短，以避免

   <form class="form-group row" action="get_update.php" method="POST" 
    enctype="multipart/form-data">
       <h4>Residential Status </h4>
       
     <div class="form-group">          
       <input type="radio" name="residency_status" value="yes" class="" 
       id=""  
    <?php 
       if ($resi_status == "yes") {  ?> checked  <?php   }
     ?>
    >yes      
    <input type="radio" name="residency_status" value="No" class="" id="" 
     <?php 
       if ($resi_status == "no" || $resi_status == "") {     ?>    checked   
       <?php     } ?> >no       
  </div>
  <br>

谢谢你。

Answer 1

您错过了mysqli_query.Your代码应该是

 <?php 
 include "dbconnection.php";

  if(isset($_POST['submit'])){
        $residency_status = $_POST['residency_status'];
  $query="UPDATE profile SET 
           residential_status='$residency_status'
  WHERE id = '".user_id."'"; 
  mysqli_query($conn,$query); // $conn must be as per your `mysqli_connect` variable
    }
    header('location:profile.php');
    ?>