我試圖創建一個更新系統,但我不知道為什么我的代碼不起作用。我的代碼是:<?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'); ?>我從 profile.php 頁面添加一些代碼:<?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; ?>表單的代碼是:我試圖保持代碼簡短以避免 <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>謝謝 。
1 回答

斯蒂芬大帝
TA貢獻1827條經驗 獲得超8個贊
你錯過了。mysqli_query你的代碼應該是
<?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');
?>
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