Question

我有一个网站，用户可以登录和注册。现在，我希望用户更新他们的个人资料。我无法更新数据库。我的表名是“成员”。每当我提交表格时都没有任何事情发生。

<?php

if(isset($_POST["submit"])){

$value1 = $_POST['year'];
$value2 = $_POST['astatus'];
$value3 = $_POST['address'];
$value4 = $_POST['phone'];
$value5 = $_POST['email'];
$newpass = $_POST['pass'];
$newpass2 = $_POST['pass2'];
$user=$_SESSION['sess_user'];



        $conn = new mysqli('localhost', 'admin', 'password') or die(mysqli_error());
        //Select DB From database
        $db = mysqli_select_db($conn, "database") or die("Database Error");
        //Selecting database

        $update_user="UPDATE '".members."' SET year='".$value1."', astatus='".$value2."', address='".$value3."', phone='".$value4."', email='".$value5."', pass='".$newpass."' WHERE '".user."'='".$user."'";

                if(isset($_SESSION["editerror"])){
                session_destroy(editerror);
                }

                session_start(editsuccess);
                $_SESSION["editsuccess"] ="Your student profile has been successfully updated.";
                header("Location:profile.php");











}
else{
echo "error";
}


?>

Answer 1

在页面顶部开始session

<?php
session_start();
if(isset($_POST["submit"])){

$value1 = $_POST['year']; 
....

您的mysqli_query()

在哪里？

运行您的query

$result = mysqli_query($conn, $update_user) or die(mysqli_error($conn));

你的query使用这个也是错误的

$update_user="UPDATE members  SET year='".$value1."', astatus='".$value2."', address='".$value3."', phone='".$value4."', email='".$value5."', pass='".$newpass."' WHERE user ='".$user."'";

Answer 2

using (HttpClient client = new HttpClient())
{
    client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
    var response = await client.PostAsJsonAsync("https://www.googleapis.com/geolocation/v1/geolocate?key=keyvalue", "" );
    //Or client.PostAsync could be used this way:
    //var response = await client.PostAsync("https://www.googleapis.com/geolocation/v1/geolocate?key=keyvalue", new StringContent("", Encoding.UTF8, "application/json"));
    return await response.Content.ReadAsStringAsync();
}

应该是

$update_user="UPDATE '".members."' SET year='".$value1."', astatus='".$value2."', address='".$value3."', phone='".$value4."', email='".$value5."', pass='".$newpass."' WHERE '".user."'='".$user."'";

此外，您不能将这些值传递给数据库。您必须正确地转义它们以防止数据库注入。

Answer 3

注意：Alwayes使用mysqli_real_escape_string - 在字符串中转义特殊字符以在SQL语句中使用，同时考虑连接的当前字符集

http://www.w3schools.com/php/func_mysqli_real_escape_string.asp

更新时使用prepare语句

http://php.net/manual/en/mysqli.prepare.php

你忘了执行查询

    $update_user="UPDATE `members` SET `year`='".$value1."', `astatus`='".$value2."', `address`='".$value3."', `phone`='".$value4."', `email`='".$value5."', `pass`='".$newpass."' WHERE `user` ='".$user."'";   


        if ($conn->query($update_user) === TRUE) {
             session_start(editsuccess);
            $_SESSION["editsuccess"] ="Your student profile has been successfully updated.";
            header("Location:profile.php");

        } else {
           echo "Error updating record: " . $conn->error;
        }

http://www.w3schools.com/php/php_mysql_update.asp

无法更新我的数据库

3 个答案: