我有一个项目,用户可以编辑自己的个人资料。我目前有一个登录/注册页面。现在是用户编辑页面的时候了。该网站与会话一起工作,因此,每当他们回来时,他们都会在导航栏中看到我的个人资料。
我已经用php测试了所有的东西,以更新帐户。
我目前有两个文件,其中更新配置文件脚本的工作。 1 is the profile.php, the other in includes/update.inc.php
在 includes/login.inc.php
我已经安排了一个会议。
if ($pwdCheck == false) {
header("Location: ../index.php?error=wrongpassword");
exit();
}
else if ($pwdCheck == true) {
session_start();
$_SESSION['userId'] = $row['idUsers'];
$_SESSION['userUid'] = $row['uidUsers'];
header("Location: ../index.php?logging=succes");
exit();
是的,它来自includes/login.inc.php。
现在,当你去你的个人资料,它检查 if $_SESSION['userId'] == true.
下面是:
这是我的profile.php
<?php
require "header.php";
require "includes/dbh.inc.php";
$sessionkk = $_SESSION['userId'];
error_reporting(E_ALL); ini_set('display_errors', 1);
$query = "SELECT idUsers, uidUsers, emailUsers,pwdUsers,Voornaam,Tussenvoegsel,Achternaam,Schooljaar,School,Opleiding,Niveau,Recht,Taal
,
printerA,printerB,printerC FROM users WHERE idUsers = '$sessionkk'";
$result = $conn->query($query) or die($conn->error);
if(mysqli_num_rows($result) > 0){
while ($row = mysqli_fetch_assoc($result)){
$id = $row['idUsers'];
$username = $row['uidUsers'];
$email = $row['emailUsers'];
$password = $row['pwdUsers'];
$voornaam = $row['Voornaam'];
$tussenvoegsel = $row['Tussenvoegsel'];
$achternaam = $row['Achternaam'];
$Schooljaar = $row['Schooljaar'];
$School = $row['School'];
$Opleiding = $row['Opleiding'];
$niveau = $row['Niveau'];
$Taal = $row['Taal'];
$printera = $row['printerA'];
$printerb = $row['printerB'];
$printerc = $row['printerC'];
}
}
?>
<div class="adminform">
<h2>gegevens aanpassen</h2><br/>
<form action="includes/update.inc.php" method="POST">
<input type="text" name="username" placeholder="Username"
value="
<?php if(isset($_GET['username'])){echo $_GET['username'];}?>"><br/><br/>
<input type="text" name="email" placeholder="E-mail" value="<?php
if(isset($_GET['email'])){echo $_GET['email'];}?>"><br/><br/>
<input type="text" name="" placeholder="Schooljaar" value="<?php
if(isset($_GET['Schooljaar'])){echo $_GET['Schooljaar'];}?>"><br/><br/>
<input type="text" name="email" placeholder="E-mail" value="<?php
if(isset($_GET['School'])){echo $_GET['School'];}?>"><br/><br/>
<button type="submit" name="adminupdate">Aanpassen</button>
</form>
<?php
echo "" . $id;
?>
</div>
<?php
?>
在这里我总结一切。表单名称等。
你也可以看到我是如何 $sessionkk = $_SESSION['userId'];
$sessionkk与
下面是include/update.inc.php
<?php
session_start();
require "dbh.inc.php";
if(isset($_POST['adminupdate'])){
$sql = "UPDATE users SET uidUsers= '?', emailUsers = '?', Schooljaar =
'?',
School = '?' WHERE idUsers = '.$sessionkk.'";
$stmt = mysqli_stmt_init($conn);
if(!mysqli_stmt_prepare($stmt, $sql)){
header("Location: ../profile.php?error=sqlerror");
exit();
} else{
mysqli_stmt_bind_param($stmt, "ssss", $username, $email,
$Schooljaar, $School);
mysqli_stmt_execute($stmt);
header("Location: ../profile.php?update=succes");
exit();
}
}
?>
我使用数据库中的id尝试了$sql查询,因此如下所示:
$sql = "UPDATE users SET uidUsers= '?', emailUsers = '?', Schooljaar = '?',
School = '?' WHERE idUsers = '4'";
这样就行了。但是我想从登录的用户那里获取id。对不起,代码是如此混乱,我不知道如何上传代码在这里正确。
谢谢您
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