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php - 如何允许登录的用户更新/编辑他们的个人资料设置/信息

转载 作者:太空宇宙 更新时间:2023-11-03 12:24:21 25 4
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手头的问题:

我如何创建 php 代码来让登录我网站的用户编辑/更新他们的个人资料设置/信息?

我有 1 个部分可以正常工作,用户可以更改他们的密码,但是,当涉及到允许登录的用户编辑/更新他们的其他设置时,我不知道从哪里开始,例如:

(1) 昵称,(2) 国家,(3) 出生日期,(4) 性别,(5) 座右铭和(6)简介


我将在下面提供我拥有的用于更改密码的 php 和 html 代码,但我知道我需要更多的代码来让用户更改/编辑/更新他们的其他信息。我尝试使用下面的内容作为引用来为其他信息创建 php 代码,但它没有用,所以我什至不知道从哪里开始!任何帮助将不胜感激...

PHP引用代码:

if($_POST['submit']=='Change')
{
$err = array();
if(!$_POST['password1'] || !$_POST['passwordnew1'])
$err[] = 'All the fields must be filled in!';
if(!count($err))
{
$_POST['password1'] = mysql_real_escape_string($_POST['password1']);
$_POST['passwordnew1'] = mysql_real_escape_string($_POST['passwordnew1']);
$row = mysql_fetch_assoc(mysql_query("SELECT id,username FROM members WHERE username='{$_SESSION['username']}' AND pass='".md5($_POST['password1'])."'"));
if($row['username'])
{
$querynewpass = "UPDATE members SET pass='".md5($_POST['passwordnew1'])."' WHERE username='{$_SESSION['username']}'";
$result = mysql_query($querynewpass) or die(mysql_error());
$_SESSION['msg']['passwordchange-success']='* You have successfully changed your password!';
}
else $err[]='Wrong password to start with!';
}
if($err)
$_SESSION['msg']['passwordchange-err'] = implode('<br />',$err);
header("Location: members.php?id=" . $_SESSION['username']);
exit;
}

HTML 引用代码:

<form action="" method="post"> 
<label class="grey" for="password1">Current Password:</label>
<input class="field" type="password" name="password1" id="password1" value="" size="23" />
<label class="grey" for="password">New Password:</label>
<input class="field" type="password" name="passwordnew1" id="passwordnew1" size="23" />
<input type="submit" name="submit" value="Change" class="bt_register" style="margin-left: 382px;" />
<div class="clear"></div>
<?php
if($_SESSION['msg']['passwordchange-err'])
{
echo '<div class="err">'.$_SESSION['msg']['passwordchange-err'].'</div>';
unset($_SESSION['msg']['passwordchange-err']);
}
if($_SESSION['msg']['passwordchange-success'])
{
echo '<div class="success">'.$_SESSION['msg']['passwordchange-success'].'</div>';
unset($_SESSION['msg']['passwordchange-success']);
}
?>
</form>

那么我将如何创建 php 代码以使用户能够从我上面提供的数字列表 (1-6) 编辑/更新他们自己的配置文件设置/信息?

而且我知道使用 mysqli/pdo 是更好的选择,但不幸的是,此时我需要为这个项目使用旧的已弃用的 mysql_* 东西......

如果您需要更多信息,请告诉我;)

编辑:附加问题,

我也假设我也需要为每一列创建变量,例如:

$昵称 = $_POST['昵称'];

$country = $_POST['country'];

等等...还是不正确?

重新编辑:

这样的事情是否适用?

$id = $_SESSION['id'];
if ($_POST['country']) {
$country = $_POST['country'];
$nickname = $_POST['nickname'];
$DOB = $_POST['DOB'];
$gender = $_POST['gender'];
$motto = $_POST['motto'];
$bio = $_POST['bio'];
$sql = mysql_query("UPDATE members SET country='$country', nickname='$nickname', DOB='$DOB', gender='$gender', motto='$motto', bio='$bio' WHERE id='$id'");

exit;
}

$sql = mysql_query("SELECT * FROM members WHERE id='$id' LIMIT 1");
while($row = mysql_fetch_array($sql)){
$country = $row["country"];
$nickname = $row["nickname"];
$DOB = $row["DOB"];
$gender = $row["gender"];
$motto = $row["motto"];
$bio = $row["bio"];
}

还是我离题太远了?

最佳答案

简短版;)

HTML 文件:

<form action="./change.php" method="post"> 
Nickname: <input type="text" name="nickname"><br />
Country: <input type="text" name="country"><br />
Date of birth: <input type="text" name="date_of_birth"><br />
Gender: <input type="text" name="gender"><br />
Motto: <input type="text" name="motto"><br />
Bio: <input type="text" name="bio"><br />
<input type="submit" value="Submit">
</form>

改变.php:

<?php
function filter($date)
{
return trim(htmlspecialchars($date));
}

$nickname = filter($_POST['nickname'])
$country = filter($_POST['country'])
$date_of_birth = filter($_POST['date_of_birth'])
$gender = filter($_POST['gender'])
$motto = filter($_POST['motto'])
$bio = filter($_POST['bio'])

if (isUserLogIn)
{
//SQL update query
}

?>

关于php - 如何允许登录的用户更新/编辑他们的个人资料设置/信息,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18344533/

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