手头的问题:
如何创建php代码以让登录我网站的用户编辑/更新其个人资料设置/信息?
我有1个部分正常工作,用户可以更改密码,但是,在允许登录的用户编辑/更新其他设置时,不知道从哪里开始,如:
(1)昵称, (2)国家/地区, (3)出生日期, (4)性别, (5)座右铭和 (6) bio
我将在下面提供我正在使用的用于更改密码的php和html代码,但我知道我需要更多信息才能让用户更改/编辑/更新其他信息。我尝试使用下面的内容作为参考来创建其他信息的PHP代码,但它没有用,所以我不知道从哪里开始!任何帮助将不胜感激......
PHP参考代码:
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=" . $_SESSION['username']);
exit;
}
HTML参考代码:
<form action="" method="post">
<label class="grey" for="password1">Current Password:</label>
<input class="field" type="password" name="password1" id="password1" value="" size="23" />
<label class="grey" for="password">New Password:</label>
<input class="field" type="password" name="passwordnew1" id="passwordnew1" size="23" />
<input type="submit" name="submit" value="Change" class="bt_register" style="margin-left: 382px;" />
<div class="clear"></div>
<?php
if($_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)中编辑/更新他们自己的配置文件设置/信息?
我知道使用mysqli / pdo是一个更好的替代方法,但是我很遗憾地需要在这个时候使用旧的弃用的mysql_ *东西...
如果您需要更多信息,请告诉我们;)
修改: 附加问题,
我也假设我需要为每个列创建变量,例如:
$ nickname = $ _POST ['nickname'];
$ country = $ _POST ['country'];
等等......或者这不正确?RE-修改:
这样的事情会适用吗?
$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 = $row["bio"];
}
还是我离开基地?
答案 0 :(得分:3)
简短版;)
HTML
档案:
<form action="./change.php" method="post">
Nickname: <input type="text" name="nickname"><br />
Country: <input type="text" name="country"><br />
Date of birth: <input type="text" name="date_of_birth"><br />
Gender: <input type="text" name="gender"><br />
Motto: <input type="text" name="motto"><br />
Bio: <input type="text" name="bio"><br />
<input type="submit" value="Submit">
</form>
change.php:
<?php
function filter($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
}
?>