希望有人可以提供帮助。我有一个配置文件页面，我想显示登录的用户详细信息。到目前为止，我在“个人资料”页面上有这个。
<?php
/* This script pulls the existing name input and displays it when the user logs in. */
session_start();
include("db.php"); ?>
<?php include("includes/header.php") ?>
<?php include("includes/nav.php") ?>
<?php
if(logged_in()) {
$result = mysqli_query($link,$query);
$row = mysqli_fetch_array($result);
if (!$_POST['name'] && $_POST['name']=="") $error.="<br />Please enter your name";
if (!$_POST['email'] && $_POST['email']=="") $error.="<br />Please enter your email";
if (!$_POST['DOB'] && $_POST['DOB']=="") $error.="<br />Please enter your date of birth";
if (!$_POST['country'] && $_POST['country']=="") $error.="<br />Please enter your country";
if ($error) {
echo '<div class="alert alert-success alert-dismissable">'.addslashes($error).'</div>';
}
if(isset($_POST['form-control'])) {
move_uploaded_file($_FILES['file']['tmp_name'],"img/".$_FILES['file']['name']);
$query = mysqli_query("UPDATE users SET image = '".$_FILES['file']['name']."'");
}
} else {
redirect("login.php");
}
?>
<Style>
.alert{
display:none;
}
#profileimg {
height: 100px;
width: auto;
}
</Style>
<div class="container">
<h1>Edit Profile</h1>
<hr>
<div class="row">
<!-- left column -->
<div class="col-md-3">
<div class="text-center">
<img src="//placehold.it/100" class="avatar img-circle" alt="avatar" id="profileimg">
<h6>Upload a different photo...</h6>
<input class="form-control" type="file" name="name">
</div>
</div>
<!-- edit form column -->
<div class="col-md-9 personal-info">
<div class="alert alert-success alert-dismissable">
<button type="button" class="close" data-dismiss="alert" aria-hidden="true">×</button>
<strong>Profile updated.</strong>
</div>
<h3>Personal info</h3>
<form class="form-horizontal" role="form" action="edit_profile.php" method="post">
<div class="form-group">
<label class="col-lg-3 control-label name">name:</label>
<div class="col-lg-8">
<input class="form-control" value="<?php echo $row['name'];?>" type="text" name="name" required>
</div>
</div>
<div class="form-group">
<label class="col-lg-3 control-label">Email:</label>
<div class="col-lg-8">
<input class="form-control" value="<?php echo $row['email'];?>" type="text" name="email" required>
</div>
</div>
<div class="form-group">
<label class="col-lg-3 control-label">DOB:</label>
<div class="col-lg-8">
<input class="form-control" value="<?php echo $row['DOB'];?>" type="date" name="DOB" required>
</div>
</div>
<div class="form-group">
<label class="col-lg-3 control-label">Country</label>
<div class="col-lg-8">
<input class="form-control" value="<?php echo $row['country'];?>" type="text" name="country" required>
</div>
</div>
<div class="form-group">
<label class="col-md-3 control-label"></label>
<div class="col-md-8">
<input class="btn btn-primary" value="Save Changes" type="submit">
<span></span>
<input class="btn btn-default" id="updated" value="Cancel" type="reset">
</div>
</div>
</form>
</div>
<!-- jQuery (necessary for Bootstrap's JavaScript plugins) -->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<!-- Include all compiled plugins (below), or include individual files as needed -->
<script src="js/bootstrap.min.js"></script>
<script>
$("#updated").click(function(){
$(".alert").hide().show('medium');
</script>
</body>
</html>
然后我有另一个用于更新的php文件：
<?php
session_start();
include("db.php");
$name = $_POST['name'];
$email = $_POST['email'];
$DOB = $_POST['DOB'];
$country = $_POST['country'];
$password = md5($salt.$_POST['password']);
$query = "UPDATE users SET name = '".$name."', email = '".$email."', DOB = '".$DOB."', country = '".$country."', password = '".$password."'";
$result = mysqli_query($link,$query);
header('Location: profile.php');
?>
所以简短的是它没有显示或更新，我不知道为什么。我是PHP的新手，所以如果这很简单，我会很容易找到，但我已经搜索过但似乎无法找到答案。
提前致谢。
答案 0 :(得分：1)
我也是新手，但通常在我检查SESSION ID是否有效时我会使用
if(isset($_SESSION['id'])) {
$query = "UPDATE users SET name = '".$name."', email = '".$email."', DOB = '".$DOB."', country = '".$country."', password = '".$password."' WHERE id='".$_SESSION['id']."'";
}
您还需要回显您尝试查询的索引行以显示结果
$name = row['username'];
echo $name;
答案 1 :(得分：0)
您的代码中存在许多错误：您尝试在同一页面上传文件，而您将表单数据发送到另一个页面。如何处理表单验证也有点开销。我在表单中做了哪些更改：我在您的提交按钮中添加了name =“save”，并添加了新的隐藏输入来存储您的用户个人资料ID。我不确定你的代码中有什么login（）函数，最好坚持if（$ id）{}。
试试这个：
<?php
/* This script pulls the existing name input
and displays it when the user logs in. */
session_start();
include("db.php");
include("includes/header.php");
include("includes/nav.php");
$id = $_SESSION['id'];
if(loginned()) {//you can do if($id){}
$query="SELECT * FROM users WHERE id='$id' LIMIT 1";
$result = mysqli_query($link,$query);
$row = mysqli_fetch_array($result);
?>
<style>
.alert{
display:none;
}
#profileimg {
height: 100px;
width: auto;
}
</style>
<div class="container">
<h1>Edit Profile</h1>
<hr>
<div class="row">
<!-- left column -->
<!-- edit form column -->
<div class="col-md-9 personal-info">
<div class="alert alert-success alert-dismissable">
<button type="button" class="close" data-dismiss="alert"
aria-hidden="true">×</button>
<strong>Profile updated.</strong>
</div>
<h3>Personal info</h3>
<form class="form-horizontal" role="form"
action="edit_profile.php" method="post">
<div class="form-group">
<label class="col-lg-3 control-label name">name:</label>
<div class="col-lg-8">
<input class="form-control" value="<?php echo $row['name'];?>"
type="text" name="name" required>
</div>
</div>
<div class="col-md-3">
<div class="text-center">
<img src="//placehold.it/100" class="avatar
img-circle" alt="avatar" id="profileimg">
<h6>Upload a different photo...</h6>
<input class="form-control" type="file" name="name">
</div>
</div>
<div class="form-group">
<label class="col-lg-3 control-label">Email:</label>
<div class="col-lg-8">
<input class="form-control" value="<?php echo $row['email'];?>"
type="text" name="email" required>
</div>
</div>
<div class="form-group">
<label class="col-lg-3 control-label">DOB:</label>
<div class="col-lg-8">
<input class="form-control" value="<?php echo $row['DOB'];?>"
type="date" name="DOB" required>
</div>
</div>
<div class="form-group">
<label class="col-lg-3 control-label">Country</label>
<div class="col-lg-8">
<input class="form-control" value="<?php echo $row['country'];?>"
type="text" name="country" required>
</div>
</div>
<div class="form-group">
<input type="hidden" name="id" value="<?php echo $row['id'];?>">
<label class="col-md-3 control-label"></label>
<div class="col-md-8">
<input class="btn btn-primary" name="save"
value="Save Changes" type="submit">
<span></span>
<input class="btn btn-default" id="updated"
value="Cancel" type="reset">
</div>
</div>
</form>
</div>
<?php }else{ redirect("login.php"); } ?>
edit_profile.php首先我们检查是否有任何名称为save的帖子，我们验证发布的数据。如果验证成功，我们继续上传您的文件，然后运行更新查询。
<?php
session_start();
include("db.php");
if(isset($_POST['save'])){
$id = isset($_POST['id'])? $_POST['id']:'';
$name = isset($_POST['name'])? $_POST['name']:'';
$email = isset($_POST['email'])? $_POST['email']:'';
$dob = isset($_POST['DOB'])? $_POST['DOB']:'';
$pass = isset($_POST['passwrd'])? md5($salt.$_POST['password']):'';
$country = isset($_POST['country'])? $_POST['country']:'';
if(empty($name)){
$error = 'Please enter your name';
}elseif(empty($email)){
$error = 'Please enter your email';
}elseif(empty($dob)){
$error = 'Please enter your date of birth';
}elseif(empty($country)){
$error = 'Please enter your country';
}elseif(empty($pass)){
$error = 'Please enter your password';
}else{
move_uploaded_file($_FILES['file']['tmp_name'],"img/".$_FILES['file']['name']);
$query = mysqli_query("UPDATE users SET image = '".$_FILES['file']['name']."'
WHERE id='$id'");
$query = "UPDATE users SET name = '$name', email = '$email',
DOB = '$DOB', country = '$country', password = '$password'
WHERE id='$id'";
$result = mysqli_query($link,$query);
header('Location: profile.php');
}
} ？＆GT;
<?php if(!empty($error)){
echo '<div class="alert alert-success
alert-dismissable">'.addslashes($error).'</div>';
}else{
echo '<div class="alert alert-success">Success</div>';
}
?>
我添加了demo here。至少这会有所帮助：