Почему закрывающая скобка черная ?
Привет,
В моем блокноте++ линия на 116 показывает закрывающую скобку ELSE black. Действительно очень загадочно!
Я знаю, что мой код не очень аккуратен, но я приведу его в порядок, основываясь на обратной связи, которую я получил до сих пор, но в то же время мой великий ум заполнен вопросом, почему строка на 116 показывает закрывающую скобку ELSE black.
<html>
<head>
<title>
$user Home Page
</title>
</head>
<body>
<body background=".png">
<?php
session_start();
require "conn.php";
/*Check if user is logged-in or not by checking if session is set or not.
If user is not logged-in then redirect to login page. Else, show user's account homepage.*/
if(!isset($_SESSION["user"]))
{
header("location:login.php");
}
else
{
$user = $_SESSION["user"];
$sql = "SELECT * FROM users WHERE Username = '".$user."'";
$result = $conn->query($sql);
while($row = mysqli_fetch_assoc($result))
{
$db_id = $row["Id"];
$db_username = $row["Username"];
$db_forename = $row["Forename"];
$db_surname = $row["Surname"];
$db_email = $row["Email"];
$db_bio = $row["Bio"];
$db_status = $row["Status"];
//Welcome user by name.
echo "<center>Welcome <h2>$db_forename $db_surname!</center>"?></h2>|
<?php
//Display log-out link.
echo "<p align='right'><a href='logout.php'>$user Log Out</a>";?>|</p><br>
<form method="post" action="">
<div class="form-group">
<center><label>Post Status To Friends:</label>
<input type="text" placeholder="Post Status To Friends" name="post_status_to_friends" /></center><br>
<center><button type="submit" class="btn btn-default" name="post_status_to_friends" />Post Status To Friends!</button></center>
</div>
<form method="post" action="">
<div class="form-group">
<center><label>Post Status To Public:</label>
<input type="text" placeholder="Post Status To Public" name="post_status_to_public" /></center><br>
<center><button type="submit" class="btn btn-default" name="post_status_to_public" />Post Status To Public!</button></center>
</div>
<?php
//Post User Status To Friends in $user table.
if (isset($_POST['post_status_to_friends']))
{
$status_to_friends = trim($_POST["post_status_to_friends"]);
$status_to_friends = mysqli_real_escape_string($conn,$status_to_friends);
$conn->query("UPDATE $user SET Status_To_Friends $status_to_friends WHERE Username = '".$user."'");
if($conn->query($sql)===TRUE)
{
echo "posted status for friends to $user table!";
$conn->close();
exit();
}
else
{
echo "posting status for friends to $user table failed!";
$conn->close();
exit();
}
//Post User Status To Friends in "users" table.
$conn->query("UPDATE users SET Status_To_Friends $status_to_friends WHERE Username = '".$user."'");
if($conn->query($sql)===TRUE)
{
echo "posted status for friends to users table!";
$conn->close();
exit();
}
else
{
echo "posting status for friends to users table failed!";
$conn->close();
exit();
}
//Display User Status To Friends. Search for User's data with $user.
$query = "SELECT * FROM $user WHERE Username = '".$user."'";
$result = mysqli_query($conn,$query);
$numrows = mysqli_num_rows($result);
if($numrows = 0)
{
echo "No data!";
$conn->close();
exit();
}
else
{
while($row = mysqli_fetch_assoc($result))
{
$db_status_to_friends = $row["Status_To_Friends"];
}
echo "<br>$user Status To Friends:<br>
$db_status_to_friends";?><br>
<br>
}
<?php
//Post User Status To Public in $user table.
if (isset($_POST['post_status_to_public']))
{
$status_to_public = trim($_POST["post_status_to_public"]);
$status_to_public = mysqli_real_escape_string($conn,$status_to_public);
$conn->query("UPDATE $user SET Status_To_Public $status_to_public WHERE Username = '".$user."'");
if($conn->query($sql)===TRUE)
{
echo "posted status for public to $user table!";
$conn->close();
exit();
}
else
{
echo "posting status for public to $user table failed!";
$conn->close();
exit();
}
//Post User Status To Public in users table.
$conn->query("UPDATE users SET Status_To_Public $status_to_public WHERE Username = '".$user."'");
if($conn->query($sql)===TRUE)
{
echo "posted status for public to users table!";
$conn->close();
exit();
}
else
{
echo "posting status for public to users table failed!";
$conn->close();
exit();
}
//Display User Status To Public. Search for User's data with $user.
$query = "SELECT * FROM $user WHERE Username = '".$user."'";
$result = mysqli_query($conn,$query);
$numrows = mysqli_num_rows($result);
if($numrows = 0)
{
echo "No data!";
$conn->close();
exit();
}
else
{
while($row = mysqli_fetch_assoc($result))
{
$db_status_to_public = $row["Status_To_Public"];
}
echo "<br>$user Status To Public:<br>
$db_status_to_public";?><br>
<br>
<?php
}
}
//Display User Bio.
echo "<br>Bio:<br>";
echo "$db_bio";?><br>
<br>
<?php
//Display User's Latest View.
echo "<br>Latest View:<br>";
echo "$db_latest_view";?><br>
<br>
<?php
//Display User's Latest Viewed Url in iFrame.?>
<iframe src="<?php $db_latest_view;?>"></iframe>
<?php
}
}
?>
</body>
</html>
Что я уже пробовал:
Пробовал менять код разными способами, но безуспешно!
Bryian Tan
не уверен, что это проблема с копией вставки, но похоже, что в коде отсутствует закрывающий тег php?
else
{
while($row = mysqli_fetch_assoc($result))
{
$db_status_to_friends = $row["Status_To_Friends"];
}
echo "<br>$user Status To Friends:<br>
$db_status_to_friends";?><br>
<br>
}
?>
<?php
//Post User Status To Public in $user table.
...
...
Member 12956789
- Вы уверены ?
Я уже добавил сюда закрывающую скобку:
$db_status_to_friends";?&ГТ;