Echo database user Name with Posted email - php

I'm trying to get the user's name from a database after posting his email on a form but its not showing the name, if I add a success message it shows it so it's working but its not showing me the name :(
<?php
$conn = mysqli_connect("xxx", "xxx", "xxxxxxxx", "xxxxxxx");
$email = $_POST['email_r'];
$sqlr = "SELECT * FROM participantes WHERE email='$email'";
$result = $conn->query($sqlr);
if(!$row = mysqli_fetch_assoc($result)) {
echo "Email Incorrecto: No se a registrado.";
} else {
echo "your name is: ->" . $row['name'] . " <- that is it.";
}
?>

If I had to guess, I'd say your table does not have a name column. Either that or the matching record has an empty name value.
This should at least help you pinpoint any potential issues as well as resolve your SQL injection vulnerability...
<?php
// show any errors
ini_set('display_errors', 'On');
// show all errors
error_reporting(E_ALL);
// make MySQLi throw exceptions
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$conn = mysqli_connect("xxx", "xxx", "xxxxxxxx", "xxxxxxx");
// safely get the request parameter
$email = isset($_POST['email_r']) ? $_POST['email_r'] : null;
// Prepare a statement with a placeholder for the "email" parameter
$stmt = $conn->prepare('SELECT `name` FROM `participantes` WHERE `email` = ?');
// bind the parameter
$stmt->bind_param('s', $email);
$stmt->execute();
// bind results. This seems easier than fetch_assoc IMHO
$stmt->bind_result($name);
// fetch records, if any
if ($stmt->fetch()) {
echo 'your name is: ->', $name, ' <- that is it.';
} else {
echo "Email Incorrecto: No se a registrado.";
}
$stmt->close();

Related

How to update status in database if status is empty without submitting a form in php?

How to update a status from database if status is empty in using php? I have this condition in php. I have this if condition that decides if $getstatus is empty it will update from database to Avail. I tried refreshing the page after querying the database. But it will not update in database. Is there anyway to update this without using form submit in php?
<?php
session_start();
include "includes/connection.php";
// Display all parking slots
$sql = $connection->prepare('SELECT * FROM parkingslot where parkingslotid = 1');
$sql->execute(); // execute query
$result = $sql->get_result(); // fetch result
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$getstatus = $row["status"];
echo $getstatus;
}
}
if (empty($getstatus)) {
$sql = $connection->prepare("UPDATE parkingslot SET status = 'Avail' where parkingslotid = 1 ");
}
?>
Codes in connection for connecting to database
connection.php
<?php
$server = "localhost";
$username = "root";
$password = "";
// create connection
$connection = mysqli_connect($server,$username,$password);
// check connection
if(!$connection)
{
die("No connection found." . mysqli_connect_error());
}
else {
// select a database
$select_db = mysqli_select_db($connection,'smartparkingsystem');
if(!$select_db)
{
$sql = 'CREATE DATABASE sample';
// create database if no db found
if(mysqli_query($connection,$sql)) {
echo "Database Created";
}
else {
echo "Database not found" . mysqli_connect_error() . '\n';
}
}
else {
// Database already existed
// do nothing...
}
}
?>
If I understand your goal of: For row(s) whereparkingslotid=1 - Update status to 'Avail' but only if status is not currently set, this might help:
<?php
session_start();
include "includes/connection.php";
$connection->prepare("UPDATE `parkingslot` SET `status`=? WHERE `parkingslotid`=? AND (`status` IS NULL OR `status`=?)");
$connection->bind_param("sis", $status, $parkingslotid, $empty_str);
$status = 'Avail';
$parkingslotid = 1;
$empty_str = '';
$connection->execute();
echo $connection->affected_rows.' rows affected';
$connection->close();
?>
This saves a bit of processing by not checking with PHP first.
You can use this query:
"UPDATE parkingslot SET status = 'Avail' where status IS NULL OR status = '' "
Edited:
#lumonald gave the right anwser in the comment. You're not executing your second SQL statement.

SQL prepared statements. PHP

May be this question will be sort of "stupid-questions", but still...
I'm new to PHP and SQL and I can't understand what I am doing wrong here:
if(isset($_POST[$logButton])) //Checking for login button pressed
{
//Retrieving information from POST method
$uid = $_POST['login'];
$upwd = $_POST['password'];
//SQL Connection
$mysqli = new mysqli('localhost', 'root', '', 'students');
if(!$mysqli)
{
echo "<h1 class='h1A'>Problem accured while connecting to the DB. " . mysqli_error($mysqli) . "</h1>"; //!!!Delete displaying error msg after dev.
}else
{
$sql = "SELECT * FROM login_data WHERE login = ? AND password = ?"; //SQL query
$stmt = $mysqli->prepare($sql) or die("error1"); //No error
$stmt->bind_param('ss', $uid, $upwd) or die("error2");//No error
$stmt->execute() or die("error3");//Giving DB query. No error
$result = $stmt->fetch() or die("error4".mysqli_error($mysqli)); //Putting query's result into assoc array. !!!Delete displaying error msg after dev. No error
echo print_r($result); //It prints out "11" ? ? ?
if(count($result['id']) < 1) //If no rows found.
{
echo "<h1 class='h1A'>Couldn't find account. Please, recheck login and password.</h1>";
die();
}elseif($result['id'] > 1)//If more then 1 row found.
{
echo "<h1 class='h1A'>Caught 9090 error. Contact the administrator, please.".mysqli_error($mysqli)."</h1>";
die();
}elseif($result['id'] == 1) //If only one row's been found.
{
$_SESSION['isLoggedIn'] = true;
redirectTo('/index.php'); //Declared function.
die();
}
}
}
Here is a part of handler function in lib.php file. This file is included to the html-page and the function is used. No errors displayed and when I print_r $result - it prints out 11. Can't get it.
Well, use print_r without echo :
print_r($result);
or pass second parameter to print_r function so it can return string:
echo print_r($result, true);
See http://php.net/manual/en/function.print-r.php for more info.

mysql error: not a valid mysql resource

I have a problem when i try to check if email is alredy registered. can someone help? I have this error:
mysql_fetch_array(): supplied argument is not a valid MySQL result resource in line...
($record =mysql_fetch_array($result); )
<?php
$nome = $_REQUEST["nome"];
$cognome = $_REQUEST["cognome"];
$psw = $_REQUEST["psw"];
$email = $_REQUEST["email"];
$nikName = $_REQUEST["nikName"];
$conn = mysql_connect("host,name","userName","Password","databaseName");
if(!$conn) {
echo "connessione non satabilita";
} else {
if(!mysql_select_db("databaseName",$conn)) {
echo "database non trovato";
} else {
$sql = "select * from utenti where User='$email'"; //costruzione comando di ricerca
$result = mysql_query($sql,$conn); //assegnazione risultati
$record =mysql_fetch_array($result); //estrazione primo risultato
if(!$record) {
$sql = "INSERT INTO User (UserId, Nome, Cognome, Email, Username, Password, TimeStamp) VALUES (NULL,'$nome','$cognome','$email','$nikName','$psw', NULL)";
$result=mysql_query($sql);
if($result) {
echo "utente registrato correttamente";
} else {
//Error
echo "errore registrazione, riprovare più tardi";
}
echo "<br />";
echo "utente registrato";
} else {
echo "utente gia registrato";
}
}
}
?>
Before this gets out of hand.
$conn = mysql_connect("host,name","userName","Password","databaseName");
You're using 4 parameters rather than 3.
Sidenote: 4 parameters is mysqli_ syntax http://php.net/manual/en/function.mysqli-connect.php
Be careful though, those different MySQL APIs do not intermix. So you cannot have mysql_ with mysqli_ should you decide to change it to that.
The manual http://php.net/manual/en/function.mysql-connect.php states:
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
the fourth is for something else.
If a second call is made to mysql_connect() with the same arguments, no new link will be established, but instead, the link identifier of the already opened link will be returned. The new_link parameter modifies this behavior and makes mysql_connect() always open a new link, even if mysql_connect() was called before with the same parameters. In SQL safe mode, this parameter is ignored.
So, just remove the 4th parameter.
Sidenote: This is questionable "host,name" (with the comma). Double check it as to what your host (if hosted) has provided you with. Most of the time, that should read as "localhost".
As stated, you're open to SQL injection.
Use a prepared statement:
https://en.wikipedia.org/wiki/Prepared_statement
As for the rest of your code:
Add error reporting to the top of your file(s) which will help find errors.
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
// rest of your code
Sidenote: Displaying errors should only be done in staging, and never production.
Also add or die(mysql_error()) to mysql_query().
About you're wanting to check if an email exists; you may be better off using mysql_num_rows().
I.e.:
$sql = "select * from utenti where User='$email'";
$result = mysql_query($sql,$conn) or die(mysql_error($conn));
if(mysql_num_rows($result) > 0)
{...}
else {...}
I noticed you may be storing passwords in plain text. If this is the case, it is highly discouraged.
I recommend you use CRYPT_BLOWFISH or PHP 5.5's password_hash() function. For PHP < 5.5 use the password_hash() compatibility pack.
Also, this doesn't help you:
if(!mysql_select_db("databaseName",$conn)){
echo "database non trovato";
}
This does:
if(!mysql_select_db("databaseName",$conn)){
die ('Can\'t use the database : ' . mysql_error());
}
In order to get the real error, should there be one.
Reference:
http://php.net/manual/en/function.mysql-select-db.php
As mentioned above there is a syntax error with mysql_connect(); where you're trying to use invalid number of params. The best way is to make a config.php file and then use it whenever you need it. This is a basic connection code in PDO.
<?php
$host = "localhost";
$database = "yourdbnamehere";
$username = "yourusernamehere";
$password = "yourpasswordhere";
try {
$dbo = new PDO('mysql:host='.$host.';dbname='.$database, $username, $password);
} catch (PDOException $e) {
print "Error!: " . $e->getMessage() . "<br/>";
die();
}
?>
You will need a solution like this. But you must switch towards PDO or MySQLi to ensure that your code stays valid in long run as well as you will be able to write secure and stable code.
Go through these at first:
http://php.net/manual/en/book.pdo.php
http://php.net/manual/en/book.mysqli.php
An example code for you solving your current problem:
<?php
$nome = $_REQUEST["nome"];
$cognome = $_REQUEST["cognome"];
$psw = $_REQUEST["psw"];
$email = $_REQUEST["email"];
$nikName = $_REQUEST["nikName"];
try{
$pdo = new PDO('mysql:dbhost=hostname; dbname=databaseName', 'userName', 'Password');
} catch (PDOException $e) {
echo "Error connecting to database with error ".$e;
die();
}
// check for email
$sql = $pdo->prepare("select * from utenti where User= ?");
$sql->bindParam('1', $email);
$sql->execute();
/* if you aren't hashing the password,
then do it first
$psw = PASSWORD_HASH($psw, PASSWORD_DEFAULT);
*/
// Insert if email not registered
if($sql->rowCount() == 0) {
$insert = $pdo->prepare("INSERT INTO User (Nome,Cognome,Email,Username,Password) VALUES (?, ?, ?, ?, ?)");
$insert->bindParam('1', $nome);
$insert->bindParam('2', $cognome);
$insert->bindParam('3', $email);
$insert->bindParam('4', $nikName);
$insert->bindParam('5', $psw);
$insert->execute();
if($insert->rowCount() > 0) {
echo "utente registrato correttamente";
} else {
echo "errore registrazione, riprovare più tardi";
}
} else {
echo "utente gia registrato";
}
?>

php num->rows not working

I keep getting an error, even though I am 100% sure I followed the example that is found in the PHP manual.
The simplified version of the code can be found below.
note: connection to the database is ok.
EDIT: I keep getting an "Catchable fatal error: Object of class mysqli_stmt could not be converted to string" error.
EDIT: Now I keep getting "Mission Failed" even though I am sure that the row count should be 1.
Here's the coode used:
# $db = new mysqli('localhost', 'USER', 'PASSWORD', 'DATABSE');
$email = $db->prepare("select * from members where email = ?");
$email->bind_param('s', $email);
$email->execute;
$email->store_result;
$email->num_rows;
if ($email > 0) {
echo "<p>This e-mail is already in use, please try again with another e-mail.</p>";
exit;
} else {
echo "mission failed";
}
exit;
EDIT:
# $db = new mysqli('localhost', 'USER', 'PASSWORD', 'DB');
if ($db->connect_errno) {
echo "<p id=\"signup_confirmed\">Error: could not connect to database. Please try again later.</p>";
exit;
}
$checkRow = $db->prepare("select * from members where email = ?");
$checkRow->bind_param('s', $email);
$checkRow->execute;
$checkRow->store_result;
if ($checkRow->num_rows > 0) {
echo "<p id=\"signup_confirmed\">This e-mail is already in use, please try again with another e-mail.</p>";
exit;
} else {
echo "<p id=\"signup_confirmed\">Row checking has failed</p>";
}
change
$email->num_rows();
to
$email->num_rows;
in your code
New edit
$count = $email->num_rows;
if ($count > 0) {
echo "<p>This e-mail is already in use, please try again with another e-mail.</p>";
exit;
} else {
echo "mission failed";
}
More edit
change this to
$email = $db->prepare("select * from members where email = ?");
$email->bind_param('s', $email);
this
// you are over riding your $email value with the query thats the reason its not working
$query= $db->prepare("select * from members where email = ?");
$query->bind_param('s', $email);
# $db = new mysqli('localhost', 'USER', 'PASSWORD?', 'DATABSE');
$query_email = "select * from members where email = ?";
$email = $db->prepare($query_email);
$email->bind_param('s', $email);
$email->execute();
$email->store_result();
if ( $email->num_rows > 0) {
echo "<p id=\"signup_confirmed\">This e-mail is already in use, please try again with another e-mail.</p>";
exit;
}
$email is a result set, an object.
It will hold data returned from SQL operation.
You need to get the num_rows() in a variable.
Corrected code:
$cnt = $email->num_rows;
if ($cnt > 0) {
OR
if ($email->num_rows > 0) {
instead of
if ($email > 0) use if ($email->num_rows() > 0)

Update query not working using PDO

I tried updating my data like so but it doesn't work
<?php
require("config.inc.php");//this piece of code us for authentication and it works fine.
if(!empty($_POST))
{
/**
the values below in the POST are valid not empty values
**/
$shell = $_POST['shell'];
$reporter = $_POST['reporter'];
//query
$query = "UPDATE `shellingdb`
SET `likes` = `likes` + 1
WHERE `shell` = :shell AND `reporter` = :reporter";
try {
$query_params = array(':shell' => $_POST['shell'], ':reporter' => $_POST['reporter']);//Updates likes
$stmt = $db->prepare($query);
$result = $stmt->execute($query_params);
$affected = $stmt->rowCount();//counts the number of affected rows during the update query
if($affected > 0)
{
$response["success"] = 1;
$response["message"] = "Updated! this number of rows were affected".$affected;
echo json_encode($response);
}else
{
$response["success"] = 2;
$response["message"] = "Not Updated! huh!".$affected;
echo json_encode($response);
}
}
catch (Exception $ex) {
$response["success"] = 0;
$response["message"] = "Database Error!".$ex->getMessage();
die(json_encode($response));
}
}
?>
the config.inc.php
<?php
// These variables define the connection information for your MySQL database
$username = "xmnj3jh0jhtheu_14265914";
$password = "jhikjskjiavethew";
$host = "sqlkjnlkkjlk101.x3kuhiu0lkj.us";
$dbname = "x3lnklj0u_1426jbkb5914_gbabbjkhjajhlert";
// UTF-8 is a character encoding scheme that allows you to conveniently store
// a wide varienty of special characters, like � or �, in your database.
// By passing the following $options array to the database connection code we
// are telling the MySQL server that we want to communicate with it using UTF-8
// See Wikipedia for more information on UTF-8:
// http://en.wikipedia.org/wiki/UTF-8
$options = array(PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES utf8');
// A try/catch statement is a common method of error handling in object oriented code.
// First, PHP executes the code within the try block. If at any time it encounters an
// error while executing that code, it stops immediately and jumps down to the
// catch block. For more detailed information on exceptions and try/catch blocks:
// http://us2.php.net/manual/en/language.exceptions.php
try
{
// This statement opens a connection to your database using the PDO library
// PDO is designed to provide a flexible interface between PHP and many
// different types of database servers. For more information on PDO:
// http://us2.php.net/manual/en/class.pdo.php
$db = new PDO("mysql:host={$host};dbname={$dbname};charset=utf8", $username, $password, $options);
}
catch(PDOException $ex)
{
// If an error occurs while opening a connection to your database, it will
// be trapped here. The script will output an error and stop executing.
// Note: On a production website, you should not output $ex->getMessage().
// It may provide an attacker with helpful information about your code
// (like your database username and password).
die("Failed to connect to the database: " . $ex->getMessage());
}
// This statement configures PDO to throw an exception when it encounters
// an error. This allows us to use try/catch blocks to trap database errors.
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
// This statement configures PDO to return database rows from your database using an associative
// array. This means the array will have string indexes, where the string value
// represents the name of the column in your database.
$db->setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE, PDO::FETCH_ASSOC);
// This block of code is used to undo magic quotes. Magic quotes are a terrible
// feature that was removed from PHP as of PHP 5.4. However, older installations
// of PHP may still have magic quotes enabled and this code is necessary to
// prevent them from causing problems. For more information on magic quotes:
// http://php.net/manual/en/security.magicquotes.php
if(function_exists('get_magic_quotes_gpc') && get_magic_quotes_gpc())
{
function undo_magic_quotes_gpc(&$array)
{
foreach($array as &$value)
{
if(is_array($value))
{
undo_magic_quotes_gpc($value);
}
else
{
$value = stripslashes($value);
}
}
}
undo_magic_quotes_gpc($_POST);
undo_magic_quotes_gpc($_GET);
undo_magic_quotes_gpc($_COOKIE);
}
// This tells the web browser that your content is encoded using UTF-8
// and that it should submit content back to you using UTF-8
header('Content-Type: text/html; charset=utf-8');
// This initializes a session. Sessions are used to store information about
// a visitor from one web page visit to the next. Unlike a cookie, the information is
// stored on the server-side and cannot be modified by the visitor. However,
// note that in most cases sessions do still use cookies and require the visitor
// to have cookies enabled. For more information about sessions:
// http://us.php.net/manual/en/book.session.php
session_start();
// Note that it is a good practice to NOT end your PHP files with a closing PHP tag.
// This prevents trailing newlines on the file from being included in your output,
// which can cause problems with redirecting users.
?>
don't know what's wrong and it gives no error it goes into the else statement, meaning the values were not updated. i tried the same code in sqlfiddle and it works but not in my PhpMyAdmin.
I know the updated value is supposed to be passed into the $query_params but am incrementing the value of likes each time it is run, and am not sure how to do that in the $query_params unless i use a seperate query to get the numberof likes and then increament it but that could be costly.
Query without PDO still it does not work this time it give update unsuccessful
<?php
$username = "x3jbhiukhkj0u426jbhjnbvh591mbhb4";
$password = "savjiuejbiuhilkmthljiew";
$host = "sqlnjhbjhnkjjjhbj";
$dbname = "x3hjbh0ukjioiuhgbjhvhgvh";
$shell = "Rustig";
$reporter = "davies";
//query
$query = "UPDATE `shellingdb`
SET `favs` = 1
WHERE `shell` = 'Rustig'";
$link = mysql_connect($host, $username, $password);
if (!$link)
{
die('Could not connect: ' . mysql_error());
}else
{
echo 'Connected successfully';
$db_selected = mysql_select_db($dbname, $link);
if (!$db_selected)
{
die ('Can\'t use foo : ' . mysql_error());
}else
{
echo 'Connected to database successfully';
if(empty($_POST))
{
$retval = mysql_query( $query, $link )or die(mysql_error($link));;
if(! $retval )
{
die('Could not query database: ' . mysql_error());
}else
{
if(mysql_affected_rows() > 0)
{
echo "Updated data successfully\n";
}else
{
//echo "shell=".$shell." reporter=".$reporter';
echo "Updated data Unsuccessfully\n";
}
}
}
}
}
mysql_close($link);
?>
The below is the output of the PDOStatement::debugDumpParams(); for the first php syntax
SQL: [124] UPDATE shellingdb SET likes = likes + 1 WHERE shell = :shell AND reporter >= :reporter Params: 2 Key: Name: [6] :shell paramno=-1 name=[6] ":shell" is_param=1 param_type=2 Key: Name: [9] :reporter paramno=-1 name=[9] ":reporter" is_param=1 param_type=2
I used bindParam. bindParam is a method on PDOStatement.
Try:
<?php
require("config.inc.php");//this piece of code us for authentication and it works fine.
if(isset($_POST))
{
/**
the values below in the POST are valid not empty values
**/
$shell = $_POST['shell'];
$reporter = $_POST['reporter'];
//query
$query = "UPDATE `shellingdb`
SET `likes` = `likes` + 1
WHERE `shell` = :shell AND `reporter` = :reporter";
try {
$stmt = $db->prepare($query);
$stmt->bindParam(":shell", $shell);
$stmt->bindParam(":reporter", $reporter);
$stmt->execute();
$affected = $stmt->rowCount();//counts the number of affected rows during the update query
if($affected > 0)
{
$response["success"] = 1;
$response["message"] = "Updated! this number of rows were affected".$affected;
echo json_encode($response);
}else
{
$response["success"] = 2;
$response["message"] = "Not Updated! huh!".$affected;
echo json_encode($response);
}
}
catch (Exception $ex) {
$response["success"] = 0;
$response["message"] = "Database Error!".$ex->getMessage();
die(json_encode($response));
}
}
?>
some how, after long hours of try and error(Brut Forcing) this finally worked
$query = "UPDATE `shellingdb` SET `likes`=`likes`+1 WHERE `shell` = :shell AND `reporter` = :reporter";
Thanks all those who tried to help. :)

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