PHP/SQL : "false" value received from database with good query - php

I have a php script who ask my database with PDO to verify if some values sent exists. If they exists, the database respond with the id of this line's value.
I tested the query on mysql and it works but the value received is false.
This code is only for personal use.
There is the code :
<?php
include("../template/pdo.php");
$query = $pdo->prepare("SELECT id_utilisateur FROM utilisateur
WHERE `mail` IN ( ':mail' )
AND `mdp` IN ( ':mdp' )");
$query->bindParam(':mail', $_GET['identifiant'], PDO::PARAM_STR);
$query->bindParam(':mdp', $_GET['mdp'], PDO::PARAM_STR);
$success = $query->execute();
if($success)
{
$result = $query->fetch();
var_dump($result); //bool(false) actually
if($result == false){
$message = "Try again.";
}
else{
$message = "Congratulation !";
}
}
I tested everything I know :
$_GET is a print/paste from my database table to my url and i have print him
Printed/pasted on phpMyAdmin the query from PDOStatement::debugDumpParams() with my $_GET values
pdo.php work and used on other scripts
No log in my logs files.
Someone can help me ?
Thanks !

If you are testing against a single value use =, not IN.
If you have a list of values, several changes are needed.
The bind code will add quotes, you already have quotes. Remove your quotes.

Related

Why does the user input not append to my SQL database?

I'm developing a login/register form for my client. Right now I am working on the registration part of the form however I seem to have encountered an issue.
I am trying to append the user's input to a database if it does not currently exist. I'm developing this functionality using PHP version 7. However, the code does not seem to append the data to the database even when telling me it has done so successfully.
Here is code:
<?php
if($_SERVER["REQUEST_METHOD"] == "POST") {
//define variables and set values to null
$email = $code = "";
//set variable values to HTML input
$email = $_POST['email'];
$code = $_POST['code'];
//check if email exists
$stmt = $conn->prepare("SELECT userEmail FROM userDetails WHERE userEmail=?");
$stmt->bind_param("s", $prepemail);
//set parameters and execute
$prepemail = $email;
$stmt->execute();
$result = $stmt->get_result();
if ($result->num_rows > 0) {
echo "email exists";
return false;
} else {
//$stmt->close(); removed as per #Akintunde-Rotimi's suggestion
//insert email into database
$stmt = $conn->prepare("INSERT INTO userDetails (userEmail) VALUES (?)");
$stmt->bind_param("s", $newemail);
//set parameters and execute
$newemail = $email;
$stmt->execute();
echo "New records created successfully";
}
}
?>
The code successfully connects to the database and even tells me if the user already exists. It just doesn't add the user's email to the database and I can't seem to figure out why.
I have researched methods on how to insert the data into the database using prepared statements as I have done here. I've used W3Schools as a reference but still no luck.
The code doesn't seem to have any obvious spelling errors, so have you tried to catch errors? Replace
$stmt->execute();
with
if(!$stmt->execute()) {
trigger_error("there was an error....".$conn->error, E_USER_WARNING);
}
You can also check how many rows are affected, -1 meaning there was an error.
printf("%d Zeile eingefügt.\n", $stmt->affected_rows);
Also, enabling more errors to be shown (at least for development)
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
// ...

Sending content from my frontend to my PHP/MySQL-backend

When I send my content from my frontend it successfully reaches my if (!empty)-statement but the table in my phpmyadmin/mysql-database does not recieve the information and does not add it.
I have two tables. One varchar (text) named "photo" and a ID called "id" which is A_I.
With my current code I only send (well attempt to send) the text about "photo" but nothing about the ID as it is A_I? Maybe I need to add some addiotional code to that as well and that might be the issue here and the reason the database does not seem to add the content that I send?
<?php
$value = json_decode(file_get_contents('php://input'));
$mysql_pekare= new mysqli ("", "","", "");
if(!empty($value)) {
echo "You reached the IF-statement";
$stmt = $mysql_pekare->prepare("INSERT INTO photoAlbum(`photo`) VALUES(?)");
$stmt->bind_param("s", $value['photo']);
$stmt->execute();
$stmt->close();
$mysql_pekare->close();
}
?>
In my frontend when I send the content I recieve this in the log:
{"photo":"test"}
And I also recieve this in the log, the echo call I did if it reaches the IF function which it successfully does:
"You reached the IF-statement"
By default, json_decode() returns an object, so your value is in $value->photo.
So your INSERT code should be -
if(!empty($value)) {
echo "You reached the IF-statement";
$stmt = $mysql_pekare->prepare("INSERT INTO photoAlbum(`photo`) VALUES(?)");
$stmt->bind_param("s", $value->photo);
$stmt->execute();
$stmt->close();
$mysql_pekare->close();
}
you must put the value for mysqli constructor required like localhost or ip , username of database , password of your database and database name then it will work fine for Example:-
$mysql_pekare=new mysqli ("localhost", "username","password", "databasename");

Trouble dealing with results from a SELECT query with MySQLI

The basic control structure I'm trying to get to work is to query the DB with the username and email, both of which are unique keys, and if either are in the DB let the user know that they have been taken and to please pick something else. The problem I'm running into is getting the result data in a usable form that I can then check the user-supplied data against.
I cut out the prepared statements for insertion from the snippit, as well as the validation routines, since both of them are working fine.
DB connection snippit
try {
if(!($dbc = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME))){ // Creates the $dbc variable object so we can
// have a connection to the database.
// uses mysqli functions.
throw new Exception;
}
}
catch (Exception $e) {
echo '<p>Could not connect to the database. Please contact the system administrator.</p>';
}
Snippit of Registration script
//before this was validation routines, if anything was wrong the script generated something into $reg_errors which is an array.
if(empty($reg_errors))
{
//queries database if there are any matches for username or email from user input.
if($stmt = $dbc->prepare("SELECT `email`, `username` FROM `users` WHERE `email` = ? OR `username` = ?"))
{
$stmt->bind_param("ss", $e, $u);
$stmt->execute();
$stmt->store_result();
$rows = $stmt->num_rows; //gives the number of rows returned from SELECT query. 0 means no dupes, 1 means one record has BOTH email and username, 2 means two different records (one with email, one with username)
##THIS IS WHERE I'M RUNNING INTO TROUBLE GETTING THE DATA IN A USABLE FORM##
$stmt->close();
} else {
echo "<p>Can't talk to database right now. Try again later, please.</p>";
}
if($rows==0) //no dupes of username or email, so let's try and add them into the DB
{
//prepared statement for insertion into DB
//also get's the count of affected rows. 1 means record inserted correctly.
//asks DB if a new row was created, and if so, thanks user for
//registration on the site & sends an email to their email.
//if query doesnt work, an error is triggered
if($count==1) {
//constructs a thank you note and emails it to the user, using the email they supplied.
exit();
} else {
echo "<p>Unable to process your registration at this time. Please try again later..</p>";
}
} else { // both username and email might be already used in DB, and error msgs are generated for array.
if($rows==2) { // this checks to make sure both entries are dupes
$reg_errors['email'] = 'This email address has already been registered. If you have forgotten your password, use the link to the right to have your password sent to you.';
$reg_errors['username'] = 'This username has already been registered. Please try another.';
} else { //this checks to see which of the two (email or username) is already in DB if both arent dupes.
if((__NEED SOMETHING HERE FROM DB QUERY___ == $_POST['email']) && (__NEED SOMETHING HERE FROM DB QUERY___ == $_POST['username'])) { //both match entries in DB
$reg_errors['email'] = 'This email address has already been registered. If you have forgotten your password, use the link to the right to have your password sent to you.';
$reg_errors['username'] = 'This username has already been registered with this email address. If you have forgotten your password, use the link to the right to have your password sent to you.';
} elseif(__NEED SOMETHING HERE FROM DB QUERY___==$_POST['email']) { // email match
$reg_errors['email'] = 'This email address has already been registered. If you have forgotten your password, use the link to the right to have your password sent to you.';
} elseif(__NEED SOMETHING HERE FROM DB QUERY___==$_POST['username']) { // username match
$reg_errors['username'] = 'This username has already been registered. Please try another one.';
}
} // end of $rows==2 ELSE
} // end of $rows == 0 IF
} else { // end of empty reg_errors conditional
//do something if the reg_error array isnt empty..
}
i'm pretty sure the answer lies in iterations and using meta_data from the result mysqli object, but after beating my head against a wall for a couple days and pouring over the mysqli php manual pages like a maniac, I'm still no closer to figuring out what I should be doing. Could anyone point me in the correct direction?
Starting from the registration script, have you tried this:
if($stmt = $dbc->prepare("SELECT `email`, `username` FROM `users` WHERE `email` = ? OR `username` = ?"))
{
$stmt->bind_param("ss", $e, $u);
$stmt->execute();
$stmt->bind_result($email, $username);
$rows = $stmt->num_rows;
//Move Conditionals Up a Little
if( $rows == 0 ) { //If No Records are Found
//Continue Registration
}
else if( $rows == 1 ) { //If One Record is Found
$stmt->fetch();
//Do Something With $email and $username from DB Here
}
else { //If More than One Record is Found
while( $stmt->fetch() ) { //Iterate Through Records
//Do Something With $email and $username from DB Here
}
}
}

mysqli - warning and error

i have this code but i got two errors. I put in the comments the errors
if(!empty($_POST['email']) && validateEmail($email)) {
$email = $_POST["email"];
if ($sql = $db->prepare("select email from users where email=?")) {
$sql->bind_param('s', $email);
$sql->execute();
$sql->bind_result($email);
while ($sql->fetch()) {
$salt = "PiuwrO1#O0rl#+luH1!froe*l?8oEb!iu)_1Xaspi*(sw(^&.laBr~u3i!c?es-l651";
$password = md5($salt . $userExists["email"]);
$pwrurl = "www.yoursite.com/reset_password.php?q=" . $password;
$mailbody = "Dear user,<br><br>If this e-mail does not apply to you please ignore it. It appears that you have requested a password reset at our website www.yoursitehere.com<br>
To reset your password, please click the link below. If you cannot click it, please paste it into your web browser's address bar.<br> <a href='$pwrurl'>$pwrurl</a> <br> <br>
Thanks,\nThe Administration";
$mail->MsgHTML($mailbody);
$mail->AddAddress("dxxb#hotmail.com","Nome da Pessoa");
$mail->IsHTML(true);
if(!$mail->Send()) {
echo "Deu erro: " . $mail->ErrorInfo;
} else {
echo "Enviado com sucesso";
}
}
$sql->close();
$db->close();
}
($sql = $db->prepare('insert into password_reset (code) values (?)')); // Warning: mysqli::prepare() [mysqli.prepare]: Couldn't fetch mysqli in
$sql->bind_param('s', $password); // Fatal error: Call to a member function bind_param() on a non-object
$sql->execute();
$sql->fetch();
$sql->close();
$db->close();
}
all code works fine, but now i need to insert the salt in the db but i can't, and i don't know why
thanks
Edited code to the last version
After you execute a query, fetch returns one result. There may be more -- there may be many, many more -- so you should be calling fetch in a loop to get them all. You aren't supposed to prepare a new query until you've finished dealing with the old one, which would usually mean fetching every row of the result and closeing (in your case) $sql. Otherwise, the database is still in the middle of answering one request when you're trying to issue another one.
The first error says it all - you can't have more than 1 prepared statement/query "in flight" at once. You've not finished fetching data from the first query (select email ...) when you tried to prepare another statement (insert into ...).

user registration php

<?php
include"include/connection.php";
$checkusername=mysql_query("SELECT * FROM employer WHERE eusername='$username'");
if (mysql_num_rows($checkusername)==1)
{
echo "username already exist";
}
else
{
$query = "insert into employer(efname,elname,egender,eemail,eusername,epwd,eadd,ephone,ecity,ecountry) values ('".$_POST['first_name']."','".$_POST['last_name']."','".$_POST['gender']."','".$_POST['email']."','".$_POST['username']."','".$_POST['password']."','".$_POST['address']."','".$_POST['phone']."','".$_POST['city']."','".$_POST['country']."')";
$result = mysql_query($query) or die (mysql_error());
echo " Thanks for registration";
}
?>
This is my code for inserting registration form data into a database. This code adds the data but also gives a parse error, but does not give the error if the username already exists.
Notice: Undefined variable: username in C:\Program Files\EasyPHP5.3.0\www\register_hirer2.php on line 6
Thanks for registration
line 6 is:
$checkusername=mysql_query("SELECT * FROM employer WHERE eusername='$username'");
Well, your $username is undefined indeed.
Most probably you want to use $_POST['username'].
And of course this obligatory XKCD comic:
If the "data source" is an html form (supposedly using method="post") you have to use $_POST['username'] when register_globals is set to off (which is the default since ...ages). see http://docs.php.net/security.globals
Also have a read of http://php.net/manual/en/security.database.sql-injection.php
<?php
include"include/connection.php";
$query = "SELECT
*
FROM
employer
WHERE
eusername='". mysql_real_escape_string($username). "'
";
$checkusername=mysql_query($query) or die(mysql_error());
if (mysql_num_rows($checkusername)==1)
{
echo "username already exist";
}
else
{
$query = "INSERT INTO employer(efname,elname,egender,eemail,eusername,epwd,eadd,ephone,ecity,ecountry) values (". same mysql_real_escape_string() thing here for each parameter .")";
$result = mysql_query($query) or die (mysql_error());
echo " Thanks for registration";
}
?>
You can also use prepared statements. This way you don't need/can't forget using an escaping function.
edit and btw: you don't need the SELECT before the INSERT in order to make the username unique. Actually it will make things even harder since now you have to deal with race conditions. You'd have to lock the table between those two queries.
If you add an unique index for the username in your table MySQL will not allow the insertion of a doublet but instead return a specific error code which your script can fetch and handle without the need of dealing with race conditions.
define('ER_DUP_ENTRY', 1062);
$mysql = mysql_connect('..', '..', '..');
mysql_select_db('..', $mysql) or die(mysql_error($mysql));
$fields = array(
'efname'=>'first_name',
'elname'=>'last_name',
'egender'=>'gender',
'eemail'=>'email',
'eusername'=>'username',
'epwd'=>'password',
'eadd'=>'address',
'ephone'=>'phone',
'ecity'=>'city',
'ecountry'=>'country'
);
$sqlparams = array();
foreach($fields as $sql=>$form) {
if ( !isset($_POST[$form]) ) {
die('missing post parameter '. $form);
}
$sqlparams[$sql] = "'".mysql_real_escape_string($_POST[$form], $mysql)."'";
}
$query = '
INSERT INTO
employer
'. join(', ', array_keys($sqlparams)) .'
VALUES
('.join(',', $sqlparams).')
';
// table:employer has been defined with "unique key idxName (eusername)"
$result = mysql_query($query, $mysql);
if ( false!==$result ) {
echo " Thanks for registration";
}
else if ( ER_DUP_ENTRY===mysql_errno($mysql) ) {
echo 'username already exists';
}
else {
echo 'an error occurred';
}
That is because you do not define $username anywhere. It looks as though you want to use $_POST['username']
mysql_query("SELECT * FROM employer WHERE eusername='{$_POST['username']}'");
Also, your code is vulnerable to a SQL Injection
You never define $usernameanywhere, so it gives that error because you are trying to use a variable that it doesn't have a value for.
This is most likely because you've not defined the '$username' variable. I presume that you're relying on this being populated from the incoming GET/POST data (most likely via the depreciated register_globals) which is bad practice.
As such, you'll need to either populate $username via $_POST or $_GET.
More importantly, you should update the insert query to escape the incoming 'untrusted' data using mysql_real_escape_string (e.g.: mysql_real_escape_string($_POST['username']), etc.)
As #Yacoby said your code is vulnerable to an SQL Injection to prevent it you can use mysqli or PDO , if you would like to use mysqli use the following code:
<?php
include"include/connection.php";
$query = "SELECT
*
FROM
employer
WHERE
eusername='". mysql_real_escape_string($username). "'
";
$checkusername=mysql_query($query) or die(mysql_error());
if (mysql_num_rows($checkusername)==1)
{
echo "username already exist";
}
else
{
$query = $conn->prepare("INSERT INTO employer(efname,elname,egender,eemail,eusername,epwd,eadd,ephone,ecity,ecountry)) values ( ? , ? , ? , ? , ? , ? , ? , ? , ? , ?)"; // preparing the insert
$query->bind_param("ssssssssss" , $variable1 , $variable2 , $variable3 , $variable4 , $variable5 , $variable6 , $variable7 , $variable8 , $variable9 , $variable10); // binding parameters
$query->execute(); // sending the parameter values
$query->close(); // closing the query
$conn->close(); // closing the connection
if ($query) { // checking if the query has been executed with no errors
echo " Thanks for registration";
}
}
?>
BE SURE TO CHANGE THE $conn AND variables to whatever you want!

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