PHP MySQL unable to jump to row 0 - php

I recently started studying webdevelopment and for the first project we have to make a very simple login and registration function. now I got somewhere but suddenly I started getting this error.
Unable to jump to row 0 on MySQL result index 4.
This is the code where I get the error.
$conn = mysql_connect($servername, $username, $password, $database) or die("!Server");
if(!$conn){
die("Connection error: " . mysql_error());
}else{
echo "connection made";
}
$select_db= mysql_select_db($database, $conn);
if(!$select_db){
die("Database selection failed:: " . mysql_error());
}else{
echo "database selected";
}
$login_user = $_GET['username'];
$select_password = "SELECT `password` FROM `members` WHERE `username` = '$login_user'";
$result = mysql_query($select_password);
if(!$result){
die("Could not fetch the data " . mysql_error());
}
$password = mysql_result($result, 0);
echo $password;
I know there are other posts asking the same question but none of them where really helpful.
hope someone is able to help me.
Thanks for reading :)

It is possible that this row:
password = mysql_result($result, 0);
Cannot get rows when it does not match anything which is why you get the error.
Try instead:
if( $password = mysql_fetch_assoc($result) ) {
echo $password['password'];
} else {
echo 'no match';
}
See if that helps.
However, you really should move from mysql_* to PDO or mysqli.

I suggest to work with PDO, instead of mysql_... functions.
Not only because it's newer and object oriented but also because mysql... is deprecated.
So my answer is with PDO
From your code example I assume you have variables of servername, database, username & password. So to connect with PDO will look like this:
$con = new PDO("mysql:host=$servername;dbname=$database", $username, $password);
Then you can take your $_GET parameter using filter_input, which is also more suggested. Simple assignment look like this:
$login_user = filter_input(INPUT_GET,'username');
and then to run & fetch query like this:
$result_query = $con->query("SELECT `password` FROM `members` WHERE `username` = '$login_user'");
$result = $result_query->fetch(PDO::FETCH_ASSOC);
Now $result is an associative array, so to get password just use:
echo $result['password'];

Thanks alot for the help :) out of these answers I got everything I needed and now I am also going to switch to PDO for the final release though I am gonna stay with MySQL for now since I have to turn it in by monday.
Thanks alot :)

Change this:
$select_password = "SELECT `password` FROM `members` WHERE `username` = '$login_user'";
Into this:
$select_password = "SELECT password FROM members WHERE username = '$login_user'";

Related

Can't fetch data from database table.... 500 error

I have tried a ton of different versions of this code, from tons of different websites. I am entirely confused why this isn't working. Even copy and pasted code wont work. I am fairly new to PHP and MySQL, but have done a decent amount of HTML, CSS, and JS so I am not super new to code in general, but I am still a beginner
Here is what I have. I am trying to fetch data from a database to compare it to user entered data from the last page (essentially a login thing). I haven't even gotten to the comparison part yet because I can't fetch information, all I am getting is a 500 error code in chrome's debug window. I am completely clueless on this because everything I have read says this should be completely fine.
I'm completely worn out from this, it's been frustrating me to no end. Hopefully someone here can help. For the record, it connects just fine, its the minute I try to use the $sql variable that everything falls apart. I'm doing this on Godaddy hosting, if that means anything.
<?php
$servername = "localhost";
$username = "joemama198";
$pass = "Password";
$dbname = "EmployeeTimesheet";
// Create connection
$conn = mysqli_conect($servername, $username, $pass, $dbname);
// Check connection
if (mysqli_connect_errno) {
echo "Failed to connect to MySQL: " . mysqi_connect_error();
}
$sql = 'SELECT Name FROM Employee List';
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
echo "Name: " . $row["Name"]. "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
?>
There be trouble here:
// Create connection
$conn = mysqli_conect($servername, $username, $pass, $dbname);
// Check connection
if (mysqli_connect_errno) {
echo "Failed to connect to MySQL: " . mysqi_connect_error();
}
There are three problems here:
mysqli_conect() instead of mysqli_connect() (note the double n in connect)
mysqli_connect_errno should be a function: mysqli_connect_errno()
mysqi_connect_error() instead of mysqli_connect_error() (note the l in mysqli)
The reason you're getting a 500 error is that you do not have debugging enabled. Please add the following to the very top of your script:
ini_set('display_errors', 'on');
error_reporting(E_ALL);
That should prevent a not-so-useful 500 error from appearing, and should instead show the actual reason for any other errors.
There might be a problem here:
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
If the query fails, $result will be false and you will get an error on the mysqli_num_rows() call. You should add a check between there:
$result = mysqli_query($conn, $sql);
if (!$result) {
die('Query failed because: ' . mysqli_error($conn));
}
if (mysqli_num_rows($result) > 0) {
The name of your database table in your select statement has a space in it. If that is intended try:
$sql = 'SELECT Name FROM `Employee List`';
i think you left blank space in your query.
$sql = 'SELECT Name FROM Employee List';
change to
$sql = "SELECT `Name` FROM `EmployeeList`";

How to SELECT column value FROM table?

Here's my code:
<?php
//recently added
$result = mysql_query("SELECT background FROM " . $shadowless_background_table . " WHERE id = 1");
if ($result == 1){
?>
<script>
jQuery(document).ready(function(){
jQuery(".eltdf-psc-slide").addClass("no-background");
});
</script>
<?php
}
//=============
?>
Basically what I'm trying to do is checking and see if the value stored in the $shadowless_background_table "DB" is == 1 and I only want that column (background). I have browse the web, but what I see are examples with while loops which I was wondering if I could do something like this instead.
If you want to fetch a single record based on a condition you can do this -
$result = mysql_query("SELECT background FROM " . $shadowless_background_table . " WHERE id = 1");
if (mysql_num_rows($result)>0){
$fetchedColum = mysql_result($result, 0, 'COLUMN_NAME');
}
There are couple of issues with your code.The first thing that i have noticed is that you are using mysql API instead of PDO.I don't blame you since the internet is full of old tutorials and you probably didn't have a chance to get some guidance.
MySql is getting old It doesn't support modern SQL database concepts such as prepared statements, stored procs, transactions etc... and it's method for escaping parameters with mysql_real_escape_string and concatenating into SQL strings is error prone and old fashioned.
Organize your project better.
As i have seen from this example you probably have a poor project organization.You should consider reading about PSR Standards
And to go back to your question ,and to update it a bit.
Instead of doing
mysql_query("SELECT background FROM " . $shadowless_background_table . " WHERE id = 1");
I would do it this way:
<?php
$host = "localhost";
$username = "user name of db";
$password = "password of db";
$dbname = "database name ";
try {
$conn = new PDO("mysql:host=$host;dbname=$dbname", $username, $password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
//your data
$id = 1; // id
$stmt = $conn->prepare("SELECT background FROM database_name WHERE id=:id");
$stmt->bindParam(':id', $id);
$stmt->execute();
$data = $stmt->fetchAll();
foreach ($data as $row) {
echo $row["row_name"];
}
}
catch(PDOException $e)
{
echo "Error: " . $e->getMessage();
}
Go read more about PHP in general ,it will help you out a lot.The biggest problem is that there are so much wrong tutorials and references or they are just old.And people learn from wrong sources.
I had the same problem ,but thanks to right people on this site i have managed to learn more.
My suggestion is that you read about PSR,PDO and PHP in general!!!
Also a thing you should consider reading about is security in php.
Good luck mate :D

PHP: json_encode() doesn't show anything with multidimensional array

I'm trying to make a simple PHP script that fetches a table from my MySQL database and encodes the results in JSON, so I can use them later in Java.
This is my code:
<?php
$servername = "localhost:3036";
$username = "example_user";
$password = "example_password";
$conn = mysql_connect($servername, $username, $password);
if(! $conn) {
die("Could not connect: " . mysql_error());
}
$sql = "SELECT * FROM table_name";
mysql_select_db("database_name");
$retval = mysql_query($sql, $conn);
if(! $retval) {
die("Could not get data: " . mysql_error());
}
while($row = mysql_fetch_assoc($retval)) {
$output[]=$row;
}
print(json_encode($output));
mysql_close($conn);
?>
This just gives a blank page as output (error messages are set to display).
However, if I change json_encode($output) to json_encode($output[0]) (or any other number within the array's bounds), the output becomes that one $row array.
This is probably a really stupid question, but after about 3 hours of research I'm at my wit's end. Thank you for any help.
User #Joni led me to the solution.
Adding mysql_set_charset("utf8") fixed my issue.
As mentioned in this post: Why is this PHP call to json_encode silently failing - inability to handle single quotes?.
Try
echo json_encode($output) ;
It seems you have some utf8 character in your result set
add this statement before running your query
mysql_query('SET CHARACTER SET utf8');
Update
"mysql"
to
"mysqli"
and add
mysqli_set_charset($variável_de _conexão, 'utf8');
below the connection variable

how to store mysql select statement inside php variable using mysqli

I know this has been asked before but I cant seem to fix my code.
What I need is to run some php code to query mysql using mysqli for a select statement to retrieve my bcrypt hashed pass so I can compare the user input with the user hashed password. NOTE: I have not yet added mysql_real_escape_string to my $POST variables.
I've changed this code a thousand times still cant get it.
Ive even copy and pasted to a new file a simple query script using num_row
and printf($row['pass']); used echo etc..... I've used fetch array ive tried almost everything I've been all via php mysql at php.net w3c.com etc etc is my system broke? Does mysqli have a bug ? and no i dont want to switch to PDO I wont stop til this is fixed and when there is no longer sql injection vulns
Heres my code:
<?php
$conn = new mysqli('localhost', 'root', '', 'social');
if (mysqli_connect_errno())
{
exit("connection failed" . mysqli_connect_error());
}
else
{
echo "connection established";
}
$db=mysqli_select_db( $conn,'social');
if ($_POST && isset($_POST['submit'], $_POST['password'], $_POST['email']))
{
$pass = ($_POST["password"]);
$email =($_POST["email"]);
$bcrypt = password_hash($pass, PASSWORD_BCRYPT, array('cost' => 12));
}
$query = "SELECT `pass` FROM `social` WHERE `email` = 'jargon#jargon'";
$fetcher = mysqli_fetch_assoc($query);
echo $fetcher;
if ($conn->query($fetcher) === TRUE)
{
echo "query has gone through now we need to store the hash<br /> for comparison";
}
else
{
echo "error did not retrieve hash info";
}
$query = "SELECT `pass` FROM `social` WHERE `email` = 'jargon#jargon'";
$fetcher = mysqli_fetch_assoc($query);
Before you can fetch records from the result of the query, you need to actually perform the query. Your code should be
$query = "SELECT `pass` FROM `social` WHERE `email` = 'jargon#jargon'";
$result = $conn->query($query); // This is where the query is executed
$fetcher = $result->fetch_assoc();
Two more points.
First, you don't need to call mysqli_select_db; you've already selected the database in your constructor call, so you only need to call mysqli_select_db if you want to access a different database.
Second, instead of calling mysql_real_escape_string you should look into using prepared statements, which do the same thing and also correctly handle type-matching and quoting.
Try fetching the value from database after executing the query
$query = "SELECT `pass` FROM `social` WHERE `email` = 'jargon#jargon'";
$executedQuery = $conn->query($query);
if($executedQuery) {
$fetcher = mysqli_fetch_assoc($executedQuery);
echo "query has gone through ---------";
} else {
echo "error did not retrieve hash info";
}
$query = "SELECT pass FROM social WHERE id = 11"; // took the (``) out of the query and added this im assuming the value is stored in the $row variable and I may be able to use $row with the user input to verify hash via bcrypt!!!
$result = $conn->query($query);
while($row = mysqli_fetch_array($result))
{
echo $row['pass'];
echo "<br />";
}

Jquery/PHP ajax login system

I'm setting up a blog type page for my business. Brand new to MySQL and PHP. Set up this login system. For some reason have no idea why the login is dropping. Suppose to check for errors then return 'good' through php if the email and password is right. If php returns good then it's suppose to redirect to the blog page.
Been dealing with this for a few months need desperate help please. Thank you.
Here is the php code that goes along with the jquery.
Link to test site is here.
test.toddprod.com/login
Would really appreciated the help.
Thanks
<?php
#fake mysql connection first
DEFINE ('DB_USER','usernamegoeshere');
DEFINE ('DB_PASSWORD','passwordhere');
DEFINE ('DB_HOST','hostnamehere');
DEFINE ('DB_NAME','andtheotherthinghere');
$dbc = mysql_connect (DB_HOST, DB_USER, DB_PASSWORD) or die ('Could not connect to MySQL');
$db = mysql_select_db(DB_NAME, $dbc) or die('Could not select database.'.mysql_error());
$e = $_POST['email'];
$pass = $_POST['pass'];
$q = 'SELECT user_id from toddprod where email="'.$e.'" and pass= SHA1("'.$pass.'")';
$r = mysql_query($db, $q);
if( mysql_num_rows($r)==1 ){
setcookie ( 'user_id', $r);
setcookie ( 'email', '$e');
setcookie ( 'logged-in', 'true');
echo 'good';
}
else if (mysql_num_rows($r)==0) {
echo 'Your '.$e.' with password '.$pass;
};
mysql_close ($db);
?>
Umm there's a number of things I see wrong here...
First of all your query should be sanitized...
$email = mysql_real_escape_string ($_POST['email']); // escape the email
$pass = SHA1(mysql_real_escape_string ($_POST['pass'])); // escape and encrypt the pass
// now you can put it into the query safely
$query = "SELECT user_id from toddprod where email = '$email' and pass = '$pass' ";
Next you're executing the query wrong, the mysql_query function takes two arguments, the query and the database connection. You're passing the wrong arguments, you're passing the query and the result of the mysql_select_db function which is just a boolean value. So you have to $dbc not $db into that query, and even then you're passing the arguments in the wrong order. The query goes first, than the connection. So it should be...
$result = mysql_query($query, $dbc);
Next you're trying to set the return value from the mysql_query function as your cookie but that value is a resource, not the userid that you need. You have to actually read the value from the resource like this.
$row = mysql_fetch_array($result);
$userid = $row["user_id"];
setcookie('user_id', $userid);
Moving on... when you're setting the email cookie, you have the variable in single quotes, so the cookie will actually contain $e and not the actual email because single quotes store strings litterly (without parsing the variables). So you should either use double quotes, or no quotes at all. So either one of the following is fine...
setcookie('email', "$e");
setcookie('email', $e);
Last but not least, you should not have the semicolon at the end of your if-statement, and you again you need to pass the connection not the database-selection result into the mysql_close function, so it should be
mysql_close($dbc);
There, hope this gets you somewhere, try out these changes and if the problem persists i'd be happy to help you further.
Here are links that will help you out:
http://www.php.net/manual/en/function.mysql-query.php
http://www.php.net/manual/en/function.mysql-fetch-array.php
http://www.php.net/manual/en/function.mysql-real-escape-string.php
Edit:
Here, I have fixed the code according to the problems I found. Try it out, I could not test it so it might have some small syntax errors here and there, but it should give you something to compare with. Also for the future, I would suggest that you name your variables semantically/properly so it's easier for others to pickup and it will also keep you from getting confused like you were passing $db instead of $dbc into a few of your functions.
<?php
// keep the function names in lowercase, no reason, just looks better to me
define('DB_USER', 'usernamegoeshere');
define('DB_PASSWORD', 'passwordhere');
define('DB_HOST', 'hostnamehere');
define('DB_NAME', 'andtheotherthinghere');
// connect to the mysql server
$conn = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD) or die ('Could not connect to MySQL');
// select the database, you don't need to store the result, it just returns true or false
mysql_select_db(DB_NAME, $conn) or die('Could not select database.' .mysql_error());
// escape the input
$email = mysql_real_escape_string($_POST['email']);
$pass = sha1(mysql_real_escape_string($_POST['pass']));
// create the query
$query = "SELECT user_id FROM toddprod WHERE email = '$email' AND pass = '$pass'";
// execute the query
$result = mysql_query($query, $conn);
$usercount = mysql_num_rows($result);
if($usercount == 1){
// read the results and get the user_id
$row = mysql_fetch_array($result);
$userid = $row['user_id'];
// set the cookies
setcookie('user_id', $userid);
setcookie('email', $email);
setcookie('logged-in', 'true');
// echo success message
echo 'good';
}elseif($usercount == 0) {
echo "You're $email with password $pass";
}
mysql_close($conn);
?>
First things first, you MUST sanitise user input with mysql_real_escape_string():
$e = mysql_real_escape_string ($_POST['email']);
$pass = mysql_real_escape_string ($_POST['pass']);
Read up a bit on SQL injection, you'll be very glad you did.
As for the main problem, could you provide a bit more context? How are you checking to see if the user is logged in?

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