Can anyone tell me what is the problem with this line of PHP code? I am trying to insert values in a database.
<?php
// Some database detail
$host = 'localhost';
$username = 'root';
$password = '';
$database = 'mywebsite';
$con;
mysqli_query($con,"INSERT INTO members (Name, Password) VALUES ('Daniel', 'abc123')");
// Making connection
$con = mysqli_connect($host, $username, $password, $database) or die(mysqli_error());
?>
THE ERROR:
Notice: Undefined variable: con in C:\xampp\htdocs\database.php on line 13
Warning: mysqli_query() expects parameter 1 to be mysqli, null given in C:\xampp\htdocs\database.php on line 13
Use Following code:
You just need to connect to database before INSERT .
<?php
// Some database detail
$host = 'localhost';
$username = 'root';
$password = '';
$database = 'mywebsite';
$con = mysqli_connect($host,$username,$password,$database) or die("Error " . mysqli_error($con));
mysqli_query($con,"INSERT INTO members (Name, Password) VALUES ('Daniel', 'abc123')");
?>
You are trying to insert a query into the database before making a connection to it. Swith lines 13 and 11.
Check this ..This will help you
$host = 'localhost';
$username = 'root';
$password = '';
$database = 'mywebsite';
$con = mysqli_connect($host, $username, $password) or die(mysqli_error());
$db_selected = mysqli_select_db("mywebsite", $con);
$check= mysqli_query("INSERT INTO members (Name, Password) VALUES ('Daniel', 'abc123')");
if($check)
echo "inserted";
else
echo "not inserted";
webserver reads PHP code line by line from the top of the page.
it displays Undefined variable error on the variable because you use it before you define $con.
so when the webserver reach line 13, it doesn't know what is variable $con and then it throws the error.
My suggestion is the same with the others.
put the
$con = mysqli_connect($host, $username, $password, $database) or die(mysqli_error());
above this
mysqli_query($con,"INSERT INTO members (Name, Password) VALUES ('Daniel', 'abc123')");
Related
public function Fetch_num1($id) {
echo "Hi";
$sql = "SELECT * FROM editors where jid='$id'";
$result = mysqli_query($connection, $sql);
$nr = mysqli_num_rows($result);
return $nr;
}
Its shows warnings like:
mysqli_query() expects parameter 1 to be mysqli, null given
mysqli_num_rows() expects parameter 1 to be mysqli_result, null given
in
As Armali says in the comments, you are not making a database connection. Here is a typical connection code copied from w3schools (https://www.w3schools.com/php/php_mysql_connect.asp):
<?php
$servername = "localhost";
$username = "username";
$password = "password";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
?>
Replace localhost, username, and password with your db name, user name, and password, then replace $connection with $conn.
Also, I think where should be all caps: WHERE.
I got the tutorial from this link
https://icreateproject.info/2014/12/14/arduino-save-data-to-database/
The tutorial is about how to save data from Arduino to xampp database over the local network.
I follow everything until step 3. This is the PHP code:
<?php
// Prepare variables for database connection
$dbusername = "arduino"; // enter database username, I used "arduino" in step 2.2
$dbpassword = "arduinotest"; // enter database password, I used "arduinotest" in step 2.2
$server = "localhost"; // IMPORTANT: if you are using XAMPP enter "localhost", but if you have an online website enter its address, ie."www.yourwebsite.com"
// Connect to your database
$dbconnect = mysql_pconnect($server, $dbusername, $dbpassword);
$dbselect = mysql_select_db("test",$dbconnect);
// Prepare the SQL statement
$sql = "INSERT INTO test.sensor (value) VALUES ('".$_GET["value"]."')";
// Execute SQL statement
mysql_query($sql);
?>
I tried to run this command in the link as mentioned in the tutorial
http://localhost/write_data.php?value=100
This is the error I get
Fatal error: Uncaught Error: Call to undefined function mysql_pconnect() in C:\xampp\htdocs\write_data.php:11 Stack trace: #0 {main} thrown in C:\xampp\htdocs\write_data.php on line 11
The mysql library that you are using is deprecated in php5.5
http://php.net/manual/en/function.mysql-connect.php
Here's an updated version with mysqli_connect
$dbusername = "arduino";
$dbpassword = "arduinotest";
$server = "localhost";
$database = 'test';
$mysqli = mysqli_connect($server, $dbusername, $dbpassword,$database);
// please validate the $_GET['value']
$sql = "INSERT INTO test.sensor (value) VALUES ('".$_GET["value"]."')";
mysqli_query($mysqli, $sql);
new mysqli http://php.net/manual/en/function.mysqli-connect.php
Somehow i managed to connect and changed the value by typing and auto insert using this code
<?php
header("Refresh:5");
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
else {
echo "hello";
}
$value = $_GET['value'];
$sql = "INSERT INTO test.sensor (value) VALUES ($value)";
$sql = "INSERT INTO test.sensor (value)
VALUES ('255')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
And without this code, which is on line 28, i will get error so i have to put this code as well
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
Thank you everyone for helping really appreciate it.
I Cannot Check whether the username already exist in database. I gone through existing questions that were answered here. None of them solved my problem. When i executes, it displays "Cannot select username from table", which i given inside die block. Code Is given below.
<?php
$username = $_POST['user_name'];
$password = $_POST['pass_word'];
$host = "localhost";
$db_username = "root";
$db_password = "";
$db_name = "my_db";
//create connection
$conn = #new mysqli($host, $db_username, $db_password, $db_name);
if (isset($_POST["submit"]))
{
# code...
//check connection established or not
if ($conn->connect_error)
{
die("Not Connected to DB");
}
else
{
$query = "SELECT 'usernamedb' FROM 'registration' WHERE usernamedb='$username'";
$result = mysqli_query($conn, $query) or die('Cannot select username from table');
if (mysqli_num_rows($result)>0)
{
$msg.="This username already exist. try Another !!";
}
else
{
$insert = "INSERT INTO 'registration'('id', 'usernamedb', 'password') VALUES ([$username],[$password])";
$insert_result = mysqli_query($conn,$insert) or die('INSERTION ERROR');
}
}
$conn->close();
}
?>
Hope someone will answer me.
First of all you should not use those unescaped queries.
But regarding your question you have an SQL error on your queries. You quoted table name. "FROM 'registration'" should be "FROM registration".
This is my CODE, id do not know where is the mistake but this code does not create any information on the database.
<?php
$username = "root";
$password = "";
$hostname = "localhost";
$dbhandle = mysql_connect($hostname, $username, $password) or die("Could not connect to database");
$selected = mysql_select_db($dbhandle);
if(isset($_POST['user']) && isset($_POST['pass'])){
$user = $_POST['user'];
$pass = $_POST['pass'];
$query = mysql_query("SELECT * FROM users WHERE Username='$user'");
if(mysql_num_rows($query) > 0 ) { //check if there is already an entry for that username
echo "Username already exists!";
}else{
mysql_query("INSERT INTO users (Username, Password) VALUES ('$user', '$pass')");
header("location:begin.html");
}
}
mysql_close();
?>
Forget database name here.
change this:
$selected = mysql_select_db($dbhandle);
With
$selected = mysql_select_db($dbname,$dbhandle);
$dbhandle = mysql_connect($hostname, $username, $password) or die("Could not connect to database");
$selected = mysql_select_db($dbhandle);
In above code you are not passing any database name to use, you should pass a database name instead of connection link to mysql_select_db($dbhandle);
like
$db_selected = mysql_select_db('foo', $link);
For reference
http://php.net/manual/en/function.mysql-select-db.php
Hi first of all Please use mysqli or PDO as mysql is depreciated and completely removed from PHP7.
Now your problem . You are not included database name in your mysql_select_db. It should be
$selected = mysql_select_db($dbhandle , $databasename) or die(mysql_error($dbhandle));
Always remember try to echo error after any query this will solve your problem in many cases
I can't seem to find out why my PHP code won't allow me to store the info in MySQL when I click the submit button. Can someone tell me whats wrong with the code below?
I want to get it to the point where users submit their registration form and their info can get moved over to a MySQL database. Then, I want the users to be taken to my index.html file.
Any help is appreciated.
Thanks
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$mysql_database = "21st";
$conn = mysqli_connect($dbhost, $dbuser, $dbpass) or die("Could not connect database");
mysqli_select_db($conn, $mysql_database) or die("Could not select database");
$username = $_POST['username'];
$email = $_POST['email'];
$password = $_POST['password'];
$password2 = $_POST['password2'];
if ($password == $password2) {
$sql = "INSERT INTO members (username, email, password) VALUES ('$userrname','$email','$password')";
} else {
echo "Your passwords must match";
}
?>
Try this using mysqli_query to actually run the query. When you set $sql all that did was set a variable named $sql to the query. Also note you set the variable $userrname in the query when it should be $username as set from the $_POST directly above. My adjusted version of your code below:
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$mysql_database = "21st";
$conn = mysqli_connect($dbhost, $dbuser, $dbpass) or die("Could not connect database");
mysqli_select_db($conn, $mysql_database) or die("Could not select database");
$username = $_POST['username'];
$email = $_POST['email'];
$password = $_POST['password'];
$password2 = $_POST['password2'];
if ($password === $password2) {
// Set the query.
$sql = "INSERT INTO members (username, email, password)"
. " VALUES (?, ?, ?)"
;
// Bind the values to the query.
mysqli_stmt_bind_param($sql, 'sss', $username, $email, $password);
// Run the query.
mysqli_query($conn, $sql);
// Free the result set.
mysqli_free_result($sql);
// Close the connection.
mysqli_close($conn);
}
else {
echo "Your passwords must match";
}