I work with the following example-code:
<?php
// Create connection
$conn = mysqli_connect('mysql3.00*****', 'a7552070******', 'fjewifn****');
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "INSERT INTO diequizapp(appid, itemid, data) VALUES ('John', 'bon', 'jovi')";
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
?>
but I get an No Database selected back.
Can somebody please tell my why ?
I took this example from a tutorial.
Auf Wiedersehen, Andre
You need to pass the database that you want to select data from as the last parameter in mysqli_connect(host, username, password, database)
You can also take the second approach and use the mysqli_select_db($connection, DATABASE) function
you did not given the database name try like this
$dbname = "****"; //database name
$dbhost = "localhost"; // host name localhost
$dbusername = "***"; // username of the mysql
$dbpassword = "***"; // password of the mysql
$link = mysqli_connect($dbhost,$dbusername,$dbpassword,$dbname);
if (mysqli_connect_errno($link))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
then run your query........
Related
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 get an "no database selected" error and i can't figure out why. Would be nice if someone could help me out. Code below. There are no typos in there and im using XAMPP/Apache as server so localhost should be right i guess?
<!--Insert in database-->
<?php
$servername = "localhost";
$dbname = "databank";
$conn = mysqli_connect($servername, $dbname);
if(!$conn)
{
die("Connection failed: " . mysqli_connect_error());
}
$Kundennummer = $_POST["id"];
$Vorname = $_POST["vorname"];
$Nachname = $_POST["nachname"];
$plz = $_POST["plz"];
$strasse = $_POST["strasse"];
$hausnummer = $_POST["hausnummer"];
$sql = "INSERT INTO kundendaten (Kundennummer, ProduktID, Vorname,Nachname, Hausnummer, Strasse, PLZ)
Values ('$Kundennummer', '0', '$Vorname', '$Nachname', '$hausnummer', '$strasse', '$plz')";
if(mysqli_query($conn, $sql))
{
echo "DONE";
}
else
{
echo "ERROR: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
?>
Learn mysqli_connect() in php
The valid syntex is
mysqli_connect(host,username,password,dbname,port,socket);
You have forgot to add username and password into the mysqli_connect.
Please check below sample.
<?php
$con = mysqli_connect("localhost","mysqli-user","mysqli-password","databank");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>
I hope this might be helpful for you to resolve your issue.
I have very strange problem. I want to run mysql query as it is shown down below, but it's not working. Connection to database is successful, INSERT query is ok too, because when I run it directly in phpmyadmin Console it works, but it's not working here in PHP code.
Could you tell me what I'm missing?
$servername = "localhost";
$username = "admin";
$password = "admin123";
$dbname = "database1";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "INSERT INTO last_visit (ip, lastvisit) VALUES ('123', '123')";
You need to run your $sql, because now your $sql is only a string, it does nothing.
Add this :
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
I have a very simple bit of code to insert data into database.
It doesn't work. I do not get any error except cannot insert. I could connect to the database so that is not the issue.
The database has 3 fields, emailadd is the 2nd field. The other 2 are auto increment id and creation date, so also a field that on add, it will add current timestamp.
$inquiry = "INSERT INTO subscribe (emailadd) VALUES ('$myemail')";
$res = mysql_query($inquiry) or die("cannot insert");
$inquiry = "INSERT INTO `subscribe` (`emailadd`) VALUES ('$myemail')";
$res = mysql_query($inquiry,$con) or die('Not Inserted : ' . mysql_error());
Some time we need $con variable to identify the database connection, let see more check here
the mysql_query is deprecated, you need to use mysqli like so :
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "UPDATE MyGuests SET lastname='Doe' WHERE id=2";
if ($conn->query($sql) === TRUE) {
echo "Record updated successfully";
} else {
echo "Error updating record: " . $conn->error;
}
$conn->close();
?>
I want to create a database. Why is not the db created with this code?
$dbname = 'regulations_db';
$con = mysql_connect("localhost","root","pass");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
if (mysql_num_rows(mysql_query("SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = '". $dbname ."'"))) {
echo "Database $dbname already exists.";
}
else {
mysql_query("CREATE DATABASE '". $dbname ."'",$con);
echo "Database $dbname created.";
}
This is working, but I think the first one is the best practice:
if (mysql_query("CREATE DATABASE IF NOT EXISTS regulations_db",$con))
{
echo "Database created";
}
else
{
echo "Error creating database: " . mysql_error();
}
Just do a simple mysql_select_db() and if the result is false then proceed with the creation.
As an example, check out the first answer here by another very smart StackOverflower.
<?php
// Connect to MySQL
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
// Make my_db the current database
$db_selected = mysql_select_db('my_db', $link);
if (!$db_selected) {
// If we couldn't, then it either doesn't exist, or we can't see it.
$sql = 'CREATE DATABASE my_db';
if (mysql_query($sql, $link)) {
echo "Database my_db created successfully\n";
} else {
echo 'Error creating database: ' . mysql_error() . "\n";
}
}
mysql_close($link);
?>
Three steps to fix this:
Don’t specify the database name when connecting.
Your SQL statement should be CREATE DATABASE IF NOT EXISTS php1.
Call mysqli_select_db($link, 'php1') to make that the default database for your connection.
If you're using MySQLi Object-oriented method, you can use following code, this code is similar to previous answer and only the method is different, I just put this because if anyone using MySQLi Object-oriented method, you can use this code directly.
$servername = "localhost";
$username = "mysql_user";
$password = "user_password";
$dbName = "databaseName";
// Connect to MySQL
$conn = new mysqli($servername, $username, $password);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// If database is not exist create one
if (!mysqli_select_db($conn,$dbName)){
$sql = "CREATE DATABASE ".$dbName;
if ($conn->query($sql) === TRUE) {
echo "Database created successfully";
}else {
echo "Error creating database: " . $conn->error;
}
}
Furthermore you can refer W3school site here.
Good Luck! :D