SERVER RESPONSE
Error: INSERT INTO epiz_19848473_Liste1 (Rowcount, Level1) VALUES (0, 'any Value')
No database selected
PHP CODE
UPDATE
<?php
$servername = "sql308.epizy.com";
$username = "epiz_19848473";
$password = "huihuibuh";
$dbname = "Liste1";<--UPDATED
// Create connection
$conn = mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO Liste1 <--UPDATED (Rowcount, Level1)
VALUES (0, 'any Value')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
This should work, right?
NEW STATUS
Connection failed: Access denied for user 'epiz_19848473'#'192.168.0.%' to database 'Liste1'
The code is perfectly all right , the problem is with mysql configuration .
Please login to mysql as **root user ** and do the following,
GRANT ALL PRIVILEGES ON . TO 'epiz_19848473'#'192.168.0.%' (use ip of php server)
FLUSH PRIVILIGES;
The above allows the user to connect to mysql table from the ip specified.
you did not defined the database name in the connection
<?php
$servername = "sql308.epizy.com";
$username = "epiz_19848473";
$password = "huihuibuh";
$dbname = "DATABASE_NAME_HERE";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
else you can define define the database name in the connection
$conn = new mysqli($servername, $username, $password, "DATABASE_NAME_HERE");
Just define the dbname,like this:
<?php
$servername = "sql308.epizy.com";
$username = "epiz_19848473";
$password = "huihuibuh";
$dbname = "test"; //You should create this db in mysql
....
You Can Use This:
CREATE TABLE [database_name].[table_name] (
[your_table_things]
);
Related
I am trying to detect whether a MySql database has been updated the second it has been updated. So far I've just found triggers but it, https://www.siteground.com/kb/mysql-triggers-use/, said I need superuser privileges and, unfortunately, the domain I use is a shared server.
So currently there is one group working on inserting all user information into the database like so.
<form method="post">
-> user post data
</form>
<?php
-> post data to varibles, $x $y as examples
$servername = "localhost";
$username = "user";
$password = "pass";
$dbname = "name";
// Check connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
if(!$conn){
die("Connection failed: " . mysqli_connect_error());
}
$sql = "INSERT INTO Test (Col1, Col2)
VALUES (". $x .",". $y .")";
mysqli_close($conn);
?>
And then my group, in a separate file, is supposed to get the data from the database when it is updated. Our file set up looks like this.
<?php
$servername = "localhost";
$username = "user";
$password = "pass";
$dbname = "name";
// Check connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
if(!$conn){
die("Connection failed: " . mysqli_connect_error());
}
->trigger on column update
$sql = "SELECT Col1, Col2, time FROM Test";
$result = mysqli_query($conn, $sql);
->etc.
mysqli_close($conn);
?>
So far I've only found triggers that are in MySql but there doesn't seem to be a way to directly do it in PHP ex.
mysql> CREATE TRIGGER agecheck BEFORE INSERT ON people FOR EACH ROW;
I am facing an issue with PHP which doesn't perform one query in my script.
The SQL query works well in my MYSQL console but nothing is happening. Year column stays NULL:
$UpdateYear='UPDATE `pat` SET `Year` = SUBSTRING(`Prepa`,7,4)';
mysqli_query($connWarehouse,$UpdateYear) or die(mysqli_error($connWarehouse));
I don't know what I am doing wrong. Here the full script:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$db="datawarehouse";
// Create connection
$connWarehouse = new mysqli($servername, $username, $password, $db);
// Check connection
if ($connWarehouse->connect_error) {
die("Connection failed: " . $connWarehouse->connect_error);
}
$UpdateYear='UPDATE `pat` SET `Year` = SUBSTRING(`Prepa`,7,4)';
mysqli_query($connWarehouse,$UpdateYear) or die(mysqli_error($connWarehouse));
mysqli_close($connWarehouse));
?>
I finally managed to solve. I don't know if it is a correct way to do it. I have to open a new connection to the database in order to perform.
alter.php:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$db="datawarehouse";
// Create connection
$connWarehouse = new mysqli($servername, $username, $password, $db);
// Check connection
if ($connWarehouse->connect_error) {
die("Connection failed: " . $connWarehouse->connect_error);
}
$AddYear='ALTER TABLE `pat` ADD COLUMN `Year` YEAR;';
mysqli_query($connWarehouse,$AddYear) or die(mysqli_error($connWarehouse));
mysqli_close($connWarehouse));
include 'uppat.php';
?>
uppat.php
<?php
require_once 'config-datawarehouse.php';
$UpdateYear='UPDATE `pat` SET `Year` = SUBSTRING(`DatePrepa`,7,4)';
mysqli_query($conn,$UpdateYear) or die(mysqli_error($conn));
mysqli_close($conn);
?>
On that way it performed the update query.
I'm working in a web app with PHP and Postgresql. I`m looking for a PHP code to create a database online using a PHP form that sends the name, template, UTF8, etc.
Something like this for MYSQL but for with Postgresql:
<?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);
}
// Create database
$sql = "CREATE DATABASE myDB";
if ($conn->query($sql) === TRUE) {
echo "Database created successfully";
} else {
echo "Error creating database: " . $conn->error;
}
$conn->close();
?>
This question already has answers here:
Connect to Multiple Databases using MySQLi
(3 answers)
Closed 6 years ago.
Ok so this is my code for the connection:
$servername = "host";
$username = "user";
$password = "pass";
$db1 = "db1";
$db2 = "db2";
// Create connection
$conn = new mysqli($servername, $username, $password, $db1); // $db1 is here as the default, is this ok?
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
Then I run a function with this connection and it works for the $db1:
$sql = "SELECT etc etc;
$result = $conn->query($sql);
etc etc
The problem is when I try to change the db to $db2, I use this:
$conn->select_db($db2);
And now it only returns the value of the 2nd Database ($db2), and the one from $db1 doesn't show up on the page anymore, it doesn't show any value.
Thanks for Reading.
You have use two different connections while connecting different databases:
Database One Connection:
$conn = new mysqli($servername, $username, $password, $db1);
$sql = "SELECT etc etc;
$result = $conn->query($sql);
Database Two Connection:
$conn1 = new mysqli($servername, $username, $password, $db2);
$sql = "SELECT etc etc;
$result = $conn1->query($sql);
The Problem is that you have already assigned the conn to DB1 and hence if you run by selecting the DB2 it will be displaying error or no results will be produced. Hence while selecting multi databases you can use different connection variables.
the one from $db1 doesn't show up on the page anymore, it doesn't show any value.
what about making it $conn->select_db($db1); back?
Create two config files
db1.php
$servername = "host1";
$username = "user1";
$password = "pass1";
$db = "db1";
db2.php
$servername = "host2";
$username = "user2";
$password = "pass2";
$db = "db2";
//connection to db1
include("db1.php");
$conn = new mysqli($servername, $username, $password, $db);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//db1 codes
--------------------
//connection to db2
include("db2.php");
$conn = new mysqli($servername, $username, $password, $db);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//db2 codes
I am trying to connect to a MySQL database through a php script. It gives me this error from mysql_error(): Access denied for user '#localhost' to database 'userinfo'
userinfo is the database.
my script is this
<?php
$servername = "localhost";
$username = "root";
$password = "'mm'";
$database = "userinfo";
$conn = mysqli_connect($servername, $username, $password);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully<br>";
mysql_select_db($database) or die(mysql_error());
echo "connected successfully to db:" . $database . "<br>";
?>
you are connecting using
mysqli_
function
and selecting data with
mysql_
avoid using both at the same time. since they're incompatible.
use the mysqli_ alternative instead
mysqli_select_db($conn, $database);
and
mysqli_error($conn)
Please keep in mind this isn't the safest way. But since you have said your learning this it is a start.
<?php
$servername = "localhost";
$username = "root";
$password = "mm";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
?>
http://www.w3schools.com/php/php_mysql_connect.asp
To select data from the database
http://www.w3schools.com/php/php_mysql_select.asp
It appeared you where combining the old mysql in php with the new mysqli