In my script I link to a page that connects to my database :
include "connect.php";
connect.php
<?php
error_reporting(E_ERROR);
/* Allows PHP to connect to your database */
// Database Variables
$Host = "myhost";
$User = "username";
$Password = "password";
$DBName = "database";
// Connect to Database
$connect = mysql_connect($Host, $User, $Password)
or die ("Could not connect to server ... \n" . mysql_error ());
mysql_select_db($DBName)
or die ("Could not connect to database ... \n" . mysql_error ());
?
Then in another script I have an insert query:
include "connect.php";
$Link = mysql_connect($Host, $User, $Password);
$Query = "INSERT INTO mytable VALUES ('0','".mysql_escape_string($forename)."','".mysql_escape_string($surname)."', '".mysql_escape_string($username)."', '".mysql_escape_string($password)."', '".mysql_escape_string($email)."')";
if(mysql_db_query ($DBName, $Query, $Link)) {
$message = "You have successfully registered";
header("Location: register.php?message=".urlencode($message));
} else {
die("Query was: $Query. Error: ".mysql_error($Link));
}
}
}
Why is this necessary :
$Link = mysql_connect($Host, $User, $Password);
Hasn't the connection already been established?
There is no point in doing this, especially as mysql_* functions will assume the last opened connection if none is given.
However, even with two calls to mysql_connect, only one connection is made. From the docs:
If a second call is made to mysql_connect() with the same arguments, no new link will be established, but instead, the link identifier of the already opened link will be returned. The new_link parameter modifies this behavior and makes mysql_connect() always open a new link, even if mysql_connect() was called before with the same parameters.
So by default, the existing connection will be returned.
Related
So this is the error message:
PHP Deprecated: mysql_connect(): The mysql extension is deprecated and will be removed in the future: use mysqli or PDO instead
This is the affected piece of code:
class wdClient {
var $dbLink; // The database link
var $prefix; // Table prefix
var $script; // The script running
/**
* Construct a new directory object.
*/
function wdClient() {
error_reporting(E_ALL ^ E_NOTICE);
$this->prefix = WDDBPREFIX;
// Connect to the database
$this->dbLink = mysql_connect(WDDBHOST, WDDBUSER, WDDBPASSWD);
// Select the database
mysql_select_db(WDDBNAME, $this->dbLink) or die('Sorry, The site is currently unavailable!');
}
where WDDBPREFIX, WDDBHOST, WDDBUSER, WDDBPASSWD, WDDBNAME are already defined in a config file.
I have tried simply using mysqli_connect instead of mysql_connect but it's not working.
Note: Never use MySQL, use this method!
//MySQLi information
$db_host = "localhost";
$db_username = "username";
$db_password = "password";
//connect to mysqli database (Host/Username/Password)
$connection = mysqli_connect($db_host, $db_username, $db_password) or die("Error " . mysqli_error());
//select MySQLi dabatase table
$db = mysqli_select_db($connection, "table") or die("Error " . mysqli_error());
Good luck!
well, as pointed in here http://php.net/manual/en/function.mysqli-connect.php , you should make something like this:
$link = mysqli_connect("127.0.0.1", "my_user", "my_password", "my_db");
Apparently, in your case it will look like this:
$link = mysqli_connect(WDDBHOST, WDDBUSER, WDDBPASSWD, WDDBNAME);
And then you can continue with your code...
if (!link){
die('Sorry, The site is currently unavailable!');
} else{
// write your SQL here and fetch it
}
I am creating a form that will take a person's name and email and see if the email exists in the database already but I can't get the database to be selected. The mysql_error() won't display the error either. Is there something wrong with my code for selecting the database? All it shows is the hardcoded text "Could not select database because" then nothing. I replaced all the variables associated with the database with random fillers so as not to give away my info but everything I have is correct regarding that.
$host = "host";
$user = "user";
$password = "pass";
$database = "db";
$port = xxxx;
$table = "table";
// Create connection
$conn = mysqli_connect($host, $user, $password, $database, $port);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connection Successful";
mysqli_select_db($database)
or die("Could not select database because".mysqli_error());
// check if the username is taken
$check = "select email from $table where email = '".$_POST['email']."';";
$qry = mysqli_query($check)
or die ("Could not match data because ".mysqli_error());
$num_rows = mysqli_num_rows($qry);
if ($num_rows != 0) {
echo "Sorry, there the username $username is already taken.";
}
Don't call mysqli_select_db. You've already selected your database with the fourth parameter to mysqli_connect.
You only need to call mysqli_select_db if you want to access a different database after the connection has been established.
Also, Raptor is correct that if you call procedural mysqli_error() you must pass it a connection handle.
For Procedural style mysqli_error(), you must supply the MySQLi DB link as parameter:
mysqli_select_db($database)
or die("Could not select database because" . mysqli_error($conn));
But it's useless to call mysqli_select_db() as you already specified the DB schema in mysqli_connect().
Additionally, your SQL is vulnerable to SQL Injection attack. Always escape the parameter before putting into SQL statement, or use prepared statements.
try this code
<?php
$username = "your_name";
$password = "your_password";
$hostname = "localhost";
//connection to the database
$dbhandle = mysqli_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
echo "Connected to MySQL<br>";
?>
So I wanted to install an acp on my homepage. I uploaded all needed files on my webserver (I use FileZilla Client) and opened phpMyAdmin with xampp.
It connects the database but then it says
"Database does not exist."
I'm fairly new to working with phpMyAdmin and mySQL as well.
this is what my db.php looks like:
<?php
$host = "localhost"; // database server
$user = "akichuchan"; // db username
$pw = "cherryjuice"; // db password
$db = "acp"; // db name
$link = mysql_connect($host, $user, $pw, $db) or die ("No connection possible.");
mysql_select_db($db, $link) or die ("Database does not exist.");
?>
What did I do wrong?
Thanks in advance. :)
Try the below working code:
<?php
$host = "localhost"; // database server
$user = "akichuchan"; // db username
$pw = "cherryjuice"; // db password
$db = "acp"; // db name
$link = mysql_connect($host, $user, $pw) or die ("No connection possible.");
mysql_select_db($db, $link) or die ("Database does not exist.");
?>
NOTE: Don't use " mysql_ ". Better to use " PDO or Mysqli_ " ( safe and secure ) .
I have a problem with my code. I'm trying to add new post to the table events. I'm confused because I have used this code in other place on the same website (but it was using mysqli_query to register new user). mysqql_error returns "No database selected"
This is the code:
<?php
$add_title = $_POST['add_title'];
$add_happen = $_POST['add_happen'];
$add_created = date('Y-m-d');
$add_content = $_POST['add_content'];
$add_author = $_POST['add_author'];
//connect to
//localhost
$db_host = "localhost";
$db_username = "root";
$db_password = "";
$db_dbname = "zhp2";
$db_con = mysql_connect($db_host, $db_username, $db_password, $db_dbname);
$query = "
INSERT INTO events ( title, happen, created, content, author )
VALUES ( '$add_title', '$add_happen', '$add_created', '$add_content', '$add_author') )
";
$retval = mysql_query($query, $db_con);
if(! $retval ){
die('Could not enter data: ' . mysql_error());
}
else{
echo "Entered data successfully\n";
}
mysql_close($db_con);
//header('Location: ../../index.php?link=events');?>
I've tried to fix it using trial and error method playing with different combinations both mysql_query and mysqli_query
You are confusing mysql_connect and mysqli_connect functions in the way you pass those parameters. In your example:
$db_con = mysql_connect($db_host, $db_username, $db_password, $db_dbname);
you are passing a fourth parameter which is the database name but that wont work as you should only pass the three first (host,username,password) and then call mysql_select_db():
$db_con = mysql_connect($db_host, $db_username, $db_password);
mysql_select_db( $db_dbname, $db_con );
In mysqli which is the BETTER way of doing it since mysql_ functions are very vulnerable and being deprecated from php you could pass four elements like here:
$db_con = mysqli_connect($db_host,$db_username, $db_password, $db_dbname) or die("Error " . mysqli_error($link));
which is close to what you are trying to do, but in a correct mysqli_ way.
Well then, you need to select the database! ;) The fourth parameter of mysql_connect() is not the database name. You need to do this separate of connecting to the MySQL server.
Using mysql_select_db() function:
$db_host = "localhost";
$db_username = "root";
$db_password = "";
$db_dbname = "zhp2";
$db_con = mysql_connect($db_host, $db_username, $db_password );
mysql_select_db( $db_dbname, $db_con );
And of course all the obligatory warnings about SQL injection, sanitizing your data, deprecation of mysql_* functions.
You need to select which database to connect to using the mysql_select_db function:
// make $db_dbname the current db
$db_selected = mysql_select_db($db_dbname, $db_con);
if (!$db_selected) {
die ("Can't use $db_dbname : " . mysql_error());
}
See the PHP manual for more info: http://php.net/manual/en/function.mysql-select-db.php
Here's a MySQL pattern in PHP:
$username="username";
$password="password";
$database="username-databaseName";
// Opens a connection to a mySQL server
$connection=mysql_connect (localhost, $username, $password);
if (!$connection) {
die("Not connected : " . mysql_error());
}
// Set the active mySQL database
$db_selected = mysql_select_db($database, $connection);
if (!$db_selected) {
die ("Can\'t use db : " . mysql_error());
}
// Search the rows in the markers table
$query = some query
$result = mysql_query($query);
I tried to replace most of it with a mysqli pattern and then stick the query part at the bottom like so:
//Database Information
$db_host = "localhost"; //Host address (most likely localhost)
$db_name = "username-databaseName"; //Name of Database
$db_user = "username"; //Name of database user
$db_pass = "password"; //Password for database user
/* Create a new mysqli object with database connection parameters */
$mysqli = new mysqli($db_host, $db_user, $db_pass, $db_name);
GLOBAL $mysqli;
if ($mysqli->connect_errno) {
echo "<p>MySQL error no {$mysqli->connect_errno} : {$mysqli->connect_error}</p>";
exit();
}
// Search the rows in the markers table
$query = some query
$result = mysql_query($query);
However, I get this error message:
Invalid query: No database selected
What am I doing wrong?
First of all, you're calling mysql_query and not mysqli_query, like you intended to.
Second, since you're using the object oriented form, you need to call mysqli_query as a method:
$result = $mysqli->query($query);