When submitting the following form I am getting this error :
Fatal error: Call to undefined function mysqli_connect() in .... mailing_list_include.php on line 7
Here's the mailing_list_include.php file - the real thing includes the correct credentials for accessing the db
<?php
function doDB() {
global $mysqli;
// connect to server and select database; you may need it
$mysqli = mysqli_connect("localhost", "username",
"password", "db");
// if connection fails, stop script execution
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
}
function emailChecker($email) {
global $mysqli, $check_res;
//check that email is not already in list
$check_sql = "SELECT id FROM SUBSCRIBERS
WHERE email = '".$email."'";
$check_res = mysqli_query($mysqli, $check_sql)
or die(mysqli_error($mysqli));
}
?>
This means your copy of PHP was not compiled with mysqli support. This doesn't mean there is anything wrong with your code, you are simply trying to use a function that PHP does not have available.
See Installation - Mysqli on PHP.net for more information on configuring it. You will have to ask your host if they can rebuild PHP and include mysqli support.
Installation of the mysqli extension
Related
Warning: mysqli::query(): Couldn't fetch mysqli in C:\Program Files (x86)\EasyPHP-DevServer-13.1VC9\data\localweb\my portable files\class_EventCalendar.php on line 43
The following is my connection file:
<?php
if(!isset($_SESSION))
{
session_start();
}
// Create array to hold error messages (if any)
$ErrorMsgs = array();
// Create new mysql connection object
$DBConnect = #new mysqli("localhost","root#localhost",
NULL,"Ladle");
// Check to see if connection errno data member is not 0 (indicating an error)
if ($DBConnect->connect_errno) {
// Add error to errors array
$ErrorMsgs[]="The database server is not available.".
" Connect Error is ".$DBConnect->connect_errno." ".
$DBConnect->connect_error.".";
}
?>
This is my class:
<?php
class EventCalendar {
private $DBConnect = NULL;
function __construct() {
// Include the database connection data
include("inc_LadleDB.php");
$this->DBConnect = $DBConnect;
}
function __destruct() {
if (!$this->DBConnect->connect_error) {
$this->DBConnect->close();
}
}
function __wakeup() {
// Include the database connection data
include("inc_LadleDB.php");
$this->DBConnect = $DBConnect;
}
// Function to add events to Zodiac calendar
public function addEvent($Date, $Title, $Description) {
// Check to see if the required fields of Date and Title have been entered
if ((!empty($Date)) && (!empty($Title))) {
/* if all fields are complete then they are
inserted into the Zodiac event_calendar table */
$SQLString = "INSERT INTO tblSignUps".
" (EventDate, Title, Description) ".
" VALUES('$Date', '$Title', '".
$Description."')";
// Store query results in a variable
$QueryResult = $this->DBConnect->query($SQLString);
I'm not great with OOP PHP and I'm not sure why this error is being raised. I pulled this code from elsewhere and the only thing I changed was the #new mysqli parameters. Can anyone help me to understand what is going wrong?
Probably somewhere you have DBconnection->close(); and then some queries try to execute .
Hint: It's sometimes mistake to insert ...->close(); in __destruct() (because __destruct is event, after which there will be a need for execution of queries)
Reason of the error is wrong initialization of the mysqli object. True construction would be like this:
$DBConnect = new mysqli("localhost","root","","Ladle");
I had the same problem. I changed the localhost parameter in the mysqli object to '127.0.0.1' instead of writing 'localhost'. It worked; I’m not sure how or why.
$db_connection = new mysqli("127.0.0.1","root","","db_name");
Check if db name do not have "_" or "-"
that helps in my case
Warning: mysqli::query(): Couldn't fetch mysqli in C:\Program Files (x86)\EasyPHP-DevServer-13.1VC9\data\localweb\my portable files\class_EventCalendar.php on line 43
The following is my connection file:
<?php
if(!isset($_SESSION))
{
session_start();
}
// Create array to hold error messages (if any)
$ErrorMsgs = array();
// Create new mysql connection object
$DBConnect = #new mysqli("localhost","root#localhost",
NULL,"Ladle");
// Check to see if connection errno data member is not 0 (indicating an error)
if ($DBConnect->connect_errno) {
// Add error to errors array
$ErrorMsgs[]="The database server is not available.".
" Connect Error is ".$DBConnect->connect_errno." ".
$DBConnect->connect_error.".";
}
?>
This is my class:
<?php
class EventCalendar {
private $DBConnect = NULL;
function __construct() {
// Include the database connection data
include("inc_LadleDB.php");
$this->DBConnect = $DBConnect;
}
function __destruct() {
if (!$this->DBConnect->connect_error) {
$this->DBConnect->close();
}
}
function __wakeup() {
// Include the database connection data
include("inc_LadleDB.php");
$this->DBConnect = $DBConnect;
}
// Function to add events to Zodiac calendar
public function addEvent($Date, $Title, $Description) {
// Check to see if the required fields of Date and Title have been entered
if ((!empty($Date)) && (!empty($Title))) {
/* if all fields are complete then they are
inserted into the Zodiac event_calendar table */
$SQLString = "INSERT INTO tblSignUps".
" (EventDate, Title, Description) ".
" VALUES('$Date', '$Title', '".
$Description."')";
// Store query results in a variable
$QueryResult = $this->DBConnect->query($SQLString);
I'm not great with OOP PHP and I'm not sure why this error is being raised. I pulled this code from elsewhere and the only thing I changed was the #new mysqli parameters. Can anyone help me to understand what is going wrong?
Probably somewhere you have DBconnection->close(); and then some queries try to execute .
Hint: It's sometimes mistake to insert ...->close(); in __destruct() (because __destruct is event, after which there will be a need for execution of queries)
Reason of the error is wrong initialization of the mysqli object. True construction would be like this:
$DBConnect = new mysqli("localhost","root","","Ladle");
I had the same problem. I changed the localhost parameter in the mysqli object to '127.0.0.1' instead of writing 'localhost'. It worked; I’m not sure how or why.
$db_connection = new mysqli("127.0.0.1","root","","db_name");
Check if db name do not have "_" or "-"
that helps in my case
I'm in the process of upgrading from mysql to mysqli.
All my mysql code was procedural, and I'd now like to convert to OOP, as most mysqli examples online are in OOP.
The problem I'm having is that, with mysql, once I had set up a connection, I never had to inject that connection into any functions as arguments for mysql to be accessible in the function.
Here is my old connection code:
$location = "localhost";
$user = "rogerRamjet";
$pass = "bestPassInTheWorld";
$dbName = "myDBName";
$link = mysql_connect($location, $user, $pass);
if (!$link) {
die("Could not connect to the database.");
}
mysql_select_db("$dbName") or die ("no database");
And an example function that has access to the mysql connection, without $link needing to be injected into the function:
function getUser($data)
{
$data=mysql_real_escape_string($data);
$error = array('status'=>false,'userID'=>-1);
$query = "SELECT `user_id`, `user_email` FROM `myTable` WHERE `data`='$data'";
if ($result = mysql_query($query))
{
$row = mysql_fetch_array($result, MYSQL_ASSOC);
if ($row['user_id']!="")
{
return array( 'status'=>true, 'userID'=>$row['user_id'], 'email'=>$row['user_email'] );
}
else return $error;
}
else return $error;
}
And here's my new mysqli connection:
$mysqli=new MySQLi($location, $user, $pass, $dbName);
So, to upgrade the first line in the above function, I'd need:
$data = $mysqli->real_escape_string($data);
But that throws the error:
Undefined variable: mysqli
Does this mean that for any function needing access to $mysqli, I need to inject $mysqli as an argument into it, or is there a way for it to be accessible the way mysql is without injection?
I know I need to move to prepared statements, but this is just so I can get my head around mysqli basics.
Making the variable global is bad practice. The singleton pattern solves the issue of needing to share one instance of an object throughout an application lifecycle. Consider using a Singleton.
The crude solution would be global $mysqli; as first line of your function. But as hsan wrote, read about PHP variable scope
Warning: mysqli::query(): Couldn't fetch mysqli in C:\Program Files (x86)\EasyPHP-DevServer-13.1VC9\data\localweb\my portable files\class_EventCalendar.php on line 43
The following is my connection file:
<?php
if(!isset($_SESSION))
{
session_start();
}
// Create array to hold error messages (if any)
$ErrorMsgs = array();
// Create new mysql connection object
$DBConnect = #new mysqli("localhost","root#localhost",
NULL,"Ladle");
// Check to see if connection errno data member is not 0 (indicating an error)
if ($DBConnect->connect_errno) {
// Add error to errors array
$ErrorMsgs[]="The database server is not available.".
" Connect Error is ".$DBConnect->connect_errno." ".
$DBConnect->connect_error.".";
}
?>
This is my class:
<?php
class EventCalendar {
private $DBConnect = NULL;
function __construct() {
// Include the database connection data
include("inc_LadleDB.php");
$this->DBConnect = $DBConnect;
}
function __destruct() {
if (!$this->DBConnect->connect_error) {
$this->DBConnect->close();
}
}
function __wakeup() {
// Include the database connection data
include("inc_LadleDB.php");
$this->DBConnect = $DBConnect;
}
// Function to add events to Zodiac calendar
public function addEvent($Date, $Title, $Description) {
// Check to see if the required fields of Date and Title have been entered
if ((!empty($Date)) && (!empty($Title))) {
/* if all fields are complete then they are
inserted into the Zodiac event_calendar table */
$SQLString = "INSERT INTO tblSignUps".
" (EventDate, Title, Description) ".
" VALUES('$Date', '$Title', '".
$Description."')";
// Store query results in a variable
$QueryResult = $this->DBConnect->query($SQLString);
I'm not great with OOP PHP and I'm not sure why this error is being raised. I pulled this code from elsewhere and the only thing I changed was the #new mysqli parameters. Can anyone help me to understand what is going wrong?
Probably somewhere you have DBconnection->close(); and then some queries try to execute .
Hint: It's sometimes mistake to insert ...->close(); in __destruct() (because __destruct is event, after which there will be a need for execution of queries)
Reason of the error is wrong initialization of the mysqli object. True construction would be like this:
$DBConnect = new mysqli("localhost","root","","Ladle");
I had the same problem. I changed the localhost parameter in the mysqli object to '127.0.0.1' instead of writing 'localhost'. It worked; I’m not sure how or why.
$db_connection = new mysqli("127.0.0.1","root","","db_name");
Check if db name do not have "_" or "-"
that helps in my case
I didn't realize there was an OO way to use mysqli, so I built a class called DB. During __construct it takes the hostname, username, password, and database name. Given the following code:
$myDB = new DB("localhost", "user", "password", "database");
$myDBConnect = $myDB->connect();
if(!$myDBConnect) {
echo "<strong>The following error has occurred: " . $myDB->getError();
}
The variable obviously contains FALSE because this if statement is currently returning TRUE. Here is the method from the DB class:
public function connect() {
// Create connection
$this->dbConnx = mysqli_connect($this->dbHost, $this->dbUsername, $this->dbPassword, $this->dbName);
if(mysqli_connect_errno($this->dbConnx)) {
$this->dbError = mysqli_error($this->dbConnx);
return false;
}
}
I'm not getting any error detail. I tried adding or die(mysqli_error()); in the connect method, but it always just outputs the text from the file that $myDB is instantiated in. I also tried variations on the error reporting code, including having no argument in mysqli_connect_errno() and using $this->dbError = mysqli_connect_error() with and without the connection argument.
Is this needlessly complicating the OO way to use mysqli? or am I missing something simple that will allow me to move on using the code I've already got?
Thanks in advance for your time.
if is not variable
if(!$myDBConnect) {
^--remove variable sign here
EDIT:
your connection should be
$myDB = new mysqli("localhost", "user", "password", "database");
Maybe you forgot the return true statement in the "connect" method?
Also you should use PDO.
Your connect() function is returning false on failure and nothing on success, so it will always fail the if. Try adding return true; at the end of that function.