I am creating a messaging website, which uses a MySQL database to store the message data. However, when there are only 5 users online at the same time, it starts failing with "MySQLi: Cannot Connect: Too many MySQL connections".
I have heard that my host, HelioHost, limits the maximum MySQL connections to four, which explains my error (see this for details).
I want to know how I can change my scripts (which currently connect when the message form is submitted, add it to the database, then disconnect) can only use a maximum of four connections over the server.
Here's some code:
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("MySQL server connection failed: " . $conn->connect_error);
}
//Process data
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$name = $_SESSION["username"];
$parent = $_POST["thread"];
$content = test_content($_POST["content"]);
$sql = "INSERT INTO `threads_replies`(`Parent`, `Author`, `Content`) VALUES ('" . $parent . "','" . $name . "','" . $content . "')";
$conn->query($sql) or die ("Failed MySQL query to save reply");
echo "Reply saved. <span class=\"fa fa-chevron-circle-left\"></span> Back to Thread";
}
function test_input($data) {
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
function test_content($data) {
$data = str_replace("'", "'", $data);
return $data;
}
Any ideas?
Right from the docs:
http://php.net/manual/en/function.mysqli-connect.php
$link = mysqli_connect("127.0.0.1", "my_user", "my_password", "my_db");
if (!$link) {
$error = mysqli_connect_error();
echo "Failed to connect to MySQL: $error";
echo 'Sorry, pal, 5 already playing around...';
exit;// or location('wait.php');
}
# ....
mysqli_close($link);
For the 5th user - he should reload the page, you can redirect him to something like wait.php:
<!-- htm, head, please add.. -->
<?php
// no need in PHP, actually
?>
<h3>Sorry, having too much users, will reload automatically...</h3>
<script>
setTimeout(
()=> location.href =
location.href + '?t=' + (new Date()).getTime(),
5000 // hope 5 seconds will be enough
);
So your code:
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
location('wait.php');
die("MySQL server connection failed: " . $conn->connect_error);
}
You can build a cache system which will be checked before accessing the database (and build up a connection). As an example, you don't need to read the news on the front page every time from the database, they don't change that often. So you read the news from the cache system (like some file in the file system) instead and display the data from it. And when the news entries has been changed you delete/invalidate the cache so the cache gets rebuild (once) on the next visit again.
Related
I use the hosting company aPlus.net, and I can't seem to get past a connection error I'm getting when trying to process some php to write database content to a webpage, and I am curious as to if this is because my database appears to not be on the same server as the entire rest of my hosting account, and if there is a way to resolve this in my code? This is my first attempt at writing PHP, and it would be good to know if my code is wrong, or if my hosting company is messing me up. (and either way, how to fix it)
Here's the code that's failing to pull from the database:
{
$con = mysql_connect("localhost","2p5dq9vxmy240651","MY_PASSWORD");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("felineasthma_2p5dq9vxmy240651", $con);
$users_name = $_POST['name'];
$users_comment = $_POST['requests'];
$users_name = mysql_real_escape_string($users_name);
$users_comment = mysql_real_escape_string($users_comment);
$inputid = $_GET['id'];
$query = "
INSERT INTO `felineasthma_2p5dq9vxmy240651`.`submissions` (`id`,
`name`, `requests`, `inputid`) VALUES (NULL,
'$users_name', '$users_comment', '$inputid');";
mysql_query($query);
echo "<h2>Your request has been processed, reload page.</h2>";
mysql_close($con);
}
and here's some screen captures from inside my hosting account (links because I don't have enough posts here yet to upload images, sorry):
felineasthma_2p5dq9vxmy240651 doesn't appear in my hosting account
yet it clearly exists in MySQL Manager, but on a different server
I was even more confused while making the user for this database, as the control panel didn't allow me to make a username, it just randomly assigned one. Help? Advice?
I found a more modern tutorial to learn PHP with, and now everything works, I just need to add security measures now. Here's the working code snippets, if anyone ever comes here asking the same questions.
here's the form action that places the entries into the database:
<?php
$servername = "sql5c40n.carrierzone.com";
$username = "my_username";
$password = "my_password";
$dbname = "my_database";
$users_name = $_POST['name'];
$users_request = $_POST['requests'];
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "INSERT INTO submissions (name, requests)
VALUES ('$users_name', '$users_request')";
if (mysqli_query($conn, $sql)) {
header("Location: clv.php");
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
?>
here's the include that puts the database entries onto the page:
<?php
$servername = "sql5c40n.carrierzone.com";
$username = "my_username";
$password = "my_password";
$dbname = "my_database";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT id, requests, name FROM submissions";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "" . $row["requests"]. " - by " . $row["name"]. "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
?>
I've looked all over here. Please be patient as I am new to php and mysql.
I got WAMPP installed & seems to be working OK. I created a simple "test" database from phpMyAdmin and "firsttable" in that. I can do a simple connect using example from w3schools, but trying to select & display data I entered only throws back errors.
<?php
$servername = "localhost";
$username = "root";
$password = "";
// Connect
$conn = mysqli_connect($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT reference, firstname, lastname, room FROM firsttable";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["reference"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "room:" . $row["room"]. "<br>";
}
} else {
echo "0 results";
}
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$conn->close();
?>
First off, I get a parse error on line 17. The one that reads:
if ($result->num_rows > 0) {
The error says: Trying to get property of non-object.
I tried wrapping the whole php code in tags and saving it as html, but then it appeared that no row data was ever found.
I am able to use very simple code that connects successfully. I can confirm the database is in there, so is the table, and the contents I added to it.
Please, what am I doing wrong?
You need to specify the database when you connect:
$database = 'test';
$conn = mysqli_connect($servername, $username, $password, $database);
where $database is the name of your database (test in this case). MySQL doesn't know which database your table resides in without you telling it.
In addition, you should always include error checking for your database connection (you have two of these, you don't need the last one) as well as any queries. Sans this, you can check your error logs for more information when something fails.
I have read a bunch of the other posts about similar issues but I am very new to this so I am likely missing something. Most of the other questions I found were a lot more complicated than mine. I am trying to follow the w3schools tutorial for this and am testing locally using XAMPP. Right now I am just trying to get this to work successfully and eventually I am planning to submit data into a mySQL database from a web form.
<?php
$servername="localhost";
$dbname="mysql";
// Create connection
$conn = mysqli_connect($servername, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "INSERT INTO owner_table (ownerFirst, ownerLast, mobile)
VALUES ('John', 'Doe', '1111111111')";
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
?>
NOTE:
The connection is not failing from what I can tell. Removing everything after the first if statement, and adding else { echo "success" } outputs success. Again though, I am new to PHP.
If someone knows that the answer has been given in another post PLEASE PLEASE comment here and ill close this out.
http://www.w3schools.com/php/php_mysql_insert.asp
You should follow the proper syntax:
$conn = mysqli_connect(<host Name>, <username>, <password>, <database name>);
Reference: http://php.net/manual/en/function.mysqli-connect.php#refsect1-function.mysqli-connect-examples
Add two more parameters- username and password here:
$username = "root"; // Default values
$password = ""; // Default values
$conn = mysqli_connect($servername, $username, $password, $dbname);
I just made my own wamp server on my computer to handle twitch donations for a friend. The friend has a program that I made using c# that connects to the my server and reads the database. This works fine, the program reads my database just fine and lists the entries. BUT I have made a php website to send the entry (a new donation) to the database but I receive a connection timed out error just seconds after I press the button... my mysql server is set to allow every user and host with the correct password. I used the same IP user and password as on the c# program!
I can't figure out why the damn php website hosted on one.com can't connect!!
this is my code:
<?php
if (!empty($_GET['act'])) {
$servername = "81.83.95.34";
$username = "root";
$password = "mypassword";
$dbname = "tbc";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("server failed: " . $conn->connect_error);
}
$sql = "INSERT INTO donations (donations)
VALUES ('1')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
} else {
?>
can anyone help me?
thank you very much!
Attempting to run a php script to update a database every hour with windows 7 Task Scheduler. The File updates when i open it up as a webpage but does not update when i run it from task scheduler . Under the trigger tab I have it set to run daily, and then repeat the task every hour for a duration of indefinetly. Under actions I have it set to Start Program. And the link to my file as C:\Users\jmac\Desktop\testdbconnection.php
Additionally under Start in box I have C:\Users\E15913\Desktop
I have no coditions that should be blcoking it from running. In addition when I test run the file it says it is completed succefully but the database is not updated. I will also post the code below, which once again works when I refresh it or pull it up in a web browser.
<?php
// Did modify login values for privacy
$servername = "10.100.";
$username = "myusername";
$password = "abc123";
$dbname = "informationdata";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$file = fopen("ALMGrade.csv","r");
while(! feof($file)){
$ar =fgetcsv($file);
$sql = "INSERT INTO gradetable_copy (Grade, Grade1, Grade2, Grade3, Grade4, Grade5, Grade6)
VALUES ('$ar[0]', '$ar[1]', '$ar[2]', '$ar[3]', '$ar[4]', '$ar[5]', '$ar[6]' )";
echo $sql;
echo "<br>";\
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
}
fclose($file);
$conn->close();
?>