I am using my own db for phpbb3 forum, and I wish to insert some data from the forum into my own tables. Now, I can make my own connection and it runs my query but in trying to use the $db variable(which I think is what you're meant to use??) it gives me an error.
I would like someone to show me the bare bones which i insert my query into to be able to run it.
Well.. You haven't given us very much information, but there are two things you need to do to connect and query to a database.
For phpbb, you may want to read the documentation they have presented:
http://wiki.phpbb.com/Database_Abstraction_Layer
Here is a general overview of how you'd execute a query:
include($phpbb_root_path . 'includes/db/mysql.' . $phpEx);
$db = new dbal_mysql();
// we're using bertie and bertiezilla as our example user credentials. You need to fill in your own ;D
$db->sql_connect('localhost', 'bertie', 'bertiezilla', 'phpbb', '', false, false);
$sql = "INSERT INTO (rest of sql statement)";
$result = $db->sql_query($sql);
I presumed that phpBB already had a connection to my database. Thus I wasnt going to use a new one? Can i make a new one and call it something else and not get an error?
And $resultid = mysql_query($sql,$db345);
Where $db345 is the name of my database connection
$db = new dbal_mysql();
// we're using bertie and bertiezilla as our example user credentials. You need to fill in your own ;D
$db->sql_connect('localhost', 'bertie', 'bertiezilla', 'phpbb', '', false, false);
$sql = "INSERT INTO (rest of sql statement)";
$result = $db->sql_query($sql);
Related
I am attempting to create new tables every time I post to this method, but for some reason I can not figure out why it dies.
<?php
$host = "127.0.0.1";
$username = 'cotten3128';
$pwd = 'pwd';
$database = "student_cotten3128";
$pin = $_REQUEST['pinSent'];
$words = $_REQUEST['resultSent'];
$tableName = $pin;
$db = new mysqli($host, $username, $pwd, $database);
if ($sql = $db->prepare("CREATE TABLE $pin (id INT(11) AUTO_INCREMENT);")) {
$sql->execute();
$sql->close();
}else{
echo $mysql->error;
die('Could not create table');
}
for($i=0;$i<count($words);$i++){
if($sql = $db->prepare("INSERT INTO ".$pin.$words[$i].";")) {
$sql->execute();
$sql->close();
}else{
echo $mysql->error;
die("Could not add data to table");
}
}
mysqli_close();
?>
Any help or insight would be greatly appreciated.
The intention of my post is to help you finding the issue by yourself. As you did not added much information I assume my post is helpful for you.
Based on the code you have shared I guess you mean one of your called die() functions is executed.
Wrong function call
As Jay Blancherd mentioned mysql_close is the wrong function. You rather have to use mysqli_close as you created a mysqli instance.
Beside of that mysql_* is deprecated and should not be used anymore.
Debugging Steps
Not only for this case but in general you should ask yourself:
Is there an error message available? (Frontend output, error log file, ...)
YES:
What's the message about?
Is it an error you can search for? E.g. via a search engine or the corresponding documentation?
Look up in the bug tracker (if available), by the software developer of the software you are using, and if it has not been reported yet report the issue.
NO: (if none error message available OR you cannot search for it as it is a custom error message)
Search in the files of the software you are using for the error message and start a core-debugging.
STILL NO SOLUTION?:
Ask on stackoverflow.com e.g. and tell your issue and the steps you have performed to find and fix the bug. Post only as much code as necessary plus use a proper format.
Debugging in your case:
In order to narrow down the scope. Which of the die() is executed? Depending on that echo the query to execute just before it actually is executed. Then copy the SQL query to an SQL editor and look at it syntax. After that you probably know the problem already.
i have two table in sql the 1st table is for account while the 2nd table is for testimonial . i am trying to update the two tables in single query. The update is successful if the account already have a testimonial but fails to update if the account has no testimonial yet .How can i fix this heres my code for the update ....
if(!$update=mysql_query(
"UPDATE
tblapplicant,
tbltestimonial
SET
tblapplicant.ImagePath='".$name."',
tbltestimonial.pic = '".$name."'
WHERE
tblapplicant.appid=tbltestimonial.appid"
)
)
1) You're working with a database, it defeats the purpose to use the same data being inserted into two different tables.
2) One gentleman also mentioned stop using MySQL... heres some reference code for you. Assuming you're using php.
3) If you want to use a single query to update 2 tables with the same info against recommendation. Use a stored procedure to update them both.
4) At which point are these account's interconnected in this query? I'm somehow intrigued if this system is in beta or testing?
With your "Where" conditions without matching a specific record, this will update every record that has a matching ID. This is highly not recommended until you add further conditions like username = .... or a condition that's specific to someone or a specific set of rows.
**I strongly advise you post the tables you're working with and what results you want achieve for the best advise. **
Can't really give a good consultation with you playing the whole overview close to the chest. Using this plain-Jane without further detail on what you're asking for is at your own risk my friend.
include/dbconnect.php optional recommended update
<?php
if (isset($mysqli)){
unset($mysqli);
}
define("HOST", "yo.ur.ip.addr"); // The host you want to connect to.
define("USER", "myfunctionalaccount"); // The database username.
define("PASSWORD", "superdoopersecurepassword!"); // The database password.
define("DATABASE", "thegoods");
$mysqli = new mysqli(HOST, USER, PASSWORD, DATABASE);
if ( $mysqli->connect_error ) {
die('Connect Error: ' . $mysqli->connect_error);
}
?>
functions.php <-- shouldn't be called functions if its going to be your form response
<?php
// SHOULD BE SOME MASSIVE LOGIC UP HERE FOR FORM DATA DECISIONING
include_once "include/dbconnect.php";
$name = addslashes($_FILES['image']['name']);
$image = mysql_real_escape_string(addslashes(file_get_contents($_FILES['image']['tmp_name'])));
if ($stmt = $mysqli->prepare("CALL UpdateTestimonials(?,?,?)"){
$stmt->bind_param($name, $image, $userid);
$stmt->execute();
// optional to show affected rows
$stmt->affected_rows
//
// use if you want to return values from DB
// $stmt->bind_result($result);
// $stmt->fetch;
}
$stmt->close
?>
MySQL build a stored procedure - fyi; definer is optional. Definer will allow you to run a query that only elevated privileges can access due to the safety of such a query. You can use create procedure w/o the definer parameter. dT is just an abbreviation for datatype. You would put varchar or int... etc..
use 'database';
DROP procedure if exists 'UpdateTestimonials';
DELIMITER $$
use 'UpdateTestimonials' $$
CREATE DEFINER='user'#'HOSTNAME/LOCALHOST/%' PROCEDURE 'mynewprocedure' (IN varINPUT varchar, IN varIMG blob, IN varAppID int)
BEGIN
UPDATE tblapplicant
SET imagepath = varINPUT,
pic = LOAD_FILE(varIMG)
WHERE appid = varAppID
END $$
DELIMITER;
Use LEFT JOIN:
if (!$update = mysql_query(
"UPDATE
tblapplicant
LEFT JOIN tbltestimonial ON tblapplicant.appid = tbltestimonial.appid
SET
tblapplicant.ImagePath = '" . $name . "',
tbltestimonial.pic = '" . $name . "'"
)
)
Also, if you need some additional filters for tbltestimonial, add them into LEFT JOIN condition
You could try with a transaction. BTW also please use prepared statements to prevent SQL Injection attacks.
<?php
// prefer mysqli over mysql. It is the more modern library.
$db = new mysqli("example.com", "user", "password", "database");
$db->autocommit(false); // begin a new transaction
// prepare statements
$update_applicant =
$db->prepare("UPDATE tblapplicant
SET tblapplicant.ImagePath = ?"));
$update_applicant->bind_param("s", $name));
$update_applicant->execute();
$update_testimonial =
$db->prepare("UPDATE tbltestimonial
SET tbltestimonial.pic = ?"));
$update_testimonial->bind_param("s", $name))
$update_testimonial->execute();
$db->commit(); // finish the whole transaction as successful,
// when everything has succeeded.
?>
Of course that would not create any testimonials, that don't exist. It just updates those, that do. When you want to insert new entries in tbltestimonial, do so explicitly with an INSERT statement inside the transaction.
MySQL does not fully support transactions. The tables have to use a table type, that can handle them, e.g. innodb. When that is the case, the transaction will make sure, that everyone else either sees all changes from the transactions, or none.
In many cases transactions allow you to perform a group of simple steps, that otherwise would need complex single queries or would not be possible without transactions at all.
Alternative Solution
Another approach of course would be an update-trigger. Create a trigger in your database, that fires whenever e.g. tblapplicant is updated and updates tbltestimonial accordingly. Then you don't have to care about that in your application code.
I'm currently doing a school project and I'm using dreamweaver along with a backend database using phpMyAdmin.
Now, what i need to do is, when I click the button, it will reduce the stock column value in the "products" table by 1.
However there are different products in the table. Shown below:
http://i.stack.imgur.com/vLZXQ.png
So lets say, A user is on the game page for "Destiny" and clicks on the Buy now button, how can i make it reduce the stock level by one, but only for the Destiny record and not for the Fifa 15 column. So Destiny stock becomes 49, but Fifa stays 50. Will i just need to make each button have a different script or?
Currently, I made a button in the page, which links to an action script, but im not sure what sort of code i will be using.
Thank you
xNeyte is giving you some good advice, but it comes across to me that you - Xrin - are completely new to programming database contents with PHP or similar?
So some step by steps:
MYSQL databases should be connected with one of two types of connection - PDO and MySQLi_ . MySQL databases will also always work using the native MySQL but as xNeyte already mentioned - this is deprecated and highly discouraged .
So what you have is you pass your information to the PHP page, so your list of games is on index.php and your working page that will update the number of games ordered would be update.php, in this example.
The Index.php file passes via anchor link and $_GET values (although I highly recommend using a php FORM and $_POST as a better alternative), to the update.php page, which needs to do the following things (in roughly this order) to work:
Update.php
Load a valid database login connection so that the page can communicate with the database
Take the values passed from the original page and check that they are valid.
establish a connection with the database and adjust the values as required.
establish the update above worked and then give the user some feedback
So, step by step we'll go through these parts:
I am going to be a pain and use MySQLi rather than PDO - xNeyte used PDO syntax in his answer to you and that is fully correct and various better than MySQLi, for the sake of clarity and your knowledge of MySQL native, it may be easier to see/understand what's going on with MySQLi.
Part 1:
Connection to the database.
This should be done with Object Orientated - Classes,
class database {
private $dbUser = "";
private $dbPass = ""; //populate these with your values
private $dbName = "";
public $dbLink;
public function __construct() {
$this->dbLink = new mysqli("localhost", $this->dbUser, $this->dbPass, $this->dbName);
}
if (mysqli_connect_errno()) {
exit('Connect failed: '. mysqli_connect_error());
}
if ( ! $this->dbLink )
{
die("Connection Error (" . mysqli_connect_errno() . ") "
. mysqli_connect_error());
mysqli_close($this->dbLink);
}
else
{
$this->dbLink->set_charset("UTF-8");
}
return true;
} //end __construct
} //end class
The whole of the above code block should be in the database.php referenced by xNeyte - this is the class that you call to interact with the database.
So using the above code in the database.php object, you need to call the database object at the top of your code, and then you need to generate an instance of your class:
include "database.php"; ////include file
$dataBase = new database(); ///create new instance of class.
Now When you write $dataBase->dbLink this is a connection to the database. If you do not know your database connection use the details PHPMyAdmin uses, it carries out its tasks in exactly the same way.
Sooo
Part 2:
That is that your database connection is established - now you need to run the update: First off you need to check that the value given is valid:
if (is_numeric($_GET['id']) && $_GET['id'] >0 ){
$id = (int)$_GET['id'];
}
This is simple code to check the value passed from the link is a integer number. Never trust user input.
It is also a good idea never to directly plug in GET and POST values into your SQL statements. Hence I've copied the value across to $id
Part 3:
$sql = "UPDATE <TABLE> SET STOCK = STOCK-1 WHERE Product_ID = ? LIMIT 1";
The table name is your table name, the LIMIT 1 simply ensures this only works on one row, so it will not effect too many stocked games.
That above is the SQL but how to make that work in PHP:
first off, the statement needs to be prepared, then once prepared, the value(s) are plugged into the ? parts (this is MySQLi syntax, PDO has the more useful :name syntax).
So:
include "database.php"; ////include file
$dataBase = new database(); ///create new instance of class.
if (is_numeric($_GET['id']) && $_GET['id'] >0 ){
$id = (int)$_GET['id'];
$sql = "UPDATE <TABLE> SET STOCK = STOCK-1 WHERE Product_id = ? LIMIT 1";
$update = $dataBase->dbLink->prepare($sql);
$update->bind_param("i",$id);
$update->execute();
$counter = $update->affected_rows;
$update->close();
//////gap for later work, see below:
}
else
{
print "Sorry nothing to update";
}
There's probably quite a lot going on here, first off the bind_param method sets the values to plug into the SQL query, replacing the ? with the value of $id. The i indicates it is meant to be an Integer value. Please see http://php.net/manual/en/mysqli-stmt.bind-param.php
The $counter value simply gets a return of the number of affected rows and then something like this can be inserted:
if ($counter > 0 ){
print "Thank you for your order. Stock has been reduced accordingly.";
}
else {
print "Sorry we could not stock your order.";
}
Part 4
And finally if you wish you can then just output the print messages or I tend to put the messages into a SESSION, and then redirect the PHP page back.
I hope this has helped a bit. I would highly recommend if you're not used to the database interactions in this way then either use PDO or MySQLi but do not combine the two, that will cause all sorts of syntax faults. Using MySQLi means that everything you know MySQL can do, is done better with the addition of the letter "i" in the function call. It is also very good for referencing the PHP.net Manual which has an excellent clear detailed examples of how to use each PHP function.
The best is to set a link on each button with the ID of your game (1 for destiny, 2 for Fifa15).
Then your script which the user will launch by clicking will be :
<?php
include('database.php'); // your database connection
if($_GET['id']) {
$id=$_GET['id'];
} else throw new Exception('Invalid parameter');
$statement = myPDO::getInstance->prepare(<<<SQL
UPDATE TABLE
SET STOCK = STOCK-1
WHERE Product_id = :id
SQL
);
$statement->execute(array(":id" => $id));
This script will do the job
I have been writing php and mySQL functions all day and as I was writing the simplest part of my project I have hit a wall.
The function should simply count how many entries are in the database and return that number (If there is a more simple way please let me know, this is my first php + mysql project)
Here is the code:
function quoteCount(){
global $db;
$totalQuoteNum = array();
$query = "SELECT * FROM Quotes";
$result_set = mysqli_query($db, $query)
or die ("Query $query failed ".mysqli_error($db)); //fails here
$totalQuoteNum = mysql_num_rows($result_set)
or die ('couldnt count rows'.mysqli_error($db));
echo 'COUNTED EVERYTHING!!!';
return $totalQuoteNum;
};
Now when the die statement prints I get the string but not the mysqli error.
Things I have tried and ruled out:
$db is correct
query works in mysql
I wasnt sure if the database was connected, so I added the connect inside this function and stil nothing.
Any ideas? From what I see it should work and its not giving me any error to work from. Please help!
Based on the comments, it seems as though $db is the database name.
Functions such as mysqli_query() expect a database link (resource), not simply the database name.
This resource is created by constructing a new mysqli object. Following your procedural style, use mysqli_connect().
I have this code:
if(!mysql_connect($host,$user,$passwd)){
die("Hoops, error! ".mysql_error());
}
...no error from here.
if(!mysql_select_db($db,$connect)){
$create_db = "CREATE DATABASE {$db}";
mysql_query($create_db,$connect);
mysql_query("USE DATABASE {$db}",$connect);
}
..."no database selected" error from here.
I would like to select database if it exists and if doesn't then create it and select it.
Why is my code not right?
Thank you in advance
Where are you saving the value returned by mysql_connect()? Don't see it here. I assume $host, $user, $password and $db are properly set ahead of time. But you're passing a param to mysql_select_db that may not be properly set.
$connect = mysql_connect($host,$user,$passwd);
if (!$connect) {
die('Could not connect: ' . mysql_error());
}
if(!mysql_select_db($db,$connect)) ...
Start by checking to see if you can select without the CREATE query first. Try a simple SELECT query to start. If you can connect, select the db, and execute a SELECT query, that's one step. Then try the CREATE query. If that doesn't work, it's almost certainly a matter of permissions.
You might need database create permissions for the user attempting to create the database.
Then you need to operate on a valid connection resource. $connect never looks to be assigned to the connection resource.
Why not simply use the CREATE DATABASE IF NOT EXISTS syntax instead?
Something like this ...
$con = mysql_connect('localhost');
$sql = 'CREATE DATABASE IF NOT EXISTS {$db}';
if (mysql_query($sql, $con)) {
print("success.\n");
} else {
print("Database {$db} creation failed.\n");
}
if(!mysql_select_db($db,$connect)){
print("Database selection failed.\n");
}
You should check the return value of mysql_query() - currently if any of those calls fail you won't know about it:
if(!mysql_select_db($db,$connect)){
if (!mysql_query("CREATE DATABASE $db", $connect)) {
die(mysql_error());
}
if (!mysql_select_db($db, $connect)) {
die(mysql_error());
}
}
Change the line
mysql_query($create_db,$connect);
mysql_query("USE DATABASE {$db}",$connect);
To
mysql_query($create_db,$connect);
mysql_select_db($db);*
and it should work.
you could try w3schools website. They have a very simple and easy to learn tutorial for selecting database. The link is : http://www.w3schools.com/php/php_mysql_select.asp
Hope this help :)
I would like to thank to all of you, however I found fault on my side. This script was in class and one of variables were not defined inside this class. So I'm really sorry.
I don't know how to consider the right answer, but I noticed my mistake after reading Clayton's answer about not properly set parameters, so I guess he is the winner ;)