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PHP - Prevent Database Connection As A Parameter To All Function
HI, i'm using a class to create a database connection etc..
So I have something like this in a common.inc.php file which is include in each page
$db_numbers_args = array();
$db_numbers_args['host'] = "localhost"; $db_numbers_args['user'] = "puser"; $db_numbers_args['pass'] = "nuser"; $db_numbers_args['db'] = "nu_name"; $db_numbers = new db($db_numbers_args,TRUE); From there I can end up making query's like $res = $db_numbers->query($uqs); So that said i'm including another file that contains all functions, when the functions need to do database transactions I always need to put $db_numbers as a paramter to the function and each function I make.. How should I make this better ? Similar TutorialsHi, So I'm trying to use another mysql connection to complete a function of mine: Code: [Select] /** * This function checks the ban status of the account. * @return 1 if banned */ function checkBan() { require("./includes/wow.php"); $result = mysql_query("SELECT * FROM wow_logon.accounts WHERE forum_acc= '.$user->data['user_id'].', $connect"); $row = mysql_fetch_array($result); if($row["banned"] == "1") { return 1; $ban_reason = $row["banreason"]; } //$user->data['user_id'] mysql_close($connect); } Inside my "wow" file, I have this: Code: [Select] <?PHP // Connect to db (edit this vars) $conf["host"] = "**"; $conf["user"] = "**"; $conf["password"] = "**"; $conf["db"] = "**"; $connect = mysql_connect($conf["host"],$conf["user"],$conf["password"]) or die(mysql_error()); mysql_select_db($conf["db"],$connect) or die(mysql_error()); ?> When executing my function I get: [function.mysql-query]: Access denied for user 'nobody'@'localhost' (using password: NO) So, I guess that my function doesn't use the ", $connect" parameter for some reason, any tips? after cloasing connection of database i still got the values form database. Code: [Select] <?php session_start(); /* * To change this template, choose Tools | Templates * and open the template in the editor. */ require_once '../database/db_connecting.php'; $dbname="sahansevena";//set database name $con= setConnections();//make connections use implemented methode in db_connectiong.php mysql_select_db($dbname, $con); //update the time and date of the admin table $update_time="update admin set last_logged_date =CURDATE(), last_log_time=CURTIME() where username='$uname'limit 3,4"; //my admin table contain 5 colums they are id, username,password, last_logged_date, last_log_time $link= mysql_query($update_time); // mysql_select_db($dbname, $link); //$con=mysql_connect('localhost', 'root','ijts'); $result="select * from admin where username='a'"; $result=mysql_query($result); mysql_close($con); //here i just check after closing data baseconnection whether i do get reselts but i do, why? echo "after the cnnection was closed"; if(!$result){ echo "cont fetch data"; }else{ $row= mysql_fetch_array($result); echo "id".$row[0]."usrname".$row[1]."passwped".$row[2]."date".$row[3]."time".$row[4]; } // echo "<html>"; //echo "<table border='1' cellspacing='1' cellpadding='2' align='center'>"; // echo "<thead>"; // echo"<tr>"; // echo "<th>"; // echo ID; // echo"</th>"; // echo" <th>";echo Username; echo"</th>"; // echo"<th>";echo Password; echo"</th>"; // echo"<th>";echo Last_logged_date; echo "</th>"; // echo "<th>";echo Last_logged_time; echo "</th>"; // echo" </tr>"; // echo" </thead>"; // echo" <tbody>"; //while($row= mysql_fetch_array($result,MYSQL_BOTH)){ // echo "<tr>"; // echo "<td>"; // echo $row[0]; // echo "</td>"; // echo "<td>"; // echo $row[1]; // echo "</td>"; // echo "<td>"; // echo $row[2]; // echo "</td>"; // echo "<td>"; // echo $row[3]; // echo "</td>"; // echo "<td>"; // echo $row[4]; // echo "</td>"; // echo "</tr>"; // } // echo" </tbody>"; // echo "</table>"; // echo "</html>"; session_destroy(); session_commit(); echo "session and database are closed but i still get values from doatabase session is destroyed".$_SESSION['admin']; ?> session is destroyed but database connection is not closed. thanks <?php class UserQuery { public function Adduser($id,$username,$email,$password) { $conn = new Config(); $sql =("INSERT INTO test.user (id, username, email, password) VALUES ('$id', '$username', '$email',$password)"); $conn->exec($sql); } }
getting an "exec doesnt exist " error, saying exec doesnt exist in my db file. it doesnt need to exist does it ? anyone any idea why ?
hey i need help im tryig to get information from my user and then process it in my database so i can use it to log them to a different web site im trying to use this method to get the information from the user but need help to get it please help me. Code: [Select] //make the database connection. $conn = mysql_connect("localhost", "Black Jack"); mysql_select_db("chaper7", $conn); //create a query $sql = "SELECT * FROM hero"; $result = mysql_query($sql, $conn); On some occasions I need to connect to a second and third database in the same script (maybe 5% of scripts have at least a second connection). Usually I would just select the new database. However, my host requires different users to be created for each database. What is the best way to do this? Close current connection (say db1) and open new (say db2) OR keep all open, creating 2nd and 3rd connections. I am happy with the design of my database, and don't want to merge all these tables into one db. Overall I am still happy with my host, so I'd rather not change. I'm connecting to my database using the following... @ $db = new mysqli('host', 'username', 'password', 'database') The .php file that is connecting to the database is in my root (htdocs) folder on the server. I know that I am not supposed to put my actual 'host', 'username', 'password', 'database' inside the mysqli function for security purposes. I know that I am supposed to put variables in instead. But here is where I am confused. Where do I set those variables? Do I set them in another file and include that file? If so, where do I store the file that holds the passwords, and what prevents a hacker from simply navigating to that file? Thanks for the help Through out my program I have used global $mysqli; as a connection to my database - this has worked fine so far. Now I have called $sql_statement = "SELECT * FROM items WHERE name='$itemName'"; $itemStats = mysqli_fetch_array(mysqli_query($mysqli, $sql_statement), MYSQLI_ASSOC); via a function and I get the following warnings - Quote Warning: mysqli_query() expects parameter 1 to be mysqli, null given in /functions/getdata.php on line 27 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in /functions/getdata.php on line 27 it still works though and returns the correct information. If I take the code out of the function and use it normally then I get no Warnings - but this defeats the whole purpose of having functions! How do I get rid of the warnings? (and no, I don't mean turn the warnings off ) Hi, I am currently trying to build an Artist's website, the artist wanted a CMS that was completely customized to the site (so no Wordpress, Joomla, Drupel, etc) - because of this I am having to create a CMS completely from scratch.
The problem I am having is with the database connection (hence the topic title), other sites that I have built with this same code work fine - however this particular site does not seem to want to play ball. It keeps giving me this error:
SQLSTATE[HY000] [1045] Access denied for user 'web113-janesart'@'10.0.44.113' (using password: YES)I have tried obvious things such as spelling mistakes, wrong password/username/db name, nothing seems to get rid of this error. Any help on what else this could be would be appreciated. The Script:
pdo_connect.php:
<?php $dsn = "mysql:host=localhost;dbname=2354"; $db = new PDO($dsn, "root", ""); ?>pdo_test.php:
<?php
try{
require_once("pdo_connect.php");
}catch(Exception $e){
$error = $e->getMessage();
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Database Connection with PDO</title>
</head>
<body>
<h1>Connection with PDO</h1>
<?php
if($db){
echo "<p>Connection successful.</p>";
}elseif(isset($error)){
echo "<p>$error</p>";
}
?>
</body>
</html>
I have just watched a tutorial and tried out this script. The issue is that I am getting the following notice alongside with the error message:
Notice: Undefined variable: db in C:\xampp\htdocs\oophp\pdo_test.php on line 18 SQLSTATE[HY000] [1049] Unknown database '2354'By the way this notice does happen in the tutorial as well. My question is: How to have this in ways, where the notice does not occur? Hey,
I'm really new to PHP and having some difficulties with $_SESSION and getting userid from the database. I've managed to put content to my database and also a login script. Though, adding sessions has been a pain. Here's what I got so far:
$sql = "SELECT username, password FROM users WHERE username = '$username' and password = '$pas'";
$query_login = $db->prepare($sql);
$query_login->execute(array('userid' => $userid, 'username' => $username, 'password' => $pas));
$result = $query_login->rowcount();
if ($result>0)
{
session_start();
$_SESSION['username'] = $username;
$_SESSION['logged'] = 1;
$_SESSION['userid'] = $result['userid'];
header('Location: ../user/user.php');
}
Hi guys, how do I use a connection file (connection.php) in multiple programs? I think include? Dear friends i,m a php beginner and i got a problem with connecting to my database i created a database called (koora) with one table called (admins) and when i tried to connect to it (database ) ; it did not connect here is the code i used for that <?php $connectdb = mysql_connect('localhost','','') or die("not connected"); $selectdb = mysql_select_db("koora", $connectdb); if(!$selectdb) { die("error connecting table" .mysql_error()); } then when refreshing my phpmyadmin page i got that message error connecting tableAccess denied for user ''@'localhost' to database 'koora' koora is the name of the database so i need your help with this problem and what is the reason not to connect to the data base Thank you I get the following error when trying to connect Warning: mysql_connect() [function.mysql-connect]: Can't connect to MySQL server on '72.18.129.104' (10061) in C:\Domains\crysvis.com\wwwroot\include\dbConnectAndSelect.php on line 8 Here's the code $host = "72.18.129.104"; $user = "deltron"; $pass = "masterconn"; $db = "customers"; $conn = mysql_connect("$host","$user","$pass"); if(!$conn) errorHandler("Msg 10:\nCould not connect to database with $host,$user,$pass in ". $_SERVER["PHP_SELF"]."\nCustId = $CustId\nmysql_error() = ".mysql_error()); if(!mysql_select_db("$db",$conn)) errorHandler("Msg 11:\nCould not select db=$db in ". $_SERVER["PHP_SELF"]."\nCustId=$CustId\nmysql_error() = ".mysql_error()); ?> What am I doing wrong? Any help will be appreciated Hi, For school, I need to build a site and this site has to have a few applications. One of the applications, I have to make is to make a script that makes it possible to upload a file and it has to put this file in a online database. Now, I'm just trying to put it in a local database. My code doesn't give an error, but it doesn't work either. The file doesn't get into the database. I can't figure out why. I hope you can help me? Thanks in advance! Lize When you open a database connection, how long does it stay open for? I have a "Change E-mail Address" script that has 4 queries in it, and I just realized that I don't create a DB Connection until Query #2 in my script, yet things work fine?! Is it possible that the DB Connection was opened earlier by another script and it is just persisting?? Debbie Hello guys. Trying to connect php with mysql database and then display results on the screen. This is my code: Code: [Select] <?php $dbhost = "localhost"; $dbuser = "username1"; $dbpass = "password1"; $db = "username1_myDB"; $connection = mysql_connect($dbhost, $dbuser, $dbpass) or die ("Could not connect"); mysql_select_db($connection, $db); $show = "SELECT Name, Description FROM people"; $result = mysql_query($show); while($show = mysql_fetch_array($result)){ $field01 = $show[Name]; $field02 = $show[Description]; echo "id: $field01<br>"; echo "description: $field02<p>"; } ?> However im getting this: Warning: mysql_select_db() expects parameter 1 to be string, resource given in /home/pain33/public_html/index.php on line 20 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/pain33/public_html/index.php on line 26 Any ideas how to fix this? Thank you. <?php require_once('upper.php'); require_once('database.php'); require_once('LoginStatement.php'); $LoginId=$_COOKIE['LoginIdCookie']; $query="select * from activity_participation where LoginId='$LoginId'"; $result=mysqli_query($dbc,$query) or die ('Not Connected'); //echo "<div class='search_output_data'> echo "<table border='0' cellspacing='0' cellpadding='0'>"; while($row=mysqli_fetch_array($result)) { echo "<tr><th align='left' valign='top'>Activity Title</th> <th align='left' valign='top'>Date of Participation</th> </tr> <tr><td colspan='4' align='left' valign='top'><table align='left' border='0' cellspacing='0' cellpadding='0'><tr><td align='left' valign='top'>".$row['Title']."</td> <td align='left' valign='top'>".$row['Date']."</td> </tr></table> </td></tr>"; } echo "</table>"; //echo "</div>"; require_once('lower.php'); ?>In table activity_participation many rows have same values. I want to display only different rows to user not to display same rows again and again. Or Anybody have idea to prevent the row duplication in database.... Help me please......... thanks in advance............. im using a function which connects to a db called 'comments' and then inside that function i again called another function that will connect to the db 'main' to get avatars.... but as i put Code: [Select] mysql_select_db("main") or die(mysql_error()); on the new function, the original function stops to fetch rows. i tried to remove "mysql_select_db("main") or die(mysql_error());" and used the new function to return some text only and it worked, so i guess the connection to the db was the problem... i tried doing Code: [Select] mysql_query("SELECT * FROM main.avatar INNER JOIN main.list ON main.avatar.title=main.list.id WHERE main.avatar.page_id='$id'"); but it also didnt work |
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