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PHP - How To Establish Database Connection Using Odbc And Php
can anyone give me an example of how to establish an ODBC connection with PHP. I have a MS ACCESS database i want to open and read from
thanks Similar TutorialsHi, I hope you guys can help me out. I only know basic PHP and also SQL. I haven't tried using connecting database yet. My problem is that I'm trying to INSERT a data into an existing table using access but when it reaches $rs = odbc_exec(); it returns an error that the query was empty. Here is the code: Code: [Select] $conn=odbc_connect('trial','',''); if (!$conn) {exit("Connection Failed: " . $conn);} $sql = "INSERT INTO Data (FirstName, LastName, BusinessName) VALUES ('$fname', '$lname', '$bname')"; $rs=odbc_exec($sql, $conn); $rs = @odbc_exec($conn,$sqlstring); if (!$rs) { echo "An error has occured. Please try again", odbc_errormsg($conn); } else { echo "The record was successfully inserted."; } odbc_close($conn); and here is the error: Quote An error has occured. Please try again[MySQL][ODBC 5.1 Driver][mysqld-5.5.20]Query was empty I really hope you could help me out. Thanks. I bought a script not so long back and I am trying to install it on my Linux hosting cPanel with GoDaddy.
I have followed the installation instructions (below) provided by the script author but I am having no luck.
1. Modify server/server.php Line: 282
$master = new ChatBot(“(author's domain)”, 10000); Hello, I'm new at PHP programming. I'm creating a simple application. I created a form where users can fill in some data. One of the fields is a username. This is a drop down list. This list is build up with data from SQL Server 2005 R2. I created the code in PHPDesigner7. The ODBC connection is working fine in PHPDesigner7. When I try to run it on the localhost I get no connection The code I used : <?php $conn=odbc_connect('MyDataBase','',''); echo "connectie ".$conn; if (!$conn) {exit("Connection Failed: ". $conn);} $sql="SELECT Id, [Name] FROM Names"; $rs=odbc_exec($conn,$sql); $options=""; echo "<br />"; echo "User : "; echo "<select name='QCT_name'>"; while ($row=odbc_fetch_array($rs)) { $id= $row["Id"]; $Controleur = $row["Name"]; echo "<option value='".$row['Id']."'>".$row['Name']."</option>"; } echo "</select>"; echo "<br />"; ?> Has anybody any clue what I'm doing wrong? So I'm doing a project and I need to make a successful login form, where it checks in MS Access if the username and password are correct, and if they are the user is taken to a new page. If they are wrong, a message comes up and they stay on the same page. The user should first just see a blank form, but after submit is pressed, it should check if the username and password are correct. IF they are should be taken to a new page. It's been a while since I used php last, so I wasn't quite sure how to tackle this issue. I was wondering if someone could please help me? here is my code. Code: [Select] <html> <head> <style type="text/css"> </style> </head> <body style="text-align:center"> <div id='title'> </div> <?php print_r ($_POST) ; if (isset($_POST['submit'])) { if(isset($_POST['username'])){ $username= $_POST['username'] ; } if(isset($_POST['password'])){ $TABLE= $_POST['password'] ; } $username = null ; $password = null ; $connection = odbc_connect('Olympics', '', ''); if (!$connection) {exit("Conection Failed: " . $connection);} $username = stripslashes($username); $password = stripslashes($password); $sql = "select * from users where users = '$username' and passwords = '$password'"; $rs=odbc_exec($connection,$sql); $count=odbc_num_rows($rs); if ($count == 1) { $_SESSION['loggedIn'] = "true"; header("Location: searchpage.php"); } else { $_SESSION['loggedIn'] = "false"; header("Location: index.php"); echo "Login failed" ; } } echo "<form action='index.php' method='post'> \n" ; echo" Please enter your username and password if you wish. <br/> \n" ; echo "Username: <input type='text' name='username' > \n " ; echo "Password: <input type='password' name='password' > \n" ; echo "<input type='submit' value='Login' name='submit'> <br/> \n" ; echo "<input type='submit' value='You may also continue you as a guest.'> \n" ; ?> </body> </html> CAN ANYONE TELL ME WHAT i AM DOING WRONG. I WANT TO RETRIEVE THE VERY FIRST RECORD IN MY DATABASE WHEN THE CODE EXECUTE IT ONLY SHOWS THE LAST RECORD IN THE DATABASE <?PHP $thisMonth = date('M'); $thisDay = date('j'); $eventMonth = array(); $eventDay = array(); $eventTime = array(); $eventName = array(); $eventLocation = array(); $dbMonth=""; $dbDay=""; $i=0; $conn = odbc_connect('eventsDB','',''); $sql= "SELECT month,day, time, event,location FROM Events"; $rs="$conn,$sql"; if (!$conn) { exit("Connection Failed: " . $conn); } $rs=odbc_exec($conn,$sql); if(!$rs) { exit("Error in SQL"); } echo "DATABASE OPEN"; while($i<3) { $dbMonth= odbc_result($rs,"month"); echo $eventMonth[$i]=odbc_result($rs,"month")."\n"; if($dbMonth<>$thisMonth) { odbc_fetch_row($rs); } echo $eventMonth[$i]=odbc_result($rs,"month")."\n"; echo $eventDay[$i]=odbc_result($rs,"day")."\n"; echo $eventTime[$i]=odbc_result($rs,"time")."\n"; echo $eventDay[$i]=odbc_result($rs,"event")."\n"; echo $eventLocation[$i]=odbc_result($rs,"location")."\n"; $i++; odbc_fetch_row($rs); echo $i; } //ends while loop odbc_close($conn); ?> Okay guys I have finished my db and I want to upload it to a live server but don't know exactly what to change in the code to get the ODBC_connect to open the path at the new server location please help. my database is located in a subfolder called Databases i.e. "www.mydomain.com/Databases/myEvents.mdb" what do i change in the connection string below to get it to open the file on the live server? Thanks in advance. $conn = odbc_connect('myEvents','',''); Hey guys, I`m having problems finding information on how to insert a image on a blob column trough odbc on a oracle database. Does anyone knows where can i find it, or can anyone help me with sample code? I really need to use odbc functions (cant use oci8 ) Tanks for your time reading this and for the possible input you may add on this. Sincerely Arestas 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. 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 guys, 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 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. 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. 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 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? 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 |
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