PHP - Moved: Connection From Php To Sql Server 2005 Database Using Sqlsrv_connect()
This topic has been moved to Installation in Windows.
http://www.phpfreaks.com/forums/index.php?topic=312024.0 Similar TutorialsI've been working on this for 8 hours and still cannot get sqlsrv_connect to work. Thank you, Microsoft. My connection strings work for Access DBs and MySQL - but I can't get this to work. For servername, I've tried: $serverName = "MYSERVER"; $serverName = "(MYSERVER\MYINSTANCE)"; $serverName = "'MYSERVER\MYINSTANCE'"; $serverName = "(MYSERVER)"; Setup: PHP Compiler = MSVC9 (Visual C++ 2008) C:\Program Files (x86)\PHP\v5.3\php.ini extension_dir C:\Program Files (x86)\PHP\v5.3\ext\ php_sqlsrv_53_nts_vc9.dll in the extension library. And I'm getting an error from sqlsrv_errors, so it looks like the DLL is working. Pulling my hair out here. Any ideas? Code: [Select] $serverName = "MYSERVER\MYINSTANCE"; $connectionInfo = array("UID" => $uid, "PWD" => $pwd, "Database"=>"$dbname"); $conn = sqlsrv_connect( $serverName, $connectionInfo); if($conn) { echo "Connection established.\n"; } else { echo "Connection could not be established.\n"; die( print_r( sqlsrv_errors(), true)); } sqlsrv_close( $conn); Connection could not be established. Array ( => Array ( => 08001 => 08001 [1] => 53 => 53 [2] => [SQL Server Native Client 10.0]Named Pipes Provider: Could not open a connection to SQL Server [53]. => [SQL Server Native Client 10.0]Named Pipes Provider: Could not open a connection to SQL Server [53]. ) [1] => Array ( => HYT00 => HYT00 [1] => 0 => 0 [2] => [SQL Server Native Client 10.0]Login timeout expired => [SQL Server Native Client 10.0]Login timeout expired ) [2] => Array ( => 08001 => 08001 [1] => 53 => 53 [2] => [SQL Server Native Client 10.0]A network-related or instance-specific error has occurred while establishing a connection to SQL Server. Server is not found or not accessible. Check if instance name is correct and if SQL Server is configured to allow remote connections. For more information see SQL Server Books Online. => [SQL Server Native Client 10.0]A network-related or instance-specific error has occurred while establishing a connection to SQL Server. Server is not found or not accessible. Check if instance name is correct and if SQL Server is configured to allow remote connections. For more information see SQL Server Books Online. ) ) Hi friends... I am using Wamp with PHP 5.3 and want to connect with SQL Server 2005 but not successful. I also tried it with PHP 5.2.8 but it still not working. I google it and found some solutions but not successsful yet. Can anyone give me detailed guide (step by step) that how i can connect PHP with Sql Server 2005??? Thanks in advance... This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=353212.0 Most of the time my localhost works like a champ but occasionally I get ... mysqli_connect() [function.mysqli-connect]: (HY000/2005): Unknown MySQL server host 'localhost' (11001) On phpfreaks I've searched for "Unknown MySQL server host localhost " and didn't see an answer to this problem Then I googled and found this http://stackoverflow...-localhost11001 However in this file the localhost ip is commented out and windows wont let me edit the file to uncomment it like it's done in the stackflow post. So, one is this the likely problem ... and if it is how do I edit a file in the Windows folder? As a note I do go into the file as the administrator using notepad Thanks Edited by floridaflatlander, 03 June 2014 - 08:21 AM. 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 I created this to test the connection to my ms sql server with a web page. I can connect with isql with out fail.
when I open this I get
PHP SQL Test
END PHP SQL Test
I was expecting to get sql data between the above lines. What am I doing wrong.
<html> <head> <title>PHP SQL Test</title> </head> <body> <p> PHP SQL Test </p> <? $conn =odbc_connect("datasource","user","password"); if(!$conn) { exit("Connection Failed: " . $conn); } $sql="SELECT top 10 * from WIP_master"; $rs =odbc_exec($conn,$sql); print_r($rs); if(!$rs){ exit("Error in SQL"); } odbc_close($conn); ?> <p> End PHP SQL Test </p> </body> </html> Here is my code: Code: [Select] <?php $con = ssh2_connect('crm.idea.com.bd','4321'); ssh2_auth_password($con, 'username', 'password'); $response = ssh2_exec($con, 'pwd'); echo $response; ?> I can connect to that server using that username, password and port address. But when I load this script it don't work. It shows loading and loading continues. No error message is given. N: B: Please help me. I am a newbie. Hi my website has been working all day now everypage is getting this error also get the same error when i try to login into phpmyadmin Lost connection to MySQL server at 'reading initial communication packet', system error: 111 is this most likely a problem with the server? On some occasions I need to connect to a second and third database in the same script (maybe 5% of scripts have at least a second connection). Usually I would just select the new database. However, my host requires different users to be created for each database. What is the best way to do this? Close current connection (say db1) and open new (say db2) OR keep all open, creating 2nd and 3rd connections. I am happy with the design of my database, and don't want to merge all these tables into one db. Overall I am still happy with my host, so I'd rather not change. <?php class UserQuery { public function Adduser($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); 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, 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 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. 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 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'); } 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 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? Hi guys, |