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PHP - After Cloasing Connection Of Database I Still Got The Values Form Database
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 Similar TutorialsHey all. I'm a noob so bare with me. Just started studying php and have an assignment due. At this phase of the project I have to create a form where a user can register data and I should retrieve that data and store it in my database. Now I have already created the form (student_reg.php) and a separate file with php code (demo.php) My database is name "registration" and I am trying to put this info in the "student" table. It has about 15 fields but for now I am only trying trying to insert the 1 field of data to test if it works. Dont know if that may cause problems? So far my demo.php connects to the database successfully. The problem I am having is the information I submit does not showing in the database field. However, I have an id field in my case a student number field which is set to auto increment so it automatically updates (1,2,3,4) with each new sumbitted info. When I submit the info the id field gets updated each time with a new row ( so i assume some information is somehow going through ) however the info I entered will remain blank in its field. :banghead: Heres my form (student_reg.php) Code: [Select] <div id="apdiv3"> <FORM action = "demo.php" method ="post "> <p>Course name:</p> <INPUT TYPE = "text" name="input"/> <INPUT TYPE = "Submit" VALUE = "Submit"/> </div> </FORM> and heres my php code to retrieve the form info Code: [Select] <?PHP include 'includes/config.php'; //connect to database $link=mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); //error check if (!$link) { die('could not connect: ' . mysql_error()); } $db_selected=mysql_select_db(DB_NAME, $link); // error check if (!$db_selected) { die('can\t use ' . DB_NAME . ': ' . mysql_error()); } // Connected to database // retrieve form data $value=(isset($_POST['input'])); $sql = "INSERT INTO student (sname) VALUES ('$value')"; if (!mysql_query($sql)) { die ('Error: ' . mysql_error()); } mysql_close(); ?> I also recieved a notice "undefined index" or something like that from this part of the code "$value=(isset($_POST['input']));" before I inserted "isset" to the code. I'm not sure if what I did fixed the problem or only hid it, or whether that notice was related to the problem I'm having of not retrieving the info. Haha sorry guys know this is probably childs play for you but I just started out and really need to fix this so I can catch up with this project. Feel free to add tips and comments on other areas. Thanks :mrgreen: Hi everyone, I'm new to this group and new to php. I have created a multi-part form that allows the user the option to add multiple input fields to a form to upload images. Here is the form structu Code: [Select] <form action="Scripts/processreports2.php" method="post" enctype="multipart/form-data" name="report_form" target="uploader" class="reportfrm"> <fieldset> <legend>Upload your images</legend> <ol id="add_images"> <li> <input type="file" class="input" name="files[]" /> </li> <li> <input type="file" class="input" name="files[]" /> </li> <li> <input type="file" class="input" name="files[]" /> </li> </ol> <input type="button" name="addFile" id="addFile" value="Add Another Image" onclick="window.addFile(this);"/> </fieldset> <p>* indicates a required field.</p> <input name="submit" type="submit" id="submit" value="Send Info!" /> </form> Through php a maximum of three input fields are fed into an array that checks to make sure that the uploaded files are images and not some other type of file. The uploading porition of this script works. Now I am trying to get the values of the input fields as a string and insert them into the database as one record. Let's say the user has three files they want to upload. I have managed to get the files as a string ie; file1.jpg, file2.jpg, file3.jpg but what is happening is that I am getting three separate records with file1.jpg, file2.jpg, file3.jpg in them. If the user has only two files to upload then I get two separate records with file1.jpg and file2.jpg in them. (Hope that makes sense). I want one record. I have been struggling with this since Monnday and while every day I get closer, this is as close as I can get. Below is the php code. #connect to the database mysql_connect("localhost", "root", ""); mysql_select_db("masscic"); //Upload Handler to check image types function is_image($file) { $file_types = array('jpeg', 'gif', 'bmp'); //acceptable file types if ($img = getimagesize($file)){ //echo '<pre>'; //print_r($_FILES); used for testing //print_r($img); used for testing if(in_array(str_replace('image/', '', $img['mime']), $file_types)) return $img; } return false; } //form submission handling if(isset($_POST['submit'])) { //file variables $fname = $_FILES['files']['name']; $ftype = $_FILES['files']['type']; $fsize = $_FILES['files']['size']; $tname = $_FILES['files']['tmp_name']; $ferror = $_FILES['files']['error']; $newDir = '../uploads/'; //relative to where this script file resides for($i = 0; $i < count($fname); $i++) { //echo 'File name ' . $fname[$i] . ' has size ' . $fsize[$i]; used for testing if ($ferror[$i] =='UPLOAD ERR OK' || $ferror[$i] ==0) { if(is_image($tname[$i])) { //append the tmp_name($tname) to the file name ($fname) and upload to the server move_uploaded_file($tname[$i], ($newDir.time().$fname[$i])); echo '<li><span class="success">'.$fname[$i].' -- image has been accepted<br></span></li>'; }else echo '<li><span class="error">'.$fname[$i].' -- is not an accepted file type<br></span></li>'; } if (is_array($fname)) $files = implode(', ',$fname); //else $files = $fname; $sqlInsert = mysql_query("INSERT INTO files (file_names) VALUES('$files')") or die (mysql_error()); } } 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); 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 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');
}
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 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? 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 how do I use a connection file (connection.php) in multiple programs? I think include? Hi guys, 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 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 ) 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 |
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