PHP - Issue Connecting To Multiple Databases
Hello,
I'm using this code in a php file to connect to 2 databases (something that I need to do): $conn_local = mysql_connect('localhost','root','',TRUE); $conn_local2 = mysql_connect('localhost','root',''); mysql_select_db('db1',$conn_local); mysql_select_db('db2',$conn_local2); I'm then trying to use this code to call it from the 1st Database: $sql = "SELECT * FROM news_items ORDER BY news_date DESC LIMIT 0,$limit"; $result = mysql_query($sql,$conn_local); if(mysql_num_rows($result)!=0){ while($row = mysql_fetch_array($result)) ...etc Although I am getting this error: Quote Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\test\includes\function.php on line 17 If I ONLY include the 1 database and don't bother putting the ',$conn_local' into the mysql_query it will work fine and return the records needed. It only seems to be when I try and include more than 1 database. Any ideas where I'm going wrong? Thanks. Similar TutorialsHello again guys! What if i have 2 databases, db1 and db2. i want to read raw data from db1 and match it's content with data from db2, then produce some information. How can i do it at the same time? I know that to be able to access mysql data, we have to use the mysql_connect() and mysql_select_db(). mysql_select_db() only allows to connect to single database. what should i do? thanks for your help in advance. Dear All, I am having 2 different DB on 2 different hosts. I am running MySQL Server on my local PC where the user is entering data in the tables. I have a website which has the identical DB on the web. I am able to connect to the database by using the codes on the server. I want to update the server DB with the local system DB by running one update command. I get the error "-SELECT command denied to user 'localusername'@'localhost' for table 'pst_data'" Given below is the code used for the process : //connecting the remote system DB $link = mysql_connect('IPAddress:3306', 'remoteusername', 'Password'); if (!$link) { die('Not connected : ' . mysql_error()); } $db_selected = mysql_select_db('remotedb', $link); if (!$db_selected) { die ('Can\'t use Remote System DB: ' . mysql_error()); } //connecting the local database on the webstite $weblink = mysql_connect("localhost","localusername","password"); if (!$weblink) { die('Not connected : ' . mysql_error()); } $webdb_selected = mysql_select_db('localwebdb', $weblink); if (!$webdb_selected) { die ('Can\'t use WebServer Database : ' . mysql_error()); } //query to fetch the records from remote ystem and insert into local website database $upd_Query=mysql_query("INSERT INTO localwebdb.`table` SELECT * FROM remotedb.`table` where field=' some condition' "); --------------------------------------------------- I have tested that I have connected the remote DB by running queries on the webserver. Could anyone bail me out so that i can copy the DB from remote to local Any help would be appreciated is it possible to have two open connections to two different mysql dbs at the same time? when i tried it, only the one on the bottom of the list was active. my config file looks like this: //---------------------------------------------// $dbname = 'xxx'; # Database Name $dbuser = 'xxx'; # Database Username $dbpass = 'xxx'; # Database Password $dbhost = 'xxx'; # Database Host $conn2 = mysql_connect($dbhost,$dbuser,$dbpass) or die ("Could not connect to $dbname: ".mysql_error()); mysql_select_db($dbname) or die ("Could not access the database: ".mysql_error()); $dbname5 = 'yyy'; # Database Name $dbuser5 = 'yyy'; # Database Username $dbpass5 = 'yyy'; # Database Password $dbhost5 = 'yyy'; # Database Host $conn = mysql_connect($dbhost5,$dbuser5,$dbpass5) or die ("Could not connect to $dbname5: ".mysql_error()); mysql_select_db($dbname5) or die ("Could not access the database: ".mysql_error()); //--------------------------------------// so i want to be able to do mysql_query($query,$conn2) when i need to access xxx db, but it doesn't seem to work that way. am i doing something incorrectly? any help would be greatly appreciated. I have done select statements where you loop through all the records. But what I need to do is pull record from one database than insert parts of that record into the second. The issue I have is no knowing how to select a different db with out stopping the loop of the first. Thanks in advance. I'm trying to select multiple peoples ID's where different values in different databases tables match. For instance, I want to make sure the person is a different gender than what the person logging in is, which would be set as 1(male) 2(female) in a Genders Column in the Membership_Profile table, while I am also checking to make sure that they are not blocked inside the Membership table and making sure they have a profile image in a user_profilepic table. How would I do this? This all needs to sort by the Date they were added inside the Membership Table and Limit it to 5 results. When I use the below code tho, it pulls up both Male and Females, but seems to look at the Membership correctly. Code: [Select] $sql2 = "SELECT Gender FROM Membership_Profile where UserID = '$user_ID'"; $sql_result2 = mysql_query($sql2); $login_row2 = mysql_fetch_assoc($sql_result2); $user_gender = $login_row2['Gender']; $true_query = mysql_query("SELECT Membership.ID, Membership_Profile.ID, user_profilepic.UserID FROM Membership, Membership_Profile, user_profilepic WHERE Membership_Profile.Gender != '$user_gender' AND Membership.ID = user_profilepic.UserID AND Membership.IsApproved='1' AND Membership.IsLockedOut = '0' AND Membership.UserLevel = '1' AND user_profilepic.Profile_Pic = '1' ORDER BY Membership.CreateDate DESC LIMIT 5;"); $is_odd_row = 1; while ($true_row = mysql_fetch_assoc($true_query)) { $right_user = $true_row['UserID']; $right_user2 = $true_row['ID']; $sql = "SELECT * FROM Users where ID = '$right_user' AND ID = '$right_user2'"; $sql_result = mysql_query($sql); $login_row = mysql_fetch_assoc($sql_result); $right_user_ID = $login_row['ID']; $rightGUID = $login_row['UserId']; $right_name = $login_row['UserName']; $sql = "SELECT DOB, City, State FROM Membership_Profile where ID = '$right_user' AND ID = '$right_user2'"; $sql_result = mysql_query($sql); $login_row = mysql_fetch_assoc($sql_result); $user_dob = $login_row['DOB']; $City = $login_row['City']; $State = $login_row['State']; if ($is_odd_row) { echo "<tr style=\"background-color: #e1ebf8\">\n"; $is_odd_row = 0; } else { echo "<tr style=\"background-color: #f3f7fc\">\n"; $is_odd_row = 1; } echo " <td align=\"left\"> <img width=\"85px\" style=\"margin-right:10px\" src=\"profilepics/$rightGUID/profile/$right_user_ID.jpg\"><br /> <span style=\"font-size:13px; font-weight:900; color:#ffae00;\">$right_name </span><br /> Location:<br /><b>$City, $State</b><br /> Age:<b>";echo CalculateAge("$user_dob"); echo "</b></td></tr>"; } I'm trying to select multiple peoples ID's where different values in different databases tables match. For instance, I want to make sure the person is a different gender than what the person logging in is, which would be set as 1(male) 2(female) in a Genders Column in the Membership_Profile table, while I am also checking to make sure that they are not blocked inside the Membership table and making sure they have a profile image in a user_profilepic table. How would I do this? This all needs to sort by the Date they were added inside the Membership Table and Limit it to 5 results. When I use the below code tho, it pulls up both Male and Females, but seems to look at the Membership correctly. Code: [Select] $sql2 = "SELECT Gender FROM Membership_Profile where UserID = '$user_ID'"; $sql_result2 = mysql_query($sql2); $login_row2 = mysql_fetch_assoc($sql_result2); $user_gender = $login_row2['Gender']; $true_query = mysql_query("SELECT Membership.ID, Membership_Profile.ID, user_profilepic.UserID FROM Membership, Membership_Profile, user_profilepic WHERE Membership_Profile.Gender != '$user_gender' AND Membership.ID = user_profilepic.UserID AND Membership.IsApproved='1' AND Membership.IsLockedOut = '0' AND Membership.UserLevel = '1' AND user_profilepic.Profile_Pic = '1' ORDER BY Membership.CreateDate DESC LIMIT 5;"); $is_odd_row = 1; while ($true_row = mysql_fetch_assoc($true_query)) { $right_user = $true_row['UserID']; $right_user2 = $true_row['ID']; $sql = "SELECT * FROM Users where ID = '$right_user' AND ID = '$right_user2'"; $sql_result = mysql_query($sql); $login_row = mysql_fetch_assoc($sql_result); $right_user_ID = $login_row['ID']; $rightGUID = $login_row['UserId']; $right_name = $login_row['UserName']; $sql = "SELECT DOB, City, State FROM Membership_Profile where ID = '$right_user' AND ID = '$right_user2'"; $sql_result = mysql_query($sql); $login_row = mysql_fetch_assoc($sql_result); $user_dob = $login_row['DOB']; $City = $login_row['City']; $State = $login_row['State']; if ($is_odd_row) { echo "<tr style=\"background-color: #e1ebf8\">\n"; $is_odd_row = 0; } else { echo "<tr style=\"background-color: #f3f7fc\">\n"; $is_odd_row = 1; } echo " <td align=\"left\"> <img width=\"85px\" style=\"margin-right:10px\" src=\"profilepics/$rightGUID/profile/$right_user_ID.jpg\"><br /> <span style=\"font-size:13px; font-weight:900; color:#ffae00;\">$right_name </span><br /> Location:<br /><b>$City, $State</b><br /> Age:<b>";echo CalculateAge("$user_dob"); echo "</b></td></tr>"; } Hi Everyone, Hopefully someone will be able to assist with my problem. Basically, my situation is that we have a server which hosts multiple websites using multiple IP address. One of the new sites we are moving to this server needs an LDAP connection outside of our network. The outside LDAP has already been enabled to accept requests from the specific IP assigned to this site. However, other sites on this server are using other IP addresses. It seems as if the LDAP authentication request is getting sent by a IP address which is not authorized by the firewall on the outside LDAP server. Thus trouble authenticating. So, my question is, is there a way to force the ldap_bind request to use a specific IP address to send the request for authentication? Supposedly this can be done using an event handler. However, I don't have much experience with event handlers so am not sure how to go about doing this. I would appreciate any help or ideas to resolve this situation. Thanks! - Jodie When a user logs into my site I was to store the user_name, user_role, and user_id in a session variable, then store this in a regular variable to make querying easier, but I am having issues with my code as its causing issues with my queries. Code: [Select] if (mysqli_num_rows($data) > 0) { //set sessions $row = mysqli_fetch_array($data); $_SESSION['username'] = $row['username']; $_SESSION['user_role'] = $row['role']; $_SESSION['user_id'] = $row['user_id']; //set variables $username = $_SESSION['username']; $user_role = $_SESSION['user_role']; $user_id = $_SESSION['user_id']; } Any ideas whats wrong?! Thanks for the help!! Hi
I am developing a website for fun to play around with, like a little fun project and am trying to work out how to upload multiple images that stores the filename in the database and the actual file on the server
I have managed to get it working if I upload just one image but can't work it out for multiple images
Can anyone point me in the right direction or advise where I need to adjust for multiple images upload
I know on the html form on the input tag needs to have multiple and then the name in the input tag be image[] but is as far as I get, it's the PHP I get stuck on
The html for the form is below
<form action="private-add-insert.php" method="post" enctype="multipart/form-data"> Car Make: <input type="text" name="make"> Car Model: <input type="text" name="model"> Exterior Colour: <input type="text" name="exteriorcolour"> Engine Size: <input type="text" name="enginesize"> Fuel Type: <input type="text" name="fueltype"> Year Registered: <input type="text" name="yearregistered"> Transmission: <input type="text" name="transmission"> Mileage: <input type="text" name="mileage"> Number of Doors: <input type="text" name="nodoors"> Body Style: <input type="text" name="bodystyle"> Price: <input type="text" name="price"> <br> <label>Upload Images</label> <input type="file" name="image[]" multiple/> <br /> <input type="hidden" name="MAX_FILE_SIZE" value="100000" /> <br /> <input type="submit" value="Submit Listing"> </form>Below is the PHP coding that processes the html form <?php ini_set('display_startup_errors',1); ini_set('display_errors',1); error_reporting(-1); ?> <?php // Start a session for error reporting session_start(); // Call our connection file require("includes/conn.php"); // Check to see if the type of file uploaded is a valid image type function is_valid_type($file) { // This is an array that holds all the valid image MIME types $valid_types = array("image/jpg", "image/jpeg", "image/bmp", "image/gif", "image/png"); if (in_array($file['type'], $valid_types)) return 1; return 0; } // Just a short function that prints out the contents of an array in a manner that's easy to read // I used this function during debugging but it serves no purpose at run time for this example function showContents($array) { echo "<pre>"; print_r($array); echo "</pre>"; } // Set some constants // This variable is the path to the image folder where all the images are going to be stored // Note that there is a trailing forward slash $TARGET_PATH = "private-listing-images/"; // Get our POSTed variables $make = $_POST['make']; $model = $_POST['model']; $exteriorcolour = $_POST['exteriorcolour']; $enginesize = $_POST['enginesize']; $fueltype = $_POST['fueltype']; $yearregistered = $_POST['yearregistered']; $transmission = $_POST['transmission']; $mileage = $_POST['mileage']; $nodoors = $_POST['nodoors']; $bodystyle = $_POST['bodystyle']; $price = $_POST['price']; $image = $_FILES['image']; // Sanitize our inputs $make = mysql_real_escape_string($make); $model = mysql_real_escape_string($model); $exteriorcolour = mysql_real_escape_string($exteriorcolour); $enginesize = mysql_real_escape_string($enginesize); $fueltype = mysql_real_escape_string($fueltype); $yearregistered = mysql_real_escape_string($yearregistered); $transmission = mysql_real_escape_string($transmission); $mileage = mysql_real_escape_string($mileage); $nodoors = mysql_real_escape_string($nodoors); $bodystyle = mysql_real_escape_string($bodystyle); $price = mysql_real_escape_string($price); $image['name'] = mysql_real_escape_string($image['name']); // Build our target path full string. This is where the file will be moved do // i.e. images/picture.jpg $TARGET_PATH .= $image['name']; // Check to make sure that our file is actually an image // You check the file type instead of the extension because the extension can easily be faked if (!is_valid_type($image)) { $_SESSION['error'] = "You must upload a jpeg, gif, bmp or png"; header("Location: private-add-listing.php"); exit; } // Lets attempt to move the file from its temporary directory to its new home if (move_uploaded_file($image['tmp_name'], $TARGET_PATH)) { // NOTE: This is where a lot of people make mistakes. // We are *not* putting the image into the database; we are putting a reference to the file's location on the server $sql = "insert into privatelistings (make, model, exteriorcolour, enginesize, fueltype, yearregistered, transmission, mileage, nodoors, bodystyle, price, filename) values ('$make', '$model', '$exteriorcolour', '$enginesize', '$fueltype', '$yearregistered', '$transmission', '$mileage', '$nodoors', '$bodystyle', '$price', '" . $image['name'] . "')"; $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error()); header("Location: private-add-listing-successfully.php?msg=Listing Added successfully"); exit; } else { // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to // Make sure you chmod the directory to be writeable $_SESSION['error'] = "Could not upload file. Check read/write persmissions on the directory"; header("Location: private-add-listing.php"); exit; } ?>I know the coding needs updating to mysqli and prevent sql injections but want to get it working first and then will do them parts Thank you in advance Kind regards Ian Just wondering, which is better php databases or sql? I have phpMyAdmin and I've that you can convert the database into php scripts? Let me explain my problem. I have an array with dates and numbers in format ($cronograma) Ex: Array ( [2020-09-21] => Array ( [0] => 2020-09-21 [1] => 2 [2] => 2 [3] => 2 ) [2020-09-28] => Array ( [0] => 2020-09-28 [1] => 2 [2] => 2 [3] => 4 ) Then i have another array with 2 ids (in this case 58,60) ($id) Finally i have a third array with numbers only (in this case 34,34) $tot So what i want is cross information beween them, for example for id 58 I must get dates (first element and last element when $tot = 34) for id 60 I must get dates (first element after $tot =34 and last element of array) Whath i have so far is this foreach ($id as $idPlan) { foreach ($cronograma as $c) { $t1 = 0; foreach ($tot as $d) { $t1 += (int)$d['tempos']; if ($c[3] == $t1) { $newAr[] =$idPlan; $newAr[] = $c[0]; } } } } My response array(8) { [0]=> string(2) "58" [1]=> string(10) "2021-02-01" [2]=> string(2) "58" [3]=> string(10) "2021-06-14" [4]=> string(2) "60" [5]=> string(10) "2021-02-01" [6]=> string(2) "60" [7]=> string(10) "2021-06-14" } null So it's clear that i have all repeated I should have a line like: 58 - 2020-09-21 -2021-02-01 Any help? Main script gets class information from a database and prints them so long as the class start date or end date is after today (actually includes today).
Main script calls "instructors.php". It queries another database based on the instructor name ($chef) and then prints the bio information for that instructor.
"instructors.php" works fine on it's own, when I add "$chef = "Chef Name" ("Chef Name" is in the Instructors database). When it's called from the main script, nothing shows up in that area - even though "Chef Name" is in the database. All of the other data is printed fine, just not anything from instructors.php. I verified that it's actually including the file, as I can add "echo "test";" to the top of instructors.php and it prints fine in the main script.
Any ideas of what I'm missing?
Main Script
<?php // Get required login info include "/path/to/login/info/file.php"; // Get required login info - changed for this post. $db = new mysqli('localhost', $username, $password, $database); // Connect to DB using required login info if($db->connect_errno > 0){ die('Unable to connect to database [' . $db->connect_error . ']'); } unset($username);// put these variables back to null unset($password);// put these variables back to null unset($database);// put these variables back to null //query db $sql = <<<SQL SELECT * FROM `ft_form_7` WHERE DATE(class_start_date) >= CURDATE() OR DATE(class_end_date) >= CURDATE() ORDER BY class_start_date ASC SQL; if(!$result = $db->query($sql)){ // if there is an error in running the query, show error message. die('There was an error running the query [' . $db->error . ']'); } while($row = $result->fetch_assoc()){ // Get start date information $start_date = $row['class_start_date']; // Get event_start_date for conversion and call it $start_date $start_date_formatted = date("l M d, Y", strtotime($start_date)); // Convert start_date $end_date = $row['class_end_date']; // Get event_end_date for conversion and call it $start_date $end_date_formatted = date("M d, Y", strtotime($end_date)); // Convert start_date // Get time information. $start_time = $row['class_start_time']; // Get event_start_time for conversion and call it $start_time $start_time_formatted = date("h:i A", strtotime($start_time)); // Convert start_time $end_time = $row['class_end_time']; // Get event_end_time for conversion and call it $end_time $end_time_formatted = date("h:i A", strtotime($end_time)); // Convert end_time // echo information... echo "<h2>" , $row['class_name'],"</h2>" ; // echo event name echo "<p><strong>",$start_date_formatted; // echo the start date if (empty($start_time)) { echo ''; } else { echo " (", $start_time; } // echo start time if (empty($end_date)) { echo ''; } else { echo " -","<br />", $end_date_formatted; } // echo end date if (empty($end_time)) { echo ')'; } else { echo " - ", $end_time, ")"; } // echo end time // if there is no start time, echo nothing. (otherwise it seems to echo 4pm). If it does contain a time, echo the time. echo "</strong><br />"; $chef = $row['Instructor']; global $chef; if ($chef != NULL) { require ('instructors.php'); } echo $row['class_description'], "<br />"; echo "<strong>" , $row['type'], " - Cost: $",$row['cost'] , " - #" , $row['course_number'] , "</strong><br />" , "</p>"; // echo class type and cost } $db->close(); $result->free(); ?>instructors.php <?php include "/path/to/login/info/file.php"; // Get required login info - changed for this post. $db_instructors = new mysqli('localhost', $username, $password, $database); // Connect to DB using required login info if($db_instructors->connect_errno > 0){ die('Unable to connect to database [' . $db_instructors->connect_error . ']'); } unset($username);// put these variables back to null unset($password);// put these variables back to null unset($database);// put these variables back to null //query db $sql_instructors = <<<SQL SELECT * FROM ft_form_8 WHERE chef_name = '$chef' SQL; if(!$result_instructors = $db_instructors->query($sql_instructors)) { // if there is an error in running the query, show error message. die('There was an error running the query [' . $db_instructors->error . ']'); } while($row_instructors = $result_instructors->fetch_assoc()) { $chef_full_name = $row_instructors['chef_name']; $chef_id = $row_instructors['submission_id']; $full_bio = $row_instructors['full_bio']; echo "<a href=\"#\" class=\"clickme\">" , $chef_full_name , "</a>"; echo "<div class=\"box\">"; echo $full_bio , "</div>"; } $db_instructors->close(); $result_instructors->free(); ?> Hi there, I have a form submitting to a mysql database. It was requested that I put a multiple file upload option on it. So I added the fields to the form and the database and added the upload script to the form processing file (and of course created the folder on the server with proper permissions to upload to). It should upload the file(s) to the folder on the server and insert the filename into the database. I keep getting: HTTP Error 500 (Internal Server Error): An unexpected condition was encountered while the server was attempting to fulfill the request. This leads me to believe the upload script isn't working properly. If someone could take a look at it I would be a very happy guy. This is the upload script: (insert.php) Code: [Select] <?php $con = mysql_connect("server","user","pass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("db", $con); $dateSubmitted = date("Y-m-d"); $target = "attachments/"; $target = $target . basename( $_FILES['doc1']['name']); $target = $target . basename( $_FILES['doc2']['name']); $target = $target . basename( $_FILES['doc3']['name']); $target = $target . basename( $_FILES['doc4']['name']); $target = $target . basename( $_FILES['doc5']['name']); $sql="INSERT INTO investments (active, project, inv_amount, account_type, prefix, first_name, last_name, address1, address2, city, province, postal_code, country, phone, email, referral_fee, ref_agent, ship_name, ship_address, ship_city, ship_province, ship_postal, ship_country, notes, dateSubmitted, doc1, doc2, doc3, doc4, doc5) VALUES ('$_POST[active]','$_POST[account_type]','$_POST[project]','$_POST[inv_amount]','$_POST[prefix]','$_POST[prefix]','$_POST[first_name]','$_POST[last_name]','$_POST[address1]', '$_POST[address2]','$_POST[city]','$_POST[province]','$_POST[postal_code]','$_POST[country]','$_POST[phone]','$_POST[email]','$_POST[referral_fee]','$_POST[ref_agent]','$_POST[ship_name]','$_POST[ship_address]','$_POST[ship_city]','$_POST[ship_province]','$_POST[ship_postal]','$_POST[ship_country]','$_POST[notes]','$_POST[dateSubmitted]',$_FILES['doc1']['name'],$_FILES['doc2']['name'],$_FILES['doc3']['name'],$_FILES['doc4']['name'],$_FILES['doc5']['name'])"; //Writes the photo to the server if(move_uploaded_file($_FILES['doc1']['tmp_name'], $target)) if(move_uploaded_file($_FILES['doc2']['tmp_name'], $target)) if(move_uploaded_file($_FILES['doc3']['tmp_name'], $target)) if(move_uploaded_file($_FILES['doc4']['tmp_name'], $target)) if(move_uploaded_file($_FILES['doc5']['tmp_name'], $target)) { //Tells you if its all ok echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory"; } else { //Gives and error if its not echo "Sorry, there was a problem uploading your file."; } if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } //echo "1 record added"; header("Location: ../forms.php"); mysql_close($con) ?> Thanks in advance. JE Hi, I am in the procress of creating discussion system however I am a bit puzzled about the best way to go about it. I am starting the discussion by creating an ID number and then match the answer to the initial ID number. However, I dont know whether if is best to put the responses into a different database. I'm a bit puzzled how ID matching systems works. Lets say: Question 1 = ID1 Question 2 = ID2 Question 3 = ID3 Question 1 Answer 1 = ID4 (How is this matched to ID1) Question 2 Answer 1 = ID5 (How is this matched to ID2) is this based on preg_match? ok , here is my mysql code to get all posts from the posts table . Code: [Select] $query = mysql_query("SELECT id,to_id,from_id,post,type,state,date FROM posts WHERE state='0' ORDER BY id DESC LIMIT 50"); and here is the code to display the users friends... Code: [Select] $sqlArray = mysql_query("SELECT friend_array FROM myMembers WHERE id='" . $logOptions_id ."' LIMIT 1"); while($row=mysql_fetch_array($sqlArray)) { $iFriend_array = $row["friend_array"]; } $iFriend_array = explode(",", $iFriend_array); if (in_array($id, $iFriend_array))see now i got as far as , if(in_array($id, $iFriend_array)) How would i put these togeather to where it would get the posts from the posts table that there friends posted? I'm trying to connect to two databases and I'm having problems with the following code. I googled to come up with this but can't figure out the errors I'm getting. Code: [Select] $connection="localhost"; $username="user"; $password="password"; $database1="dbone"; $database2="dbtwo"; $db1 = mysql_connect($connection,$username,$password) or die(mysql_error()); $sel1 = mysql_select_db($database1, $db1); $query1 = "SELECT * FROM TBLUSERS"; $result1 = mysql_query($query1, $db1); while($nt1 = mysql_fetch_array($result1, $db1)) { } $db2 = mysql_connect($connection,$username,$password) or die(mysql_error()); $sel2 = mysql_select_db($database2, $db2); $query2 = "SELECT * FROM TBLPD20101101"; $result2 = mysql_query($query2, $db2) or die(mysql_error()); while($nt2 = mysql_fetch_array($result2, $db2)) { } The error I get is Quote Warning: mysql_fetch_array() expects parameter 2 to be long, resource given in C:\xampp\htdocs\HighVisibility\DashBoard2.php on line 13 Warning: mysql_fetch_array() expects parameter 2 to be long, resource given in C:\xampp\htdocs\HighVisibility\DashBoard2.php on line 22 I have just read my upcoming modules for my final year at uni and 'multimedia databases' is one of them. I am just wondering if any of you had any clue on what a multimedia database is? I am guessing it's a database populated with directory data, but that would be to simple... I have a site where I need to have lets call it image1 displayed, then I want to change this image based on a php if statement, for instance: if $var == $var2 change the image ....blah blah so I was also going to have the names of my images stored in my database, i.e. image1.jpg and image2.jpg in my database. the image is in its own div tag set as the background image of the div tag if that makes any diference. Thanks What would be the fastest way to search 2+ databases with the same search information? Each database is different, and may return different information. What's the best way for putting actual quotations into a database? I was using... $quote = htmlspecialchars(mysqli_real_escape_string($dbc, $_POST['quote'])); Should I be? |