PHP - Join Query Slowing Down The Server
I was running this query and feel this is slowing down the server:
Code: [Select] $agent_query=$this->db->query(" SELECT u.id,u.name,u.team_id,l.agent_id from users AS u,leads AS l WHERE u.id=l.agent_id AND u.team_id IS NOT NULL AND u.is_active='1' ORDER BY u.name ASC "); Is there a better way to write the above code. Similar TutorialsHi guys. I have a SQL query running within my php. But if i add and inner join to link another table, it slows the performance(refresh, loading time) of the page by at least 2-3 seconds... I am working from localhost wampp serever. Should this happen when using inner Join? Code: [Select] $sql="SELECT * FROM cars INNER JOIN postcodes ON postcodes.Pcode=cars.Location where id>='1'" Thanks in anticipation Hi guys,
I’m really hoping someone can help with this query. I'm sure it must use join somehow but i cant work out exactly how to do it.
I have two tables (‘booking_slots’ and ‘booking_reservation’). ‘booking_slots’ has ‘slot_date’ and ‘slot_id’ fields. ‘booking_reservation’ has a number of fields including ‘slot_id’ but not ‘slot_date’.
I want to run a query to delete all records between a certain date range in BOTH tables (say for example 01 Jan 2012 to 01 Jan 2013). To do this I want to find and delete all records using ‘slot_date’ in ‘booking_slots’ and use the corresponding ‘slot_id’ of the deleted records to delete the records with the same ‘slot_id’ in ‘booking_reservation’.
Any help with this would be very greatly appreciated.
Thanks
Hello, The query below works well. It pulls information from 3 MySQL tables: login, submission, and comment. It creates a value called totalScore2 based on calculation of values pulled from these three tables. The MySQL tables "comment" and "submission" both have the following fields: Code: [Select] loginid submissionid In the table "submission," each "submissionid" has only one entry/row, and thus only one "loginid" associated with it. In the table "comment," the field "submissionid" could have several entries/rows, and could be associated with multiple "loginid"s. Each time one of the "submissionid"s in "comment" is associated with the same "loginid" that it has in the table "submission," I would like to add this as a factor to the equation below. I would like to multiple instances like this times (-10). How could I do this? Thanks in advance, John $sqlStr2 = "SELECT l.loginid, l.username, l.created, DATEDIFF(NOW(), l.created) + COALESCE(s.total, 0) * 5 + COALESCE(scs.total, 0) * 10 + COALESCE(c.total, 0) AS totalScore2 FROM login l LEFT JOIN ( SELECT loginid, COUNT(1) AS total FROM submission GROUP BY loginid ) s ON l.loginid = s.loginid LEFT JOIN ( SELECT loginid, COUNT(1) AS total FROM comment GROUP BY loginid ) c ON l.loginid = c.loginid LEFT JOIN ( SELECT S2.loginid, COUNT(1) AS total FROM submission S2 INNER JOIN comment C2 ON C2.submissionid = S2.submissionid GROUP BY S2.loginid ) scs ON scs.loginid = l.loginid GROUP BY l.loginid ORDER BY totalScore2 DESC LIMIT 25"; Hi.. I think I have bad code in three different similar query's in my code. The first query is: $sql="SELECT * FROM invoice as d INNER JOIN members as c ON d.buyer=c.usernum order by " . $orderBy . " " . $order; $result = $con->query($sql); Actually in this one I didn't even know I had a problem other than it was slow working until I put this error trap in:
if (!$check1_res) { The other two pass $id from the previous script they a $sql=mysqli_query($con,"SELECT * FROM invoice_items as d inner JOIN items as c ON d.itemnum=c.itemnum where invnumber = '$id'"); $row = mysqli_fetch_array($sql); and $sql=mysqli_query($con,"SELECT * FROM invoice as d inner JOIN members as c ON d.buyer=c.usernum where invnumber = '$id'"); $row = mysqli_fetch_array($sql); To be honest I didn't know I had a problem with the first two until The third would not return the correct data. I only got partial or none of the invoice data I was expecting. But I realized all three have problems when I used the error trap which come back with this:
Notice: Undefined variable: check1_res in C:\Apache24\htdocs\choo\tc_invoice.php on line 37 The error I get is identical for the first query. At first I thought it was the way that I was passing the variable, but $id is valid when echoed just below the query. So I am hoping that this is going to be something I did wrong with the querys or possibly the fetch statement after the query is wrong??? Really appreciate any help you can give me. If you need to see more of the code on either script please let me know. Hi, I have a fantasy football website, and on a user account page I want to display fixtures that are coming up that include teams that the current user has chosen. My test_teams table stores all the team names and their teamid. The test_selections table is where each users team selections are stored, it has two columns, userid and teamid. The test_fixtures table has two columns, hometeam and awayteam, these two cloumns hold the teamid of the teams that are playing. The code below correctly displays the fixtures that contain any of the current users team selections. However, it is only displaying the teamid of the teams that are playing as they have not been matched to the test_teams table to get the team name. Does anybody now how I can do this? I believe it can be done using a left join but so far I just keep getting errors when i try to write the code. Any help would be very much appreciated. Code: [Select] <table width="380" border="0"> <?php $query = "SELECT test_fixtures.competition, test_fixtures.date, test_fixtures.hometeam, test_fixtures.awayteam FROM test_fixtures, test_selections WHERE test_selections.userid = '{$_SESSION['userid']}' AND (test_selections.teamid = test_fixtures.hometeam OR test_selections.teamid = test_fixtures.awayteam)"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_assoc($result)) { ?> <tr> <td width="85" class="fixtures_date"><?php echo $row['date']; ?></td> <td width="30" class="fixtures_comp"><?php echo $row['competition']; ?></td> <td width="135" class="fixtures_home_teams"><?php echo $row['hometeam']; ?></td> <td width="25" class="fixtures_center">v</td> <td width="135" class="fixtures_away_teams"><?php echo $row['awayteam']; ?></td> </tr> <?php } ?> </table> Sorry, I originally posted part of this in MYSQL Help, but I was advised to post it here. I'm hoping someone can help me out. I have 2 tables. "categories" and "subcats" (sub categories). I used a JOIN query for both.. Code: [Select] SELECT categories.*, subcats.* FROM categories JOIN subcats on (categories.cat_id = subcats.cat_id) ORDER BY cat_name Then I used a while loop to echo all the categories from the "categories" table and their respective sub-categories from the "subcats" table. $sql = "SELECT categories.*, subcats.* FROM categories JOIN subcats on (categories.cat_id = subcats.cat_id) ORDER BY cat_name"; $cats_result = $connection->query($sql) or die(mysqli_error($connection)); while ($row = $cats_result->fetch_assoc()) { $cat_id = $row['cat_id']; $cat_name = $row['cat_name']; $subcat_name = $row['subcat_name']; //should output all categories once and nest their subcategories echo "<strong>".$cat_name."</strong>"; echo "<br />"; echo $subcat_name; echo "<br />"; } No MYSQL errors but unfortunately I get this output.. Technology Computers Technology Gadgets Technology Robots Health Fitness Health Diet What sort of code can use in the loop to echo categories just once instead of multiple times for each topic? I am having a warning which indicates there is a not valid mysql result, I think the problems lay down at the WHERE clause, but I am not sure. Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /storeprueba/sidebar.php on line 21 Code: [Select] $categoryurl = $_GET['categoryurl']; $sql= mysql_query("SELECT * FROM products, categories WHERE products.category = '$categoryurl' DESC LIMIT 6"); $productCount = mysql_num_rows($sql); // line 21 if ($productCount>0 ) { while($row = mysql_fetch_array($sql)) { $id= $row["id"]; $product_name= $row["product_name"]; $price = $row["price"]; $category = $row["category"]; $subcategory = $row["subcategory"]; $location = $row["location"]; $date_added = strftime("%b %d, %Y", strtotime($row["date_added"])); thanks. I am using two seperate queries to calculate a streak, but the queries must be grouped together to find the actual streak? Query 1: SELECT COUNT(matchID) as streak, clan1 FROM webs_cup_matches WHERE ladID='17' AND clan1='2630' AND score1 > score2 AND confirmscore='1' AND einspruch='0' GROUP BY clan1 ORDER BY streak DESC LIMIT 1 Query 2: SELECT COUNT(matchID) as streak, clan2 FROM webs_cup_matches WHERE ladID='17' AND clan2='2630' AND score1 < score2 AND confirmscore='1' AND einspruch='0' GROUP BY clan2 ORDER BY streak DESC LIMIT 1 is it possible someone can join these queries together? This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=342696.0 It's been a while since I sat down to build some pages and teach myself php. So now that I've started back up, I'm at a loss for what I've done. I deleted a file, and have to rebuild from an old broken version: I have a form that submits a query to the database, but the results pages is giving me this error: Code: [Select] Oops, my query failed. The query is: SELECT COUNT 'descriptors'.* ,'plantae'.* FROM 'descriptors' LEFT JOIN 'plantae' ON ('descriptors'.'plant_id' = 'plantae'.'plant_name') WHERE 'leaf_shape' LIKE '%auriculate%' AND 'leaf_venation' LIKE '%%' AND 'leaf_margin' LIKE '%%' The error is: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '.* ,'plantae'.* FROM ' at line 2 But I'm not seeing what the syntax error is. Here's the code: Code: [Select] <?php require ('connection.php'); $display = 2; // it's intentionally only 2 for the moment if (isset($_GET['np'])) { $num_pages = $_GET['np']; } else { $data = "SELECT COUNT 'descriptors'.* ,'plantae'.* FROM 'descriptors' LEFT JOIN 'plantae' ON ('descriptors'.'plant_id' = 'plantae'.'plant_name') WHERE 'leaf_shape' LIKE '%$s1%' AND 'leaf_venation' LIKE '%$s3%' AND 'leaf_margin' LIKE '%$s4%'"; $result = mysql_query ($data); if (!$result) { die("Oops, my query failed. The query is: <br>$data<br>The error is:<br>".mysql_error()); } $row = mysql_fetch_array($result, MYSQL_NUM); $num_records = $row[0]; if ($num_records > $display) { $num_pages = ceil ($num_records/$display); } else { $num_pages = 1; } } if (isset($_GET['s'])) { $start = $_GET['s']; } else { $start = 0; } if(isset($_POST[submitted])) { // Now collect all info into $item variable $shape = $_POST['s1']; $color = $_POST['s2']; $vein = $_POST['s3']; $margin = $_POST['s4']; // This will take all info from database where row tutorial is $item and collects it into $data variable $data = mysql_query("SELECT 'descriptors'.* ,'plantae'.* FROM 'descriptors' LEFT JOIN 'plantae' ON ('descriptors'.'plant_id' = 'plantae'.'plant_name') WHERE 'leaf_shape` LIKE '%$s1%' AND 'leaf_venation' LIKE '%$s3%' AND 'leaf_margin' LIKE '%$s4%' ORDER BY 'plantae'.'scientific_name` ASC LIMIT $start, $display"); //chs added this in... echo '<table align="center" cellspacing="0" cellpading-"5"> <tr> <td align="left"><b></b></td> <td align="left"><b></b></td> <td align="left"><b>Leaf margin</b></td> <td align="left"><b>Leaf venation</b></td> </tr> '; //end something chs added in // This creates a loop which will repeat itself until there are no more rows to select from the database. We getting the field names and storing them in the $row variable. This makes it easier to echo each field. while($row = mysql_fetch_array($data)){ echo '<tr> <td align="left"> <a href="link.php">View plant</a> </td> <td align="left"> <a href="link.php">unknown link</a> </td> <td align="left">' . $row['scientific_name'] . '</td> <td align="left">' . $row['common_name'] . '</td> <td align="left">' . $row['leaf_shape'] . '</td> </tr>'; } echo '</table>'; // row 95 } if ($num_pages > 1) { echo '<br /><p>'; $current_page = ($start/$display) + 1; // row 100 if ($current_page != 1) { echo '<a href="leafsearch2a.php?s=' . ($start - $display) . '&np=;' . $num_pages . '">Previous</a> '; } for ($i = 1; $i <= $num_pages; $i++) { if($i != $current_page) { echo '<a href="leafsearch2a.php?s=' . (($display * ($i - 1))) . '$np=' . $num_pages . '">' . $i . '</a>'; } else { echo $i . ' '; } } if ($current_page != $num_pages) { echo '<a href="leafsearch2a.php?s=' . ($start + $display) . '$np=' . $num_pages . '"> Next</a>'; } } //added curly ?> Is there a cleaner way to do this? $query = "SELECT branches.Language FROM eua_users, branches WHERE eua_users.AssignedBranch = branches.country" ; include("dbconnectlocal.php") ; $result = mysql_query($query) ; $row = mysql_fetch_object($result) ; $usrlang = $row->Language ; $query = "SELECT UserName, Email FROM eua_users WHERE UserName = '$user'" ; include("dbconnectlocal.php") ; $result = mysql_query($query) or die(mysql_error()); $row = mysql_fetch_object($result) ; $branchmail = $row->Email ; $query = "SELECT $usrlang FROM autoreplies WHERE ReplyID = '0'" ; include("dbconnectlocal.php") ; $result = mysql_query($query) or die(mysql_error()) ; $row = mysql_fetch_object($result) ; $message = $row->$usrlang ; $query = "SELECT $usrlang FROM autoreplies WHERE ReplyID = '1'" ; include("dbconnectlocal.php") ; $result = mysql_query($query) or die(mysql_error()) ; $row = mysql_fetch_object($result) ; $url = $row->$usrlang ; $query = "SELECT $usrlang FROM autoreplies WHERE ReplyID = '2'" ; include("dbconnectlocal.php") ; $result = mysql_query($query) or die(mysql_error()) ; $row = mysql_fetch_object($result) ; $subject = $row->$usrlang ; This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=313679.0 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=308855.0 I have a PHP web system that store in a windows server. In the system, there is a function for user to upload files to another server (Shared server in Unix). When i try to upload a file, it gives warning: Warning: move_uploaded_file(\\unixserver/sharedfolder/upload/test.txt) [function.move-uploaded-file]: failed to open stream: Permission denied in C:\wamp\www\upload\index.php on line 40 For your information, my username has been assigned in xxx's group that has access to read and write on that folder. Besides, i'm able to open,create and delete files on that folder's server manually (samba). The safe mode setting is off. Does anybody has any idea why this thing happen? Hi, I am not a PHP programmer. I took on a new client with a simple PHP site, without any databases. The site is up and running on the web. I would like to get it running on my local machine for further development. I have latest version of WAMP installed, running Apache version 2.2.11 and PHP version 5.3.0 I created a directory in the WAMP "www" project directory and it shows up there like it's supposed to when I browse to "localhost" Problem: The home page of website displays text but no, images, styles, footer, header, nav links, etc. Here is the code for the home page: <? define("NAV","home"); require_once('local/local.php'); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>TITLE</title> <meta name="keywords" content=""> <meta name="Description" content=""> <? include("common/dochead.php"); ?> </head> <body onLoad="<? include('common/preloads.php'); ?>"> <!-- ============================ main ============================= --> <div id="main-frame"><div id="main" class="noCollapse"> <? include("common/sign.php"); ?> <div id="right-frame"> <? include("common/navigation.php"); ?> <div id="content-frame"> <div id="content"> <h1>Welcome</h1> <p>This is the content area. This is the content area. This is the content area. </p> </div><!-- end content --> </div><!-- end content-frame --> </div><!-- end right-frame --> <div class="clearFloats"></div> </div><!-- end main --></div><!-- end main-frame --> <? include("common/footer.php"); ?> </body> </html> Any help would be greatly appreciated. I have spent many hours on this. Regards I'm trying to make a simple website where people register to my website. When the user doesn't fill anything inside the boxes they get a message "Please fill all required fields" on the register.php page On my local host require_once works good. It shows up.
But when i upload the files to my sever the require_once does not show up on the register.php It just refreshes and i dont get the message "Please fill all required fields"
This is the code that works in local host but not in a live server <?php require_once 'messages.php'; ?>
Here is my full code
Register page: <html> <?php require_once 'messages.php'; ?> <br><br> <form action="register-clicked.php" method="POST"> Username:<br> <input type="text" name="usernamebox" placeholder="Enter Username Here"> <br><br> Email:<br> <input type="text" name="emailbox" placeholder="Enter email here"> <br><br> Password:<br> <input type="password" name="passwordbox" placeholder="Enter password here"> <br><br> Confirm Password:<br> <input type="password" name="passwordconfirmbox" placeholder="Re-enter password here"> <br><br> <input type="submit" name="submitbox" value="Press to submit"> <br><br> </form> </html>
Register clicked <?php session_start(); $data = $_POST; if( empty($data['usernamebox']) || empty($data['emailbox']) || empty($data['passwordbox']) || empty($data['passwordconfirmbox'])) { $_SESSION['messages'][] = 'Please fill all required fields'; header('Location: register.php'); exit; } if ($data['passwordbox'] !== $data['passwordconfirmbox']) { $_SESSION['messages'][] = 'Passwords do not match'; header('Location: register.php'); exit; } $dsn = 'mysql:dbname=mydatabase;host=localhost'; $dbUser='myuser'; $dbPassword= 'password'; try{ $connection = new PDO($dsn, $dbUser, $dbPassword); } catch (PDOException $exception){ $_SESSION['messages'][] = 'Connection failed: ' . $exception->getMessage(); header('Location: register.php'); exit; }
messages.php <?php session_start(); if (empty($_SESSION['messages'])){ return; } $messages = $_SESSION['messages']; unset($_SESSION['messages']); ?> <ul> <?php foreach ($messages as $message): ?> <li><?php echo $message; ?></li> <?php endforeach; ?> </ul> Edited Wednesday at 12:49 AM by bee65 Hi folks, I'm curious if I can for example, save a file from my server and it will save to all other servers - obviously if they accepted the connection first. It's for a software I developed and is almost complete and know there will be frequent updates to it. Instead of users downloading upates, I want the update files from my server to somehow synchronize to their server automatically? Anything called this?? Thanks for info. This isn't exactly an application design question, but rather a system design one.
I am about to install an Inventory Control System inside this store I work in.
The store itself also owns a Linode VPS running Centos 6.4 which hosts our website.
This new Inventory System will come built in with a Microsoft SQL Server, and supposedly it is a SQL Anywhere database, but I'm not too sure what that means.
I need to make this database publicly accessible, but only via the Linode VPS. Surely, setting restrictions is easy enough to address that issue. That isn't my question.
My first idea is to put this server into the DMZ, easy. But it doesn't exactly sound safe. So my next idea was to put a middleman server in the DMZ, this way the Linode can send queries to that middleman server and it will send that data to the SQL Server and back. This is very vaguely described I know, but I don't want to get too much into details, but rather, understand how I can create that middleman server, and what could Install onto it that would allow me to securely process queries?
My first thought was to install a webservice, that accepts an XML/JSON request and returns an XML/JSON response.
Then, I realized directly afterwards that I don't have any experience setting up a webservice like that.
What kind of options are there out there? Ultimately, my question is, should I just put the Server in the DMZ or should I create the middleman, and if so, can someone point me in the right direction as to getting a webservice set up? Edited by Zane, 15 July 2014 - 11:28 PM. Following a tutorial on udemy, i tried to learn the very basics of mvc structure. I built the same project on my local server and it worked without giving me any error. but when i tried it on live server. its not working as it should. not showing any error. I tried to figure out the problem and found that for every page loading, it stops at the same line in my main.php file. <?php require($view); ?> starting from the above line. it stops. i came here to share my problem but i am unable to upload my files here. if there is a way to upload and share my files, please guide. zip file size of the whole project is 31.6 kb |