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  • Retrieving data from MySQL in one SQL statement

    - by james.ingham
    Hi all, If I'm getting my data from Mysql like so: $result = $dbConnector->Query("SELECT * FROM branches, businesses WHERE branches.BusinessId = businesses.Id ORDER BY businesses.Name"); $resultNum = $dbConnector->GetNumRows($result); if($resultNum > 0) { for($i=0; $i < $resultNum; $i++) { $row = $dbConnector->FetchArray($result); // $row['businesses.Name']; // $row['branches.Name']; echo $row['Name']; } } Does anyone know how to print the field Name in businesses and how to print the name from branches? My only other alternative is to rename the fields or to call Mysql with two seperate queries. Thanks in advance

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  • Mod redirct error.

    - by user150253
    when I used the SSL fot my website i got following error. in error log. and site shows me default page instead of actual page. I got following error. .htaccess: RewriteEngine not allowed here

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  • Complicated MySQL query?

    - by Scott
    I have two tables: RatingsTable that contains a ratingname and a bit whether it is a positive or negative rating: Good 1 Bad 0 Fun 1 Boring 0 FeedbackTable that contains feedback on things...the person rating, the rating and the thing rated. The feedback can be determined if it's a positive or negative rating based on RatingsTable. Jim Chicken Good Jim Steak Bad Ted Waterskiing Fun Ted Hiking Fun Nancy Hiking Boring I am trying to write an efficient MySQL query for the following: On a page, I want to display the the top 'things' that have the highest proportional positive ratings. I want to be sure that the items from the feedback table are unique...meaning, that if Jim has rated Chicken Good 20 times...it should only be counted once. At some point I will want to require a minimum number of ratings (at least 10) to be counted for this page as well. I'll want to to do the same for highest proportional negative ratings, but I am sure I can tweak the one for positive accordingly.

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  • nginx multiple domain virtual host configuration

    - by Poe
    I'm setting up nginx with multiple domain or wildcard support for convenience sake, rather than setting up 50+ different sites-available/* files. Hopefully this is enough to show you what I'm trying to do. Some are static sites, some are dynamic with usually wordpress installed. If an index.php exists, everything works as expected. If a file is requested that does not exist (missing.html), a 500 error is given due to the rewrite. The logged error is: *112 rewrite or internal redirection cycle while processing "/index.php/index.php/index.php/index.php/index.php/index.php/index.php/index.php/index.php/index.php/index.php/missing.html" The basic nginx configuration I'm currently using is: ` listen 80 default; server _; ... location / { root /var/www/$host; if (-f $request_filename) { expires max; break; } # problem, what if index.php does not exist? if (!-e $request_filename) { rewrite ^/(.*)$ /index.php/$1 last; } } ... ` If an index.php does not exist, and the file also does not exist, I would like it to error 404. Currently, nginx does not support multiple condition if's or nested if so I need a workaround.

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  • I'm working on a website that sells different artwork, what's the best way to handle different image

    - by ThinkingInBits
    I'm working on a website that will allow users to upload and sell their artwork in different sizes. I was wondering what the best way would be to handle the different file sizes automatically. A few points I was curious on: How to define different size categories (small, medium, large) in such a way that I'll be able to dynamically re-size images with proportional dimensions. Should I store actual jpegs of the different sizes for download? Or would it be easier to generate these different sizes for download on the fly My thumbnails will be somewhat larger than your average thumbnails, should I store a second 'thumbnail image' with the sites watermark overlaying it? Or once again, generate this on the fly? All opinions, advice are greatly appreciated!

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  • VB - Server communication

    - by Bubby4j
    How would I make a Visual Basic application talk to a web server? Someone will press something on the site and I want the application to display the information that the web site passes along to it.

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  • How to optimize this user ranking query

    - by James Simpson
    I have 2 databases (users, userRankings) for a system that needs to have rankings updated every 10 minutes. I use the following code to update these rankings which works fairly well, but there is still a full table scan involved which slows things down with a few hundred thousand users. mysql_query("TRUNCATE TABLE userRankings"); mysql_query("INSERT INTO userRankings (userid) SELECT id FROM users ORDER BY score DESC"); mysql_query("UPDATE users a, userRankings b SET a.rank = b.rank WHERE a.id = b.userid"); In the userRankings table, rank is the primary key and userid is an index. Both tables are MyISAM (I've wondered if it might be beneficial to make userRankings InnoDB).

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  • Doctrine enum type by value

    - by MitMaro
    I have a column in a table defined as following in my yaml file: myTable: columns: value: type: enum length: 2 values: ['yes', 'no'] In the code I am trying to insert data into this table but I can't figure out a way to insert the data using the enum text value (ie. 'yes' or 'no'). What I was trying was is something like this: $obj = new myTable(); // the model for this table $obj->value = 'yes'; // if I use the numerical value for this it works I am using Doctrine 1.1.0.

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  • How to set the content of the $script_for_layout in cakephp?

    - by sipiatti
    Hi, I am learning cakephp in an autodidact manner and I am a complete newbie :) I am creating a simple application. Some logic work so now I going to dive into designing views and layouts. I read the docs and the tutorial but I could not find where to set the content of $script_for_layout. Especially I want to set a $html-css to create link in pages to the stylesheet. I found out that I could do it in directly in the layout template but I would like to create the same link in all pages/views/layouts to the stylesheet so I hope there is a simple way, and avoid to do this in all layouts and/or controllers

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  • CakePHP: How can I change this find call to include all records that do not exist in the associated

    - by Stephen
    I have a few tables with the following relationships: Company hasMany Jobs, Employees, and Trucks, Users I've got all my foreign keys set up properly, along with the tables' Models, Controllers, and Views. Originally, the Jobs table had a boolean field called "assigned". The following find operation (from the JobsController) successfully returns all employees, all trucks, and any jobs that are not assigned and fall on a certain day for a single company (without returning users by utilizing the containable behavior): $this->set('resources', $this->Job->Company->find('first', array( 'conditions' => array( 'Company.id' => $company_id ), 'contain' => array( 'Employee', 'Truck', 'Job' => array( 'conditions' => array( 'Job.assigned' => false, 'Job.pickup_date' => date('Y-m-d', strtotime('Today')); ) ) ) ))); Now, since writing this code, I decided to do a lot more with the job assignments. So I've created a new model "Assignment" that belongsTo Truck and belongsTo Job. I've added the hasMany Assignments to both the Truck model and the Jobs Model. I have both foreign keys in the assignments table, along with some other assignment fields. Now, I'm trying to get the same information above, only instead of checking the assigned field from the job table, I want to check the assignments table to ensure that the job does not exist there. I can no longer use the containable behavior if I'm going to use the "joins" feature of the find method due to mysql errors (according to the cookbook). But, the following query returns all jobs, even if they fall on different days. $this->set('resources', $this->Job->Company->find('first', array( 'joins' => array( array( 'table' => 'employees', 'alias' => 'Employee', 'type' => 'LEFT', 'conditions' => array( 'Company.id = Employee.company_id' ) ), array( 'table' => 'trucks', 'alias' => 'Truck', 'type' => 'LEFT', 'conditions' => array( 'Company.id = Truck.company_id' ) ), array( 'table' => 'jobs', 'alias' => 'Job', 'type' => 'LEFT', 'conditions' => array( 'Company.id = Job.company_id' ) ), array( 'table' => 'assignments', 'alias' => 'Assignment', 'type' => 'LEFT', 'conditions' => array( 'Job.id = Assignment.job_id' ) ) ), 'conditions' => array( 'Job.pickup_date' => $day, 'Company.id' => $company_id, 'Assignment.job_id IS NULL' ) )));

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  • I made a horrible loop.... help fix my logic please

    - by Webnet
    I know I'm doing this a bad way... but I'm having trouble seeing any alternatives. I have an array of products that I need to select 4 of randomly. $rawUpsellList is an array of all of the possible upsells based off of the items in their cart. Each value is a product object. I know this is horribly ugly code but I don't see an alternative now.... someone please put me out of my misery so this code doesn't make it to production..... $rawUpsellList = array(); foreach ($tru->global->cart->getItemList() as $item) { $product = $item->getProduct(); $rawUpsellList = array_merge($rawUpsellList, $product->getUpsellList()); } $upsellCount = count($rawUpsellList); $showItems = 4; if ($upsellCount < $showItems) { $showItems = $upsellCount; } $maxLoop = 20; $upsellList = array(); for ($x = 0; $x <= $showItems; $x++) { $key = rand(0, $upsellCount); if (!array_key_exists($key, $upsellList) && is_object($rawUpsellList[$key])) { $upsellList[$key] = $rawUpsellList[$key]; $x++; } if ($x == $maxLoop) { break; } } Posting this code was highly embarassing...

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  • Checking for valid email addresses

    - by Roland
    I'm running a website with more than 60 000 registered users. Every week notifications are send to these users via email, now I've noticed some of the mail addresses do not exists anymore eg. the domain address is valid but the email name en asdas@ is not valid anymore since person does not work at a company anymore etc. Now I'm looping through the database and doing some regular expression checks and checking if the MX records exist with the following two functions function verify_email($email){ if(!preg_match('/^[_A-z0-9-]+((\.|\+)[_A-z0-9-]+)*@[A-z0-9-]+(\.[A-z0-9-]+)*(\.[A-z]{2,4})$/',$email)){ return false; } else { return true; } } // Our function to verify the MX records function verify_email_dns($email){ list($name, $domain) = split('@',$email); if(!checkdnsrr($domain,'MX')){ return false; } else { return true; } } If the email address is in an invalid format or the domain does not exists I delete the users account. Are there any methods I could use to check if the email address still exists or not if the domain name is valid and the email address is in the correct format? For example [email protected] does not exist anymore but test.com is a valid domain name.

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  • GeSHi single like link

    - by justme
    Is it possible to instruct GeSHi (http://qbnz.com/highlighter/) to generate link to single line. For example if I show highlighted code on 'example.com/my-code' URL, I would like to be able to have link like: 'example.com/my-mode#line-69' or something like that...

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  • Mysql syntax using IN help!

    - by Axel
    Hi, i have a pictures table : pictures(articleid,pictureurl) And an articles table : articles(id,title,category) So, briefly, every article has a picture, and i link pictures with article using articleid column. now i want to select 5 pictures of articles in politic category. i think that can be done using IN but i can't figure out how to do it. Note: Please only one query, because i can do it by selecting articles firstly then getting the pictures. Thanks

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  • How to Sync CI (Hudson) Activity into an existing automated Build Process (phing, svn)?

    - by maraspin
    OUR CURRENT BUILD PROCESS We're a small team of developers (2 to 4 people depending on project) who currently use Phing to deploy code to a staging environment, before going live. We keep our code in a SVN repo, where the trunk holds current active development and, at certain times, we do make branches that we test and then (if successful), tag and export to the staging env. If everything goes well there too, we finally deploy'em in production servers. Actions are highly automated, but always triggered by human intervention. THE DOUBT We'd now like to introduce Continuous Integration (with Hudson) in the process; unfortunately we have a few doubts about activity syncing, since we're afraid that CI could somewhat interfere with our build process and cause certain problems. Considering that an automated CI cycle has a certain frequency of automatically executed actions, we see 2 possible cases for "integration", each with its own problems: Case A: each CI cycle produces a new branch with its own name; we do use such a name to manually (through phing as it happens now) export the code from the SVN to the staging env. The problem I see here is that (unless specific countermeasures are taken - IE deletion) the number of branches we have can easily grow out of control (let's suppose we commit often, so that we have a fresh new build/branch every N minutes). Case B: each CI cycle creates a new branch named 'current', which is then tagged with a unique name only when we manually decide to export it to staging; the current branch, at any case is then deleted, as soon as the next CI cycle starts up. The problem we see here is that a new cycle could kick in while someone is tagging/exporting the 'current' branch to staging thus creating an inconsistent build (but maybe here I'm just too pessimist, since I confess I don't know whether SVN offers some built-in protection against this). With all this being said, I was wondering if anyone with similar experiences could be so kind to give us some hints on the subject, since none of the approaches depicted above looks completely satisfing to us. Is there something important we just completely left off in the overall picture? Thanks for your attention & (in advance) for your help!

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  • rookie MySql question about paging; Is one query enough?

    - by Camran
    I have managed to get paging to work, almost. I want to display to the user, total nr of records found, and the currently displayed records. Ex: 4000 found, displaying 0-100. I am testing this with the nr 2 (because I don't have that many records, have like 20). So I am using LIMIT $start, $nr_results; Do I have to make two queries in order to display the results the way I want, one query fetching all records and then make a mysql_num_rows to get all records, then the one with the LIMIT ? I have this: mysql_num_rows($qry_result); $total_pages = ceil($num_total / $res_per_page); //$res_per_page==2 and $num_total = 2 if ($p - 10 < 1) { $pagemin=1; } else { $pagemin = $p - 10; } if ($p + 10 $total_pages) { $pagemax = $total_pages; } else { $pagemax = $p + 10; } Here is the query: SELECT mt.*, fordon.*, boende.*, elektronik.*, business.*, hem_inredning.*, hobby.* FROM classified mt LEFT JOIN fordon ON fordon.classified_id = mt.classified_id LEFT JOIN boende ON boende.classified_id = mt.classified_id LEFT JOIN elektronik ON elektronik.classified_id = mt.classified_id LEFT JOIN business ON business.classified_id = mt.classified_id LEFT JOIN hem_inredning ON hem_inredning.classified_id = mt.classified_id LEFT JOIN hobby ON hobby.classified_id = mt.classified_id ORDER BY modify_date DESC LIMIT 0, 2 Thanks, if you need more input let me know. Basically Q is, do I have to make two queries?

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  • Preventing the opening of a form on an ad

    - by Jonathan
    Hey guys, Did you guys know how to prevent the open of a Form when I click on a add button? Maybe using beforeShowForm? function(formid) { if(jQuery('#gridap').getGridParam('selrow')) { idgridap=jQuery('#gridap').getGridParam('selrow'); jQuery('#FK_numerocontrato_ap',formid).val(idgridap).attr('readonly','readonly'); } else { // I want to prevent the openning of the add form here and maybe show an alert using the "alertcap" } } CHECAROW; $grid->setNavEvent('add','beforeShowForm',$checarowid); BTW, there's a way to call the alertmod of jqgrid and add a custom message to it? tks!

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  • Is it possible to combine these 3 mySQL queries?

    - by Greenie
    I know the $downloadfile - and I want the $user_id. By trial and error I found that this does what I want. But it's 3 separate queries and 3 while loops. I have a feeling there is a better way. And yes, I only have a very little idea about what I'm doing :) $result = pod_query("SELECT ID FROM wp_posts WHERE guid LIKE '%/$downloadfile'"); while ($row = mysql_fetch_assoc($result)) { $attachment = $row['ID']; } $result = pod_query("SELECT pod_id FROM wp_pods_rel WHERE tbl_row_id = '$attachment'"); while ($row = mysql_fetch_assoc($result)) { $pod_id = $row['pod_id']; } $result = pod_query("SELECT tbl_row_id FROM wp_pods_rel WHERE tbl_row_id = '$pod_id' AND field_id = '28'"); while ($row = mysql_fetch_assoc($result)) { $user_id = $row['pod_id']; }

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  • to take values of checkbox in table attributes

    - by mwj
    i have a database patient with 3-4 tables n each table has about 8 attributes.... i have a table medical history which has attribute additional info ... under which i have 5 checkboxes.... all the values entered are taken up except the chekbox values..... plz help

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  • User accounts in Symfony?

    - by gruner
    I'm new to Symfony. Is my understanding correct that the User class is actually for controlling sessions? But is there built-in login and account creation? I'm not finding it. But if there's an admin backend generator, how can it function without user logins?

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  • Connection for control user as defined in your configuration failed. xampp

    - by Zann
    when i tried to uninstall xampp and reinstall xampp.I received below error message when i go phpmyadmin Need help and guide to solve it .thanks Error MySQL said: Documentation 1045 - Access denied for user 'root'@'localhost' (using password: NO) Connection for controluser as defined in your configuration failed. phpMyAdmin tried to connect to the MySQL server, and the server rejected the connection. You should check the host, username and password in your configuration and make sure that they correspond to the information given by the administrator of the MySQL server.

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