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  • Ubuntu terminal questions

    - by Camran
    I am wondering about my VPS providers ubuntu terminal. Are all these terminals the same? I think they are so user-UN-friendly. I can't copy-paste into the terminal, when I try opening textfiles, I can't scroll up and down easily. I cant save easily. Nothing is easy... Is it always like this with Ubuntu? Is there any way to make it easier? I use windows but I login to my vps provider with login details and then simply click "terminal" to open the terminal. Please help me out here

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  • Turning Floats into Their Closest (UTF-8 Character) Fraction.

    - by Mark Tomlin
    I want to take any real number, and return the closest number, with the closest fraction as available in the UTF-8 character set, appropriate. 0/4 = 0.00 = # < .125 1/4 = 0.25 = ¼ # > .125 & < .375 2/4 = 0.50 = ½ # > .375 & < .625 3/4 = 0.75 = ¾ # > .625 & < .875 4/4 = 1.00 = # > .875 I made this function to do that task: function displayFraction($realNumber) { if (!is_float($realNumber)) { return $realNumber; } list($number, $decimal) = explode('.', $realNumber); $decimal = '.' . $decimal; switch($decimal) { case $decimal < 0.125: return $number; case $decimal > 0.125 && $decimal < 0.375: return $number . '¼'; # 188 ¼ &#188; case $decimal > 0.375 && $decimal < 0.625: return $number . '½'; # 189 ½ &#189; case $decimal > 0.625 && $decimal < 0.875: return $number . '¾'; # 190 ¾ &#190; case $decimal < 0.875: return ++$number; } } What are the better / diffrent way to do this? echo displayFraction(3.1) . PHP_EOL; # Outputs: 3 echo displayFraction(3.141593) . PHP_EOL; # Outputs: 3¼ echo displayFraction(3.267432) . PHP_EOL; # Outputs: 3¼ echo displayFraction(3.38) . PHP_EOL; # Outputs: 3½ Expand my mind!

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  • function with array variable

    - by mtokoly
    How do I pass an array through a function, for example: $data = array( 'color' => 'red', 'height' => 'tall' ); something($data); function something($data) { if ($data['color'] == 'red') { // do something } } how can I get the function to recognize $data[color] and $data[height]?

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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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  • 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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  • 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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  • 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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  • 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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  • Cakephp change model's title without rename model/dbtable

    - by Vincent
    I have a table that store all information about class, but because Class is reserved for class name I had to rename the table from classes to types. But in the view section I need it to be displayed as "Class", including the Paginator links. Anyway to achieve this by adding something in the Type model, without completely customize Paginator and and all view compoenents?

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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • From where to send mails in a MVC framework, so that there is no duplication of code?

    - by Sabya
    It's a MVC question. Here is the situation: I am writing an application where I have "groups". You can invite other persons to your groups by typing their email and clicking "invite". There are two ways this functionality can be called: a) web interface and b) API After the mail sending is over I want to report to the user which mails were sent successfully (i.e., if the SMTP send succeeded. Currently, I am not interested in reporting mail bounces). So, I am thinking how should I design so that there is no code duplication. That is, API and web-interface should share the bulk of the code. To do this, I can create the method "invite" inside the model "group". So, the API and and the Web-interface can just call: group-invite($emailList); This method can send the emails. But the, problem is, then I have to access the mail templates, create the views for the mails, and then send the mails. Which should actually be in the "View" part or at least in the "Controller" part. What is the most elegant design in this situation? Note: I am really thinking to write this in the Model. My only doubt is: previously I thought sending mails also as "presentation". Since it is may be considered as a different form of generating output.

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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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