Search Results

Search found 22756 results on 911 pages for 'power query'.

Page 476/911 | < Previous Page | 472 473 474 475 476 477 478 479 480 481 482 483  | Next Page >

  • Problems while installing Ubuntu 12.04 64 & 32 bits from LiveCd

    - by user76104
    I am trying to install Ubuntu on my new pc but I am encountering problems. When I boot from the installation CD, everything runs fine until the "window" when the user has to decide to Try Ubuntu or Install it in the machine. Well that "window" appears in blank, and my mouse and keyboard behaves really slow. I can't do anything , so i have to shut it down by pressing power button. The specification of my pc are this Motherboard: Gigabyte ex58 ud7 CPU: i7 950 HD: Western digital caviar black Int. memory: 6gb memory corsair Graphics: evga gtx580 I really need to install Ubuntu or another Linux distribution, i am using de seismic Unix program {Edit : Guess it is this} on my laptop. I hope someone can help me.

    Read the article

  • Can not login to Normal mode

    - by Technology is good
    I installed ubuntu 12 in my old PC Acer Power Series. After installation i got the Login Screen which appears unusally big. when i logged-in in some fail-safe mode. I got the normal ubuntu running. but if i restart the system i can login using fail-safe mode but i can't login normally. I think the problem must be i don't have a graphic card. Even if i do have one that must be very very low configuration one. Kindly help me with graphic card config if that is the problem with my PC. I just want to use ubuntu for normal documentation purpose. I am new to ubuntu so techcies help me with basic guidance. Thank you. Karthik Muralitharan

    Read the article

  • jquery ajax form function(data)

    - by RussP
    Can some one please tell me where I have gone wrong. What ever I do I get the answer "no" JQuery to send data to php query $j.post("logincheck.php",{ username:$j('#username').attr('value'), password:$j('#password').attr('value'), rand:Math.random() } , function(data) { if(data=='yes') {alert('yes');} else {alert('no');} } ); Here is the php query if(isset($_POST['username'])): $username = $_POST['username']; $password = $_POST['password']; $posts = mysql_query("SELECT * FROM users WHERE username='$username'"); $no_rows = mysql_num_rows($posts ); while($row = mysql_fetch_array($posts)): print 'yes'; endwhile; else: print 'no'; //header('location: index.php'); endif; endif; Thank in adance

    Read the article

  • Simulating python's With statement in java

    - by drozzy
    Is there something like Python with context manager in Java? For example say I want to do something like the following: getItem(itemID){ Connection c = C.getConnection(); c.open(); try{ Item i = c.query(itemID); }catch(ALLBunchOfErrors){ c.close(); } c.close(); return c; } where in python I just have: with( C.getConnection().open() as c): Item i = c.query(itemID); return i;

    Read the article

  • Please help me understand this PHP script.Noob here

    - by NissGTR
    I'm learning PHP,MySQL and came across this function today function get_director($director_id) { global $db; $query = 'SELECT people_fullname FROM people WHERE people_id = ' . $director_id; $result = mysql_query($query, $db) or die(mysql_error($db)); $row = mysql_fetch_assoc($result); extract($row); return $people_fullname; } I understand what functions are and I've created a few while learning PHP.But this one is a bit more complicated.I can't understand the WHERE people_id = ' . $director_id I'm new to MySQL.I guess the single quote ends the MySQL statement?And then it is concatenated with the argument? Please help me out.

    Read the article

  • Getting Popular Topics on a Custom Made Forum

    - by Navarr
    For this website we're working on, we're trying to get the most popular topics (based on how many posts have been made in them within the last 24 hours). We have a medium to large based forum, and the current MySQL query looks like this: SELECT `forums_topics`.*, ( SELECT COUNT(`id`) FROM `forums_posts` WHERE `postdate` > (UNIX_TIMESTAMP()-60*60*24) AND `topicid`=`forums_topics`.`id` ) AS `trendy_threads` FROM `forums_topics` WHERE `deleted`=0 ORDER BY `trendy_threads` DESC,`postdate` DESC LIMIT 3 This unfortunately causes an SQL timeout with just one query and Error 500's. How can we get this information as quickly and as efficiently as possible?

    Read the article

  • A look at an example of anti-spam algorithm

    - by pragmaticCamel
    What is a good approach to an anti-spam algorithm for a website similar to reddit? Their anti-spam algorithm seems awfully broken (banning on words in the title and doing a horrible job for that matter). Considering a post spam because it has the word 'spam' in the title is really not a wise choice. Anyway, how can one approach such problem ? Are there any tools that help in such cases? Also, what are the /technical/ reasons behind reddit's choice not using reCAPTCHA on every post submission? It seems like a much better solution than what they have right now. Since reddit is basically a community-driven website why not give such power to the communities' trusted members?

    Read the article

  • Optimizing PHP code (trying to determine min/max/between case)

    - by Swizzh
    I know this code-bit does not conform very much to best coding practices, and was looking to improve it, any ideas? if ($query['date_min'] != _get_date_today()) $mode_min = true; if ($query['date_max'] != _get_date_today()) $mode_max = true; if ($mode_max && $mode_min) $mode = "between"; elseif ($mode_max && !$mode_min) $mode = "max"; elseif (!$mode_max && $mode_min) $mode = "min"; else return; if ($mode == "min" || $mode == "between") { $command_min = "A"; } if ($mode == "max" || $mode == "between") { $command_max = "B"; } if ($mode == "between") { $command = $command_min . " AND " . $command_max; } else { if ($mode == "min") $command = $command_min; if ($mode == "max") $command = $command_max; } echo $command;

    Read the article

  • How do check if PDO object is connected properly inside a different class?

    - by tgun926
    I want to fetch some information from my mysql database in a class, so I'm passing in the PDO object into a __construct function, and working from there. However, what's an elegant way of checking to see if the PDO object was correctly created, and that the connection is open when the Table class is instantiated? class Table{ public function __construct(PDO $db, $week){ try{ $query = $db -> query ("SELECT * FROM `table1` where `day` = 'monday'"); } catch(PDOExeption $e){ echo 'error: '. $e->getMessage(); //die(); } } } I don't think this code does what I want.

    Read the article

  • Help with MySQL and CASE WHEN with a range of values

    - by kickdaddy
    I have an accounts table and a records table where accounts have multiple records. I would like to break down the account totals by "count of records" range. i.e. show the breakdown of Count of Records | Count 0-25 | 100 25 - 50 | 122 50 - 100 | 300 etc. I am using the following query, but I can't get it to group by "grp" which is what I want, any help on the best way to modify query? Thanks! SELECT count(*) as ct, CASE WHEN COUNT(*) < 25 THEN '1-25' WHEN COUNT(*) >= 25 < 50 THEN '25-50' WHEN COUNT(*) >= 50 < 100 THEN '50-100' WHEN COUNT(*) >= 100 < 250 THEN '100-250' WHEN COUNT(*) >= 250 < 500 THEN '250-500' WHEN COUNT(*) >= 500 < 1000 THEN '500-1000' ELSE '1000+' END AS grp FROM records r,accounts a WHERE r.account_id=a.id ORDER BY ct

    Read the article

  • Count number of messages per user

    - by Pr0no
    Consider the following tables: users messages ----------------- ----------------------- user_id messages msg_id user_id content ----------------- ----------------------- 1 0 1 1 foo 2 0 2 1 bar 3 0 3 1 foobar 4 3 baz 5 3 bar I want to count the number of messages per user and insert the outcome into users.messages, like this: users ----------------- user_id messages ----------------- 1 3 2 0 3 2 I could use PHP to perform this operation, pseudo: foreach ($user_id in users) { $count = select count(msg_id) from messages where user_id = $user_id update users set messages = $count } But this is probably very inefficient as compared to one query executed in MySQL directly: UPDATE users SET messages = ( SELECT COUNT(msg_id) FROM messages ) But I'm sure this is not a proper query. Therefore, any help would be appreciated :-)

    Read the article

  • Separate Datasource for Gridview Footer

    - by Mike Keller
    I need to able to display the results from a seperate query in the footer of a gridview. What is the best way to go about this? I did some google searches and what I found pretty much revolves using the original datasource and doing something like addition to display the information in the footer. However I really need to be able to display this single line of information from a seperate query. Any direction given would be appreciated. Links, tips, tricks, etc. Thanks.

    Read the article

  • What should I do when Ubuntu freezes?

    - by ændrük
    All operating systems freeze sometimes, and Ubuntu is no exception. What should I do to regain control when... just one program stops responding? nothing at all responds to mouse clicks or key presses? the mouse stops moving entirely? In what order should I try various solutions before deciding to pull the power plug? What should I do when starting up Ubuntu fails? Is there a diagnostic procedure I can follow?

    Read the article

  • Broken count(*) after adding LEFT JOIN

    - by Iain Urquhart
    Since adding the LEFT JOIN to the query below, the count(*) has been returning some strange values, it seems to have added the total rows returned in the query to the 'level': SELECT `n`.*, exp_channel_titles.*, round((`n`.`rgt` - `n`.`lft` - 1) / 2, 0) AS childs, count(*) - 1 + (`n`.`lft` > 1) + 1 AS level, ((min(`p`.`rgt`) - `n`.`rgt` - (`n`.`lft` > 1)) / 2) > 0 AS lower, (((`n`.`lft` - max(`p`.`lft`) > 1))) AS upper FROM `exp_node_tree_6` `n` LEFT JOIN `exp_channel_titles` ON (`n`.`entry_id`=`exp_channel_titles`.`entry_id`), `exp_node_tree_6` `p`, `exp_node_tree_6` WHERE `n`.`lft` BETWEEN `p`.`lft` AND `p`.`rgt` AND ( `p`.`node_id` != `n`.`node_id` OR `n`.`lft` = 1 ) GROUP BY `n`.`node_id` ORDER BY `n`.`lft` I'm totally stumped... Thank you!

    Read the article

  • Select where a value present

    - by Roy
    First a database example: id, product_id, cat, name, value -------------------------------- 1,1,Algemeen,Processor,2 Ghz 2,1,Algemeen,Geheugen,4 GB 3,2,Algemeen,Processor,3 Ghz 4,2,Algemeen,Geheugen,4 GB 5,3,Beeldscherm,Inch,22" 6,3,Beeldscherm,Kleur,Zwart 7,3,Algemeen,Geheugen,3 GB 8,3,Algemeen,Processor,3 Ghz I want with one query to select the follow id's: 1,2,3,4,7,8 Because the cat = algemeen and the name = processor by these products. ID 5,6 are only present by product 3. So, the entry's (cat and name) which are present by all products (product_id) have to be selected. The database contains 80.000 entry's with a lot of diffrent cat's, name's and value's. Is this possible with one query or is some php necessary? How do I do this? My apologies for the bad English.

    Read the article

  • What will happen if I change the type of a column from int to year?

    - by MachinationX
    I have a table in MySQL 4.0 which currently has a year field as a smallint(6) type. What will happen if I convert it directly to a Year type with a query like the following: ALTER TABLE t MODIFY y YEAR(4) NOT NULL DEFAULT CURRENT_TIMESTAMP; When the current members of column y have values like 2010? I assume that because the year type is technically values from 1-255, values above that will be truncated or broken. So if MySQL isn't smart enough to realize that 2010(int) = 110(year), what would be the simplest query or queries to convert the values? Thanks for your help!

    Read the article

  • GROUP BY on multiple columns

    - by Tams
    I have a table that looks like the following - Id Reference DateAttribute1 DateAttribute2 1 MMM005 2011-09-11 2012-09-10 2 MMM005 2012-06-13 2012-09-10 3 MMM006 2012-08-22 2012-09-10 4 MMM006 2012-08-22 2012-09-11 I have handle to the id values. I would like to query such that I get the following result Id Reference DateAttribute1 DateAttribute2 2 MMM005 2012-06-13 2012-09-10 4 MMM006 2012-08-22 2012-09-11 I would like my result to be grouped by reference and then 'DateAttribute1' and then 'DateAttribute2' as such - DateAttribute1 has a priority over DateAttribute2 as you can see above in the result. How should I write my query to fetch the results in the above manner? Any solution?

    Read the article

  • problem in using while loop in php&mysql

    - by Mac Taylor
    hey guys im using a while loop to show my latest forum topics now i need to count some fields either I'm trying to do it in one query and here is my code : $result = $db->sql_query("SELECT t.*,p.*, SUM(t.topic_approved='1') AS Amount_Of_Topics, SUM(t.topic_views) AS Amount_Of_Topic_Views, SUM(t.topic_replies) AS Amount_Of_Topic_Replies, SUM(p.post_approved ='1') AS Amount_Of_Posts FROM bb3topics t left join bb3posts p ON t.topic_id=p.topic_id ORDER BY t.topic_last_post_id DESC LIMIT 10 " ); while( $row = $db->sql_fetchrow($result) ) { problem : this code shows only one forum topic and not the rest , but if i remove sum() part from it , then it shows the rest is there anything wrong with my query code ?!

    Read the article

  • Why is my left join not returning nulls?

    - by Griz
    In sql server 2008, I have the following query: select c.title as categorytitle, s.title as subcategorytitle, i.title as itemtitle from categories c join subcategories s on c.categoryid = s.categoryid left join itemcategories ic on s.subcategoryid = ic.subcategoryid left join items i on ic.itemid = i.itemid where (ic.isactive = 1 or ic.isactive is null) and i.siteid = 132 order by c.title, s.title I am trying to get items in their subcategories, but I still want to return a record if there are no items in the category or subcategory. Subcategories that have no items are never returned. What am I doing wrong? Thank you EDIT Modified query with a second left join and where clause, but it's still not returning nulls. :/

    Read the article

  • Calling Google's Custom Search API via Python

    - by user353829
    I am writing in Python a module that will query Google's Custom Search API and return all listings of domain 'example.com' I Have been reading instructions at https://code.google.com/apis/customsearch/v1/getting_started.html and am a little stumped at the moment. Are my assumptions listed below correct? For example, to search for results that has 'example.com' in the URL, the query is: *'https://www.googleapis.com/customsearch/v1?key=my_key&cx=017576662512468239146:omuauf_lfve&q=site:example.com*' *key=my_key*: value of key given by google cx=017576662512468239146: name of the search engine (google)? *omuauf_lfve*: I have no idea what this is q=site:example.com: This should return all results with 'example.com'; e.g. www.a.example.com, b.example.com, example .com

    Read the article

  • Help with PHPExcel Library and mySQL data from a table

    - by user547363
    Hello! Everyone.. I have this script $query = "SELECT id,last_name,first_name FROM users WHERE tmima_id='6'"; $result = @mysql_query($query); while($row = mysql_fetch_array($result)) { $i = 3; $emp_id = $row['id']; $cell = 'A'.$i; $objPHPExcel->setActiveSheetIndex(0) ->setCellValue($cell, $row['last_name']. $row['first_name']); $i++; } But in the .xls file it prints only one user. Why id doesnt print all of the users ? W Thanks in advance.

    Read the article

  • Manually re-enable wireless and bluetooth

    - by tacozmeister
    So, today, I decided to turn off bluetooth to save battery power. When I did so, it did everything like normal. Whatever. But whenever I want to re-enable Wifi and Bluetooth, it just has "Enable Wireless" and "Enable bluetooth" grayed out in the respective things on the top-right of the desktop. I am using 12.04 Precise Pangolin on a Dell laptop that has never had wireless problems before. Clicking the grayed-out part does nothing. Re-logging and restarting has no effect, and I'm on an account that has full administrative permission. My attempts: ~$ sudo dmesg | grep wlan0 [19.960211] ADDRCONF(NETDEV_UP): wlan0: link is not ready . ~$ sudo dmesg | grep blue [[nothing]] What do I do to manually re-enable wireless and bluetooth?

    Read the article

< Previous Page | 472 473 474 475 476 477 478 479 480 481 482 483  | Next Page >