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  • SQL: Gather right hand values from a join

    - by Max Williams
    Let's say a question has many tags, via a join table called taggings. I do a join thus: SELECT DISTINCT `questions`.id FROM `questions` LEFT OUTER JOIN `taggings` ON `taggings`.taggable_id = `questions`.id LEFT OUTER JOIN `tags` ON `tags`.id = `taggings`.tag_id I want to order the results according to a particular tag name, eg 'piano', so that piano is at the top, then by all the other tags in alphabetical order. Currently i'm using this order clause: ORDER BY (tags.name = 'piano') desc, tags.name Which is going completely wrong - the first results i get back aren't even tagged with 'piano' at all. I think my problem is that i need to group the tag names somehow and do my ordering test against that: i think that doing it against the straight tags.name isn't working due to the structure of the resultant join table (it does work if i just do a simple select on the tags table) but i can't get my head around how to fix it. grateful for any advice, max

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  • Getting libjpeg/libTIFF.dylib error after recompiling php in OS X 10.6.3 Server

    - by leggo-my-eggo
    I've recompiled php 5.3.0 under OS X Server (10.6.3) in order to gain Freetype support in GD. As part of the process, I recompiled libjpeg-8a (and before that I tried libjpeg-7). But when I run apachectcl configtest, I get the following error: httpd: Syntax error on line 155 of /private/etc/apache2/httpd.conf: Cannot load /usr/libexec/apache2/mod_auth_user_host_apple.so into server: dlopen(/usr/libexec/apache2/mod_auth_user_host_apple.so, 10): Symbol not found: __cg_jpeg_resync_to_restart\n Referenced from: /System/Library/Frameworks/ApplicationServices.framework/Versions/A/Frameworks/ImageIO.framework/Versions/A/Resources/libTIFF.dylib\n Expected in: /usr/lib/libJPEG.dylib\n in /System/Library/Frameworks/ApplicationServices.framework/Versions/A/Frameworks/ImageIO.framework/Versions/A/Resources/libTIFF.dylib I don't really understand the error. Is there anyone who could help me figure out how to fix it? In php the gd library now seems unaware of libjpeg support.

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  • recursive function to get all the child categories

    - by user253530
    Here is what I'm trying to do: - i need a function that when passed as an argument an ID (for a category of things) will provide all the subcategories and the sub-sub categories and sub-sub-sub..etc. - i was thinking to use a recursive function since i don't know the number of subcategories their sub-subcategories and so on so here is what i've tried to do so far function categoryChild($id) { $s = "SELECT * FROM PLD_CATEGORY WHERE PARENT_ID = $id"; $r = mysql_query($s); if(mysql_num_rows($r) > 0) { while($row = mysql_fetch_array($r)) echo $row['ID'].",".categoryChild($row['ID']); } else { $row = mysql_fetch_array($r); return $row['ID']; } } If i use return instead of echo, i won't get the same result. I need some help in order to fix this or rewrite it from scratch

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  • php smart learner

    - by rajesh1984
    Is there a possibility to create a html or a php page that will count redirects? I mean, let's say you have a page with a link in it. I want the page to count how many times the link is clicked per ip adress or username. The counting would be reported into a log file or text document.

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  • Rows dropping when I try to join data from two tables

    - by blcArmadillo
    I have a fairly simple query I'm try to write. If I run the following query: SELECT parts.id, parts.type_id FROM parts WHERE parts.type_id=1 OR parts.type_id=2 OR parts.type_id=4 ORDER BY parts.type_id; I get all the rows I expect to be returned. Now when I try to grab the parent_unit from another table with the following query six rows suddenly drop out of the result: SELECT parts.id, parts.type_id, sp.parent_unit FROM parts, serialized_parts sp WHERE (parts.type_id=1 OR parts.type_id=2 OR parts.type_id=4) AND sp.parts_id = parts.id ORDER BY parts.type_id In the past I've never really dealt with ORs in my queries so maybe I'm just doing it wrong. That said I'm guessing it's just a simple mistake. Let me know if you need sample data and I'll post some. Thanks.

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  • Error installing RedCloth

    - by meta
    I'm getting this erron when trying to install RedCloth on openSuse: sudo gem install RedCloth Building native extensions. This could take a while... ERROR: Error installing RedCloth: ERROR: Failed to build gem native extension. /usr/bin/ruby extconf.rb creating Makefile make sh: make: nie znaleziono polecenia Gem files will remain installed in /usr/lib/ruby/gems/1.8/gems/RedCloth-4.2.3 for inspection. Results logged to /usr/lib/ruby/gems/1.8/gems/RedCloth-4.2.3/ext/redcloth_scan/gem_make.out I tried to google this out and triend everything. So I need help with that.

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  • Query Not Working

    - by John
    Hello, The simple query below is not working. Any idea why? When I echo the three variables, the correct values are returned, so I know I have variables. Thanks in advance, John $comment = $_POST['comment']; $uid = $_POST['uid']; $subid = $_POST['submissionid']; echo $comment; echo $uid; echo $subid; mysql_connect("mysqlv12", "username", "password") or die(mysql_error()); mysql_select_db("database") or die(mysql_error()); $query = sprintf("INSERT INTO comment VALUES (NULL, '%s', '%s', '%s', NULL, NULL)", $uid, $subid, $comment); mysql_query($query);

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  • Scala type conversion error, need help!

    - by Mansoor Ashraf
    Hello I am getting a weird error when trying to use a Java map in Scala. This is the snippet of code val value:Double = map.get(name) if (value eq null) map.put(name, time) else map.put(name, value + time) the map is defined as val map=new ConcurrentHashMap[String,Double] and this is the error I am getting error: type mismatch; found : Double required: ?{val eq: ?} Note that implicit conversions are not applicable because they are ambiguous: both method double2Double in object Predef of type (Double)java.lang.Double and method doubleWrapper in object Predef of type (Double)scala.runtime.RichDouble are possible conversion functions from Double to ?{val eq: ?} if (value eq null) map.put(name, time) I am new to Scala so I am having a hard time parsing the stacktrace. Any help would be appreciated

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  • Login failed for user 'NT AUTHORITY\NETWORK SERVICE'.

    - by kumar
    Hi Guys, i tried to open the website from broswer project is deployed at IIS i am getting this exception Exception information: Exception type: SqlException Exception message: Cannot open database "TestDB" requested by the login. The login failed. Login failed for user 'NT AUTHORITY\NETWORK SERVICE'. any solution? Regards kumar

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  • SQL: Need to SUM on results that meet a HAVING statement

    - by Wasauce
    I have a table where we record per user values like money_spent, money_spent_on_candy and the date. So the columns in this table (let's call it MoneyTable) would be: UserId Money_Spent Money_Spent_On_Candy Date My goal is to SUM the total amount of money_spent -- but only for those users where they have spent more than 10% of their total money spent for the date range on candy. What would that query be? I know how to select the Users that have this -- and then I can output the data and sum that by hand but I would like to do this in one single query. Here would be the query to pull the sum of Spend per user for only the users that have spent 10% of their money on candy. SELECT UserId, SUM(Money_Spent), SUM(Money_Spent_On_Candy) / SUM(Money_Spent) AS PercentCandySpend FROM MoneyTable WHERE DATE >= '2010-01-01' HAVING PercentCandySpend > 0.1;

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  • "Find nearest location" by Zip/Postal Code?

    - by codemonkey613
    I need a "find nearest location" on our website. Where visitor enters their zip/postal code, then they are redirected to specific webpage for our nearest location. We have forty USA and Canada locations. How can I build something like this? Could I do this with the Google Maps API? I already have a custom map on Google Maps. It's plotted with our locations. It would be nice to send Google Maps a command to say "what's our nearest location at __ zip code". Any suggestions?

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  • Single value data to multiple values of data in database relation

    - by Sofiane Merah
    I have such a hard time picturing this. I just don't have the brain to do it. I have a table called reports. --------------------------------------------- | report_id | set_of_bads | field1 | field2 | --------------------------------------------- | 123 | set1 | qwe | qwe | --------------------------------------------- | 321 | 123112 | ewq | ewq | --------------------------------------------- I have another table called bads. This table contains a list of bad data. ------------------------------------- | bad_id | set_it_belongs_to | field2 | field3 | ------------------------------------- | 1 | set1 | qwe | qwe | ------------------------------------- | 2 | set1 | qee | tte | ------------------------------------- | 3 | set1 | q44w | 3qwe | ------------------------------------- | 4 | 234 | qoow | 3qwe | ------------------------------------- Now I have set the first field of every table as the primary key. My question is, how do I connect the field set_of_bads to set_it_belongs_to in the bads table. This way if I want to get the entire set of data that is set1 by calling on the reports table I can do it. Example: hey reports table.. bring up the row that has the report_id 123. Okay thank you.. Now get all the rows from bads that has the set_of_bads value from the row with the report_id 123. Thanks.

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  • PHP/SQL/Wordpress: Group a user list by alphabet

    - by rayne
    I want to create a (fairly big) Wordpress user index with the users categorized alphabetically, like this: A Amy Adam B Bernard Bianca and so on. I've created a custom Wordpress query which works fine for this, except for one problem: It also displays "empty" letters, letters where there aren't any users whose name begins with that letter. I'd be glad if you could help me fix this code so that it only displays the letter if there's actually a user with a name of that letter :) I've tried my luck by checking how many results there are for that letter, but somehow that's not working. (FYI, I use the user photo plugin and only want to show users in the list who have an approved picture, hence the stuff in the SQL query). <?php $alphabet = range('A', 'Z'); foreach ($alphabet as $letter) { $user_count = $wpdb->get_results("SELECT COUNT(*) FROM wp_users WHERE display_name LIKE '".$letter."%' ORDER BY display_name ASC"); if ($user_count > 0) { $user_row = $wpdb->get_results("SELECT wp_users.user_login, wp_users.display_name FROM wp_users, wp_usermeta WHERE wp_users.display_name LIKE '".$letter."%' AND wp_usermeta.meta_key = 'userphoto_approvalstatus' AND wp_usermeta.meta_value = '2' AND wp_usermeta.user_id = wp_users.ID ORDER BY wp_users.display_name ASC"); echo '<li class="letter">'.$letter.''; echo '<ul>'; foreach ($user_row as $user) { echo '<li><a href="/author/'.$user->user_login.'">'.$user->display_name.'</a></li>'; } echo '</ul></li>'; } } ?> Thanks in advance!

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  • find elements of a varchar in another varchar

    - by Luca Romagnoli
    hi, i have a varchar field with the content like these: a,b,c,d e,d,a,c b,q,d,e i need to do a query that select only the rows with the field that has elements equals with an input string. ex input: c,a rows selected: a,b,c,d e,d,a,c is possible without use the OR (field like '%a%' OR field like '%c%') ? thanks

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  • Error With Foundation.h

    - by Nathan Campos
    Hello, I learning Objective-C in Linux(Ubuntu), but when i tryed to compile my application that needs the Foundation headers i got an error saying that the file cannot be found, but i have installed the GNUstep development package(gnustep-devel). Here is my code: // Fraction.h #import <Foundation/NSObject.h> @interface Fraction: NSObject { int numerator; int denominator; } - (void) print; - (void) setNumerator: (int) n; - (void) setDenominator: (int) d; - (void) numerator; - (void) denominator; @end And here is the console log: ubuntu@eeepc:~$ gcc main.m -o frac -lobjc In file included from main.m:3: Fraction.h:2:26: error: objc/NSObject.h: No such file or directory In file included from main.m:3: Fraction.h:4: error: cannot find interface declaration for ‘NSObject’, superclass of ‘Fraction’ ubuntu@eeepc:~$ What i need to do?

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  • Architectural advice on connecting multiple diverse sites into a single community.

    - by Aleksandar
    Hi SO, I've been given a task to connect multiple sites of the same client into a single network. So i would like to hear an architectural advice on connecting these sites into a single community. These sites include: 1. Invision Power Board Forum (the most important site) 2. 3 custom made cms-s (changes to code allowable) 3. 1 drupal site 4. 3-4 wordpress blogs Requirements are as follows: 1. Connecting all users of all sites into a single administrable entity. With permissions changing ability, users banning etc. 2. Later on, based on this implementation I have to implement "facebook like" chat, which will be available to all users regardless of place of login. I have few ideas on my mind on how to go with this, but would like to hear some people with more experience and expertize than my self. Cheers!

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  • Could you help me write a proper query in rails for accessing the following information?

    - by aditi-syal
    @workname = [] @recos = [] @bas = [] if current_user.recommendations.size != 0 current_user.recommendations.each do |r| if r.work_type == 'J' @job = Job.find_by_id(r.work_id) @workname.push "#{@job.title} at #{@job.company.name}" else @qualification = Qualification.find_by_id(r.work_id) @workname.push "Student at #{@qualification.school_name}" end @recommender = User.find_by_id(r.recommender_id) if r.recommender_work_type == 'J' @job = Job.find_by_id(r.recommender_work_id) @recos.push "#{@recommender.first_name} #{@recommender.last_name}" @bas.push "#{r.basis.gsub("You","#{@job.title} at #{@job.company.name}")}" else @qualification = Qualification.find_by_id(r.recommender_work_id) @recos.push "#{@recommender.first_name} #{@recommender.last_name} as " @bas.push "#{r.basis.gsub("You","Student at #{@qualification.school_name}")}" end end end

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  • drawRect context Error

    - by user361228
    Hi, I heard that a lot of people get a context error by not using drawRect Now I have this: - (void)drawRect:(CGRect)rect { NSLog(@"drawRect: Starts"); CGContextRef context = UIGraphicsGetCurrentContext(); CGContextSetRGBStrokeColor(context, 1.0, 1.0, 1.0, 1.0); CGContextSetLineWidth(context, 3.0); CGContextMoveToPoint(context, lineStart.x, lineStart.y); CGContextAddLineToPoint(context, lineEnd.x, lineEnd.y); CGContextStrokePath(context); } Error: <Error>: CGContextSetRGBStrokeColor: invalid context Which had work on previous programs, but not on this one. Whats different: I have a view controller, which calls this UIView: -(void)createLine:(CGPoint)start:(CGPoint)end { NSLog(@"createLine: Starts"); lineEnd = start; lineStart = end; self = [super initWithFrame:drawRect:CGRectMake(fmin(lineStart.x, lineEnd.x), fmin(lineStart.y, lineEnd.y), fabs(lineStart.x - lineEnd.x), fabs(lineStart.y - lineEnd.y))]; } This is my first question, and I am not sure how much info code I should be putting here so be easy on me :D Thanks

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  • I am getting an error when trying to use melt() on a dataframe containing Dates

    - by Dan
    I'd like to melt the dataframe so that in one column i have dates in a second i have username as the variable and finally the value. I'm getting this error: Error in as.Date.numeric(value) : 'origin' must be supplied and while I understand the error I'm not exactly sure how to get around it. A small sample of the data is: structure(list(created_at = structure(c(14007, 14008, 14009, 14010, 14011, 14012), class = "Date"), benjamin = c(16, 0, 0, 0, 0, 0), byron = c(0, 0, 0, 0, 0, 0), cameronc = c(0, 0, 0, 0, 0, 0), daniel = c(0, 0, 0, 0, 0, 0), djdiaz = c(0, 0, 0, 0, 0, 0), gene = c(16, 77, 64, 38, 72, 36), joel = c(0, 0, 0, 0, 0, 2), kerem = c(0, 0, 0, 0, 0, 0), sophia = c(0, 0, 0, 0, 0, 0), SuperMoonMan = c(0, 0, 0, 0, 0, 0)), .Names = c("created_at", "benjamin", "byron", "cameronc", "daniel", "djdiaz", "gene", "joel", "kerem", "sophia", "SuperMoonMan"), row.names = c(NA, 6L), class = c("cast_df", "data.frame")) Thanks for your help.

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  • Sql simple query

    - by Josemalive
    Hello, I have the following table Persons_Companies that shows a relation between persons and companies knowns by these persons: PersonID | CompanyID 1 1 2 1 2 2 3 2 4 2 Imaging that company 1="Google" and company 2 is ="Microsoft", i would like to know the query to have the following result: PersonID | Microsoft | Google 1 0 1 2 1 1 3 1 0 4 1 0 Until this moment i have something similar: select PersonID, case when CompanyID=1 then 1 else 0 end as Google, case when EmpresaID=2 then 1 else 0 end as Microsoft from Persons_Companies My problem is with the persons that knows both companies, i cant imagine how could this query be. Could you give me a hand? Thanks in advance. Best Regards. Josema.

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  • ext error with zend framework mvc

    - by terrani
    Hi, I am trying to setup ext javascript grid within zend framework mvc. I included ext css and js using the following code. $this->headScript() ->appendFile('/Resource/scripts/ext/jquery-1.4.2.js') ->appendFile('/Resource/scripts/ext/jquery/ext-jquery-adapter.js') ->appendFile('/Resource/scripts/ext/jquery/ext-all.js'); $this->headLink() ->appendStylesheet('/Layouts/admin/css/content.css') ->appendStylesheet('/Layouts/admin/css/ui.css') ->appendStylesheet('/Layouts/admin/css/button.css') ->appendStylesheet('/Layouts/admin/css/moon.css') ->appendStylesheet('/Resource/scripts/ext/css/ext-all.css'); when I run the code, I get the following error message from firefox. syntax error [Break on this error] \n What should I do to fix this?

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  • Pagination links broken - php/jquery

    - by ClarkSKent
    Hey, I'm still trying to get my pagination links to load properly dynamically. But I can't seem to find a solution to this one problem. vote down star Hi everyone, I am still trying to figure out how to fix my pagination script to work properly. the problem I am having is when I click any of the pagination number links to go the next page, the new content does not load. literally nothing happens and when looking at the console in Firebug, nothing is sent or loaded. I have on the main page 3 links to filter the content and display it. When any of these links are clicked the results are loaded and displayed along with the associated pagination numbers for that specific content. I believe the problem is coming from the sql query in generate_pagination.php (seen below). When I hard code the sql category part it works, but is not dynamic at all. This is why I'm calling $ids=$_GET['ids']; and trying to put that into the category section but then the numbers don't display at all. If I echo out the $ids variable and click on a filter it does display the correct name/id, so I don't know why this doesn't work Here is the main page so you can see how I am including and starting the function(I'm new to php): <?php include_once('generate_pagination.php'); ?> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.1/jquery.min.js"></script> <script type="text/javascript" src="jquery_pagination.js"></script> <div id="loading" ></div> <div id="content" data-page="1"></div> <ul id="pagination"> <?php //Pagination Numbers for($i=1; $i<=$pages; $i++) { echo '<li class="page_numbers" id="page'.$i.'">'.$i.'</li>'; } ?> </ul> <br /> <br /> <a href="#" class="category" id="marketing">Marketing</a> <a href="#" class="category" id="automotive">Automotive</a> <a href="#" class="category" id="sports">Sports</a> Here is the generate pagination where the problem seems to occur: <?php $ids=$_GET['ids']; include_once('config.php'); $per_page = 3; //Calculating no of pages $sql = "SELECT COUNT(*) FROM explore WHERE category='$ids'"; $result = mysql_query($sql); $count = mysql_fetch_row($result); $pages = ceil($count[0]/$per_page); ?> I thought I might as well post the jquery script if someone wants to see: $(document).ready(function(){ //Display Loading Image function Display_Load() { $("#loading").fadeIn(900,0); $("#loading").html("<img src='bigLoader.gif' />"); } //Hide Loading Image function Hide_Load() { $("#loading").fadeOut('slow'); }; //Default Starting Page Results $("#pagination li:first").css({'color' : '#FF0084'}).css({'border' : 'none'}); Display_Load(); $("#content").load("pagination_data.php?page=1", Hide_Load()); // Editing below. // Sort content Marketing $("a.category").click(function() { Display_Load(); var this_id = $(this).attr('id'); $.get("pagination.php", { category: this.id }, function(data){ //Load your results into the page var pageNum = $('#content').attr('data-page'); $("#pagination").load('generate_pagination.php?category=' + pageNum +'&ids='+ this_id ); $("#content").load("filter_marketing.php?page=" + pageNum +'&id='+ this_id, Hide_Load()); }); }); //Pagination Click $("#pagination li").click(function(){ Display_Load(); //CSS Styles $("#pagination li") .css({'border' : 'solid #dddddd 1px'}) .css({'color' : '#0063DC'}); $(this) .css({'color' : '#FF0084'}) .css({'border' : 'none'}); //Loading Data var pageNum = $(this).attr("id").replace("page",""); $("#content").load("pagination_data.php?page=" + pageNum, function(){ $(this).attr('data-page', pageNum); Hide_Load(); }); }); }); If any could assist me on solving this problem that would be great, thanks.

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  • relational database: how to design this table

    - by donpal
    I'm a database newbie designing a database. I'll use SO to ask my question because it's easier to ask it on something that you can see already, but it's not the same, it will just help me understand the right approach. As you can see, there are many questions here and each can have many answers. How should I store the answers in a table? Should I store all the answers in the SAME table with a unique id (make it the key) and just a new field for the question id? What if there are 100,000 answers like there is here? Do I still store them in 1 table? What keys should I use to minimize search time when I want to search for the answers of a specific question? The database is both read and write if that makes any difference in this case.

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