Search Results

Search found 9894 results on 396 pages for 'outer space'.

Page 10/396 | < Previous Page | 6 7 8 9 10 11 12 13 14 15 16 17  | Next Page >

  • SUM of column with Left Outer Join

    - by Matt
    I am trying to get the Count of all records that have at least on person who is authorized on the record. Basically, a Record can have more than one person associated with it. I want to return the count of Total Records, a count of total Authorized Records where at least 1 person is authorized, and a count of total NotAuthorized records where no person associated with record is authorized. It doesn't matter if one person is authorized per Record or if 3 people are authorized for that record, that should add 1 to the Authorized counter. The current query is incrementing Auth and Non auth for each person added per record rather, than one per record. If no people are assigned to the record that should also count towards Not Auth. SELECT Count(DISTINCT Record.RecordID) AS TotalRecords, SUM(CASE WHEN People.PersonLevel = 1 THEN 1 ELSE 0 END) AS Authorized, SUM(CASE WHEN People.PersonLevel <> 1 THEN 1 ELSE 0 END) AS NotAuthorized FROM Record LEFT OUTER JOIN RecordPeople ON Record.RecordID = RecordPeople.RecordID LEFT OUTER JOIN People ON RecordPeople.PersonID = People.PersonID

    Read the article

  • Left outer joins that don't return all the rows from T1

    - by Summer
    Left outer joins should return at least one row from the T1 table if it matches the conditions. But what if the left outer join performs a join successfully, then finds that another criterion is not satisfied? Is there a way to get the query to return a row with T1 values and T2 values set to NULL? Here's the specific query, in which I'm trying to return a list of candidates, and the user's support for those candidates IF such support exists. SELECT c.id, c.name, s.support FROM candidates c LEFT JOIN support s on s.candidate_id = c.id WHERE c.office_id = 5059 AND c.election_id = 92 AND (s.user_id = 2 OR s.user_id IS NULL) --This line seems like the problem ORDER BY c.last_name, c.name The query joins the candidates and support table, but finds that it's a different user who supported this candidate (user_id=3, say). Then the candidate disappears entirely from the result set.

    Read the article

  • Slow queries in Rails- not sure if my indexes are being used.

    - by Max Williams
    I'm doing a quite complicated find with lots of includes, which rails is splitting into a sequence of discrete queries rather than do a single big join. The queries are really slow - my dataset isn't massive, with none of the tables having more than a few thousand records. I have indexed all of the fields which are examined in the queries but i'm worried that the indexes aren't helping for some reason: i installed a plugin called "query_reviewer" which looks at the queries used to build a page, and lists problems with them. This states that indexes AREN'T being used, and it features the results of calling 'explain' on the query, which lists various problems. Here's an example find call: Question.paginate(:all, {:page=>1, :include=>[:answers, :quizzes, :subject, {:taggings=>:tag}, {:gradings=>[:age_group, :difficulty]}], :conditions=>["((questions.subject_id = ?) or (questions.subject_id = ? and tags.name = ?))", "1", 19, "English"], :order=>"subjects.name, (gradings.difficulty_id is null), gradings.age_group_id, gradings.difficulty_id", :per_page=>30}) And here are the generated sql queries: SELECT DISTINCT `questions`.id FROM `questions` LEFT OUTER JOIN `taggings` ON `taggings`.taggable_id = `questions`.id AND `taggings`.taggable_type = 'Question' LEFT OUTER JOIN `tags` ON `tags`.id = `taggings`.tag_id LEFT OUTER JOIN `subjects` ON `subjects`.id = `questions`.subject_id LEFT OUTER JOIN `gradings` ON gradings.question_id = questions.id WHERE (((questions.subject_id = '1') or (questions.subject_id = 19 and tags.name = 'English'))) ORDER BY subjects.name, (gradings.difficulty_id is null), gradings.age_group_id, gradings.difficulty_id LIMIT 0, 30 SELECT `questions`.`id` AS t0_r0 <..etc...> FROM `questions` LEFT OUTER JOIN `answers` ON answers.question_id = questions.id LEFT OUTER JOIN `quiz_questions` ON (`questions`.`id` = `quiz_questions`.`question_id`) LEFT OUTER JOIN `quizzes` ON (`quizzes`.`id` = `quiz_questions`.`quiz_id`) LEFT OUTER JOIN `subjects` ON `subjects`.id = `questions`.subject_id LEFT OUTER JOIN `taggings` ON `taggings`.taggable_id = `questions`.id AND `taggings`.taggable_type = 'Question' LEFT OUTER JOIN `tags` ON `tags`.id = `taggings`.tag_id LEFT OUTER JOIN `gradings` ON gradings.question_id = questions.id LEFT OUTER JOIN `age_groups` ON `age_groups`.id = `gradings`.age_group_id LEFT OUTER JOIN `difficulties` ON `difficulties`.id = `gradings`.difficulty_id WHERE (((questions.subject_id = '1') or (questions.subject_id = 19 and tags.name = 'English'))) AND `questions`.id IN (602, 634, 666, 698, 730, 762, 613, 645, 677, 709, 741, 592, 624, 656, 688, 720, 752, 603, 635, 667, 699, 731, 763, 614, 646, 678, 710, 742, 593, 625) ORDER BY subjects.name, (gradings.difficulty_id is null), gradings.age_group_id, gradings.difficulty_id SELECT count(DISTINCT `questions`.id) AS count_all FROM `questions` LEFT OUTER JOIN `answers` ON answers.question_id = questions.id LEFT OUTER JOIN `quiz_questions` ON (`questions`.`id` = `quiz_questions`.`question_id`) LEFT OUTER JOIN `quizzes` ON (`quizzes`.`id` = `quiz_questions`.`quiz_id`) LEFT OUTER JOIN `subjects` ON `subjects`.id = `questions`.subject_id LEFT OUTER JOIN `taggings` ON `taggings`.taggable_id = `questions`.id AND `taggings`.taggable_type = 'Question' LEFT OUTER JOIN `tags` ON `tags`.id = `taggings`.tag_id LEFT OUTER JOIN `gradings` ON gradings.question_id = questions.id LEFT OUTER JOIN `age_groups` ON `age_groups`.id = `gradings`.age_group_id LEFT OUTER JOIN `difficulties` ON `difficulties`.id = `gradings`.difficulty_id WHERE (((questions.subject_id = '1') or (questions.subject_id = 19 and tags.name = 'English'))) Actually, looking at these all nicely formatted here, there's a crazy amount of joining going on here. This can't be optimal surely. Anyway, it looks like i have two questions. 1) I have an index on each of the ids and foreign key fields referred to here. The second of the above queries is the slowest, and calling explain on it (doing it directly in mysql) gives me the following: +----+-------------+----------------+--------+---------------------------------------------------------------------------------+-------------------------------------------------+---------+------------------------------------------------+------+----------------------------------------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+----------------+--------+---------------------------------------------------------------------------------+-------------------------------------------------+---------+------------------------------------------------+------+----------------------------------------------+ | 1 | SIMPLE | questions | range | PRIMARY,index_questions_on_subject_id | PRIMARY | 4 | NULL | 30 | Using where; Using temporary; Using filesort | | 1 | SIMPLE | answers | ref | index_answers_on_question_id | index_answers_on_question_id | 5 | millionaire_development.questions.id | 2 | | | 1 | SIMPLE | quiz_questions | ref | index_quiz_questions_on_question_id | index_quiz_questions_on_question_id | 5 | millionaire_development.questions.id | 1 | | | 1 | SIMPLE | quizzes | eq_ref | PRIMARY | PRIMARY | 4 | millionaire_development.quiz_questions.quiz_id | 1 | | | 1 | SIMPLE | subjects | eq_ref | PRIMARY | PRIMARY | 4 | millionaire_development.questions.subject_id | 1 | | | 1 | SIMPLE | taggings | ref | index_taggings_on_taggable_id_and_taggable_type,index_taggings_on_taggable_type | index_taggings_on_taggable_id_and_taggable_type | 263 | millionaire_development.questions.id,const | 1 | | | 1 | SIMPLE | tags | eq_ref | PRIMARY | PRIMARY | 4 | millionaire_development.taggings.tag_id | 1 | Using where | | 1 | SIMPLE | gradings | ref | index_gradings_on_question_id | index_gradings_on_question_id | 5 | millionaire_development.questions.id | 2 | | | 1 | SIMPLE | age_groups | eq_ref | PRIMARY | PRIMARY | 4 | millionaire_development.gradings.age_group_id | 1 | | | 1 | SIMPLE | difficulties | eq_ref | PRIMARY | PRIMARY | 4 | millionaire_development.gradings.difficulty_id | 1 | | +----+-------------+----------------+--------+---------------------------------------------------------------------------------+-------------------------------------------------+---------+------------------------------------------------+------+----------------------------------------------+ The query_reviewer plugin has this to say about it - it lists several problems: Table questions: Using temporary table, Long key length (263), Using filesort MySQL must do an extra pass to find out how to retrieve the rows in sorted order. To resolve the query, MySQL needs to create a temporary table to hold the result. The key used for the index was rather long, potentially affecting indices in memory 2) It looks like rails isn't splitting this find up in a very optimal way. Is it, do you think? Am i better off doing several find queries manually rather than one big combined one? Grateful for any advice, max

    Read the article

  • Problems with texture orientation in space

    - by frankie
    I am currently drawing texture in 3D space and have some problems with it's orientation. I'd like me textures always to be oriented with front face to user. My desirable result looks like Note, that text size stay without changes when we rotating world and stay oriented with front face to user. Now I can draw text in 3D space, but it is not oriented with front but rotating with world. Such results I got with following shaders: Vertex Shader uniform vec3 Position; void main() { gl_Position = vec4(Position, 1.0); } Geometry Shader layout(points) in; layout(triangle_strip, max_vertices = 4) out; out vec2 fsTextureCoordinates; uniform mat4 projectionMatrix; uniform mat4 modelViewMatrix; uniform sampler2D og_texture0; uniform float og_highResolutionSnapScale; uniform vec2 u_originScale; void main() { vec2 halfSize = vec2(textureSize(og_texture0, 0)) * 0.5 * og_highResolutionSnapScale; vec4 center = gl_in[0].gl_Position; center.xy += (u_originScale * halfSize); vec4 v0 = vec4(center.xy - halfSize, center.z, 1.0); vec4 v1 = vec4(center.xy + vec2(halfSize.x, -halfSize.y), center.z, 1.0); vec4 v2 = vec4(center.xy + vec2(-halfSize.x, halfSize.y), center.z, 1.0); vec4 v3 = vec4(center.xy + halfSize, center.z, 1.0); gl_Position = projectionMatrix * modelViewMatrix * v0; fsTextureCoordinates = vec2(0.0, 0.0); EmitVertex(); gl_Position = projectionMatrix * modelViewMatrix * v1; fsTextureCoordinates = vec2(1.0, 0.0); EmitVertex(); gl_Position = projectionMatrix * modelViewMatrix * v2; fsTextureCoordinates = vec2(0.0, 1.0); EmitVertex(); gl_Position = projectionMatrix * modelViewMatrix * v3; fsTextureCoordinates = vec2(1.0, 1.0); EmitVertex(); } Fragment Shader in vec2 fsTextureCoordinates; out vec4 fragmentColor; uniform sampler2D og_texture0; uniform vec3 u_color; void main() { vec4 color = texture(og_texture0, fsTextureCoordinates); if (color.a == 0.0) { discard; } fragmentColor = vec4(color.rgb * u_color.rgb, color.a); } Any ideas how to get my desirable result? EDIT 1: I make edit in my geometry shader and got part of lable drawn on screen at corner. But it is not rotating. .......... vec4 centerProjected = projectionMatrix * modelViewMatrix * center; centerProjected /= centerProjected.w; vec4 v0 = vec4(centerProjected.xy - halfSize, 0.0, 1.0); vec4 v1 = vec4(centerProjected.xy + vec2(halfSize.x, -halfSize.y), 0.0, 1.0); vec4 v2 = vec4(centerProjected.xy + vec2(-halfSize.x, halfSize.y), 0.0, 1.0); vec4 v3 = vec4(centerProjected.xy + halfSize, 0.0, 1.0); gl_Position = og_viewportOrthographicMatrix * v0; ..........

    Read the article

  • Not enough free disk space

    - by carmatt95
    I'm new to Ubuntu and I'm getting an error in software updater. When I try and do my daily updates, it says: The upgrade needs a total of 25.3 M free space on disk /boot. Please free at least an additional 25.3 M of disk space on /boot. Empty your trash and remove temporary packages of former installations using sudo apt-get clean. I tried typing in sudo apt-get clean into the terminal but I still get the message. All of the pages I read seem to be for experianced Ubuntuers. Any help would be appreciated. I'm running Ubuntu 12.10. I want to upgrade to 13.04 but understand I have to finish these first. EDIT: @Alaa, This is the output from typing in cat /etc/fstab into the terminal: # /etc/fstab: static file system information. # # Use 'blkid' to print the universally unique identifier for a # device; this may be used with UUID= as a more robust way to name devices # that works even if disks are added and removed. See fstab(5). # # <file system> <mount point> <type> <options> <dump> <pass> /dev/mapper/ubuntu-root / ext4 errors=remount-ro 0 1 # /boot was on /dev/sda1 during installation UUID=fa55c082-112d-4b10-bcf3-e7ffec6cebbc /boot ext2 defaults 0 2 /dev/mapper/ubuntu-swap_1 none swap sw 0 0 /dev/fd0 /media/floppy0 auto rw,user,noauto,exec,utf8 0 0 matty@matty-G41M-ES2L:~$ df -h: Filesystem Size Used Avail Use% Mounted on /dev/mapper/ubuntu-root 915G 27G 842G 4% / udev 984M 4.0K 984M 1% /dev tmpfs 397M 1.1M 396M 1% /run none 5.0M 0 5.0M 0% /run/lock none 992M 1.8M 990M 1% /run/shm none 100M 52K 100M 1% /run/user /dev/sda1 228M 222M 0 100% /boot matty@matty-G41M-ES2L:~$ dpkg -l | grep linux-image: ii linux-image-3.5.0-17-generic 3.5.0-17.28 i386 Linux kernel image for version 3.5.0 on 32 bit x86 SMP ii linux-image-3.5.0-18-generic 3.5.0-18.29 i386 Linux kernel image for version 3.5.0 on 32 bit x86 SMP ii linux-image-3.5.0-19-generic 3.5.0-19.30 i386 Linux kernel image for version 3.5.0 on 32 bit x86 SMP ii linux-image-3.5.0-21-generic 3.5.0-21.32 i386 Linux kernel image for version 3.5.0 on 32 bit x86 SMP ii linux-image-3.5.0-22-generic 3.5.0-22.34 i386 Linux kernel image for version 3.5.0 on 32 bit x86 SMP ii linux-image-3.5.0-23-generic 3.5.0-23.35 i386 Linux kernel image for version 3.5.0 on 32 bit x86 SMP ii linux-image-3.5.0-24-generic 3.5.0-24.37 i386 Linux kernel image for version 3.5.0 on 32 bit x86 SMP ii linux-image-3.5.0-25-generic 3.5.0-25.39 i386 Linux kernel image for version 3.5.0 on 32 bit x86 SMP ii linux-image-3.5.0-26-generic 3.5.0-26.42 i386 Linux kernel image for version 3.5.0 on 32 bit x86 SMP iF linux-image-3.5.0-28-generic 3.5.0-28.48 i386 Linux kernel image for version 3.5.0 on 32 bit x86 SMP

    Read the article

  • TSQL Conditionally Select Specific Value

    - by Dzejms
    This is a follow-up to #1644748 where I successfully answered my own question, but Quassnoi helped me to realize that it was the wrong question. He gave me a solution that worked for my sample data, but I couldn't plug it back into the parent stored procedure because I fail at SQL 2005 syntax. So here is an attempt to paint the broader picture and ask what I actually need. This is part of a stored procedure that returns a list of items in a bug tracking application I've inherited. There are are over 100 fields and 26 joins so I'm pulling out only the mostly relevant bits. SELECT tickets.ticketid, tickets.tickettype, tickets_tickettype_lu.tickettypedesc, tickets.stage, tickets.position, tickets.sponsor, tickets.dev, tickets.qa, DATEDIFF(DAY, ticket_history_assignment.savedate, GETDATE()) as 'daysinqueue' FROM dbo.tickets WITH (NOLOCK) LEFT OUTER JOIN dbo.tickets_tickettype_lu WITH (NOLOCK) ON tickets.tickettype = tickets_tickettype_lu.tickettypeid LEFT OUTER JOIN dbo.tickets_history_assignment WITH (NOLOCK) ON tickets_history_assignment.ticketid = tickets.ticketid AND tickets_history_assignment.historyid = ( SELECT MAX(historyid) FROM dbo.tickets_history_assignment WITH (NOLOCK) WHERE tickets_history_assignment.ticketid = tickets.ticketid GROUP BY tickets_history_assignment.ticketid ) WHERE tickets.sponsor = @sponsor The area of interest is the daysinqueue subquery mess. The tickets_history_assignment table looks roughly as follows declare @tickets_history_assignment table ( historyid int, ticketid int, sponsor int, dev int, qa int, savedate datetime ) insert into @tickets_history_assignment values (1521402, 92774,20,14, 20, '2009-10-27 09:17:59.527') insert into @tickets_history_assignment values (1521399, 92774,20,14, 42, '2009-08-31 12:07:52.917') insert into @tickets_history_assignment values (1521311, 92774,100,14, 42, '2008-12-08 16:15:49.887') insert into @tickets_history_assignment values (1521336, 92774,100,14, 42, '2009-01-16 14:27:43.577') Whenever a ticket is saved, the current values for sponsor, dev and qa are stored in the tickets_history_assignment table with the ticketid and a timestamp. So it is possible for someone to change the value for qa, but leave sponsor alone. What I want to know, based on all of these conditions, is the historyid of the record in the tickets_history_assignment table where the sponsor value was last changed so that I can calculate the value for daysinqueue. If a record is inserted into the history table, and only the qa value has changed, I don't want that record. So simply relying on MAX(historyid) won't work for me. Quassnoi came up with the following which seemed to work with my sample data, but I can't plug it into the larger query, SQL Manager bitches about the WITH statement. ;WITH rows AS ( SELECT *, ROW_NUMBER() OVER (PARTITION BY ticketid ORDER BY savedate DESC) AS rn FROM @Table ) SELECT rl.sponsor, ro.savedate FROM rows rl CROSS APPLY ( SELECT TOP 1 rc.savedate FROM rows rc JOIN rows rn ON rn.ticketid = rc.ticketid AND rn.rn = rc.rn + 1 AND rn.sponsor <> rc.sponsor WHERE rc.ticketid = rl.ticketid ORDER BY rc.rn ) ro WHERE rl.rn = 1 I played with it yesterday afternoon and got nowhere because I don't fundamentally understand what is going on here and how it should fit into the larger context. So, any takers? UPDATE Ok, here's the whole thing. I've been switching some of the table and column names in an attempt to simplify things so here's the full unedited mess. snip - old bad code Here are the errors: Msg 102, Level 15, State 1, Procedure usp_GetProjectRecordsByAssignment, Line 159 Incorrect syntax near ';'. Msg 102, Level 15, State 1, Procedure usp_GetProjectRecordsByAssignment, Line 179 Incorrect syntax near ')'. Line numbers are of course not correct but refer to ;WITH rows AS And the ')' char after the WHERE rl.rn = 1 ) Respectively Is there a tag for extra super long question? UPDATE #2 Here is the finished query for anyone who may need this: CREATE PROCEDURE [dbo].[usp_GetProjectRecordsByAssignment] ( @assigned numeric(18,0), @assignedtype numeric(18,0) ) AS SET NOCOUNT ON WITH rows AS ( SELECT *, ROW_NUMBER() OVER (PARTITION BY recordid ORDER BY savedate DESC) AS rn FROM projects_history_assignment ) SELECT projects_records.recordid, projects_records.recordtype, projects_recordtype_lu.recordtypedesc, projects_records.stage, projects_stage_lu.stagedesc, projects_records.position, projects_position_lu.positiondesc, CASE projects_records.clientrequested WHEN '1' THEN 'Yes' WHEN '0' THEN 'No' END AS clientrequested, projects_records.reportingmethod, projects_reportingmethod_lu.reportingmethoddesc, projects_records.clientaccess, projects_clientaccess_lu.clientaccessdesc, projects_records.clientnumber, projects_records.project, projects_lu.projectdesc, projects_records.version, projects_version_lu.versiondesc, projects_records.projectedversion, projects_version_lu_projected.versiondesc AS projectedversiondesc, projects_records.sitetype, projects_sitetype_lu.sitetypedesc, projects_records.title, projects_records.module, projects_module_lu.moduledesc, projects_records.component, projects_component_lu.componentdesc, projects_records.loginusername, projects_records.loginpassword, projects_records.assistedusername, projects_records.browsername, projects_browsername_lu.browsernamedesc, projects_records.browserversion, projects_records.osname, projects_osname_lu.osnamedesc, projects_records.osversion, projects_records.errortype, projects_errortype_lu.errortypedesc, projects_records.gsipriority, projects_gsipriority_lu.gsiprioritydesc, projects_records.clientpriority, projects_clientpriority_lu.clientprioritydesc, projects_records.scheduledstartdate, projects_records.scheduledcompletiondate, projects_records.projectedhours, projects_records.actualstartdate, projects_records.actualcompletiondate, projects_records.actualhours, CASE projects_records.billclient WHEN '1' THEN 'Yes' WHEN '0' THEN 'No' END AS billclient, projects_records.billamount, projects_records.status, projects_status_lu.statusdesc, CASE CAST(projects_records.assigned AS VARCHAR(5)) WHEN '0' THEN 'N/A' WHEN '10000' THEN 'Unassigned' WHEN '20000' THEN 'Client' WHEN '30000' THEN 'Tech Support' WHEN '40000' THEN 'LMI Tech Support' WHEN '50000' THEN 'Upload' WHEN '60000' THEN 'Spider' WHEN '70000' THEN 'DB Admin' ELSE rtrim(users_assigned.nickname) + ' ' + rtrim(users_assigned.lastname) END AS assigned, CASE CAST(projects_records.assigneddev AS VARCHAR(5)) WHEN '0' THEN 'N/A' WHEN '10000' THEN 'Unassigned' ELSE rtrim(users_assigneddev.nickname) + ' ' + rtrim(users_assigneddev.lastname) END AS assigneddev, CASE CAST(projects_records.assignedqa AS VARCHAR(5)) WHEN '0' THEN 'N/A' WHEN '10000' THEN 'Unassigned' ELSE rtrim(users_assignedqa.nickname) + ' ' + rtrim(users_assignedqa.lastname) END AS assignedqa, CASE CAST(projects_records.assignedsponsor AS VARCHAR(5)) WHEN '0' THEN 'N/A' WHEN '10000' THEN 'Unassigned' ELSE rtrim(users_assignedsponsor.nickname) + ' ' + rtrim(users_assignedsponsor.lastname) END AS assignedsponsor, projects_records.clientcreated, CASE projects_records.clientcreated WHEN '1' THEN 'Yes' WHEN '0' THEN 'No' END AS clientcreateddesc, CASE projects_records.clientcreated WHEN '1' THEN rtrim(clientusers_createuser.firstname) + ' ' + rtrim(clientusers_createuser.lastname) + ' (Client)' ELSE rtrim(users_createuser.nickname) + ' ' + rtrim(users_createuser.lastname) END AS createuser, projects_records.createdate, projects_records.savedate, projects_resolution.sitesaffected, projects_sitesaffected_lu.sitesaffecteddesc, DATEDIFF(DAY, projects_history_assignment.savedate, GETDATE()) as 'daysinqueue', projects_records.iOnHitList, projects_records.changetype FROM dbo.projects_records WITH (NOLOCK) LEFT OUTER JOIN dbo.projects_recordtype_lu WITH (NOLOCK) ON projects_records.recordtype = projects_recordtype_lu.recordtypeid LEFT OUTER JOIN dbo.projects_stage_lu WITH (NOLOCK) ON projects_records.stage = projects_stage_lu.stageid LEFT OUTER JOIN dbo.projects_position_lu WITH (NOLOCK) ON projects_records.position = projects_position_lu.positionid LEFT OUTER JOIN dbo.projects_reportingmethod_lu WITH (NOLOCK) ON projects_records.reportingmethod = projects_reportingmethod_lu.reportingmethodid LEFT OUTER JOIN dbo.projects_lu WITH (NOLOCK) ON projects_records.project = projects_lu.projectid LEFT OUTER JOIN dbo.projects_version_lu WITH (NOLOCK) ON projects_records.version = projects_version_lu.versionid LEFT OUTER JOIN dbo.projects_version_lu projects_version_lu_projected WITH (NOLOCK) ON projects_records.projectedversion = projects_version_lu_projected.versionid LEFT OUTER JOIN dbo.projects_sitetype_lu WITH (NOLOCK) ON projects_records.sitetype = projects_sitetype_lu.sitetypeid LEFT OUTER JOIN dbo.projects_module_lu WITH (NOLOCK) ON projects_records.module = projects_module_lu.moduleid LEFT OUTER JOIN dbo.projects_component_lu WITH (NOLOCK) ON projects_records.component = projects_component_lu.componentid LEFT OUTER JOIN dbo.projects_browsername_lu WITH (NOLOCK) ON projects_records.browsername = projects_browsername_lu.browsernameid LEFT OUTER JOIN dbo.projects_osname_lu WITH (NOLOCK) ON projects_records.osname = projects_osname_lu.osnameid LEFT OUTER JOIN dbo.projects_errortype_lu WITH (NOLOCK) ON projects_records.errortype = projects_errortype_lu.errortypeid LEFT OUTER JOIN dbo.projects_resolution WITH (NOLOCK) ON projects_records.recordid = projects_resolution.recordid LEFT OUTER JOIN dbo.projects_sitesaffected_lu WITH (NOLOCK) ON projects_resolution.sitesaffected = projects_sitesaffected_lu.sitesaffectedid LEFT OUTER JOIN dbo.projects_gsipriority_lu WITH (NOLOCK) ON projects_records.gsipriority = projects_gsipriority_lu.gsipriorityid LEFT OUTER JOIN dbo.projects_clientpriority_lu WITH (NOLOCK) ON projects_records.clientpriority = projects_clientpriority_lu.clientpriorityid LEFT OUTER JOIN dbo.projects_status_lu WITH (NOLOCK) ON projects_records.status = projects_status_lu.statusid LEFT OUTER JOIN dbo.projects_clientaccess_lu WITH (NOLOCK) ON projects_records.clientaccess = projects_clientaccess_lu.clientaccessid LEFT OUTER JOIN dbo.users users_assigned WITH (NOLOCK) ON projects_records.assigned = users_assigned.userid LEFT OUTER JOIN dbo.users users_assigneddev WITH (NOLOCK) ON projects_records.assigneddev = users_assigneddev.userid LEFT OUTER JOIN dbo.users users_assignedqa WITH (NOLOCK) ON projects_records.assignedqa = users_assignedqa.userid LEFT OUTER JOIN dbo.users users_assignedsponsor WITH (NOLOCK) ON projects_records.assignedsponsor = users_assignedsponsor.userid LEFT OUTER JOIN dbo.users users_createuser WITH (NOLOCK) ON projects_records.createuser = users_createuser.userid LEFT OUTER JOIN dbo.clientusers clientusers_createuser WITH (NOLOCK) ON projects_records.createuser = clientusers_createuser.userid LEFT OUTER JOIN dbo.projects_history_assignment WITH (NOLOCK) ON projects_history_assignment.recordid = projects_records.recordid AND projects_history_assignment.historyid = ( SELECT ro.historyid FROM rows rl CROSS APPLY ( SELECT TOP 1 rc.historyid FROM rows rc JOIN rows rn ON rn.recordid = rc.recordid AND rn.rn = rc.rn + 1 AND rn.assigned <> rc.assigned WHERE rc.recordid = rl.recordid ORDER BY rc.rn ) ro WHERE rl.rn = 1 AND rl.recordid = projects_records.recordid ) WHERE (@assignedtype='0' and projects_records.assigned = @assigned) OR (@assignedtype='1' and projects_records.assigneddev = @assigned) OR (@assignedtype='2' and projects_records.assignedqa = @assigned) OR (@assignedtype='3' and projects_records.assignedsponsor = @assigned) OR (@assignedtype='4' and projects_records.createuser = @assigned)

    Read the article

  • SQL SERVER – Powershell – Get a List of Fixed Hard Drive and Free Space on Server

    - by pinaldave
    Earlier I have written this article SQL SERVER – Get a List of Fixed Hard Drive and Free Space on Server. I recently received excellent comment by MVP Ravikanth. He demonstrated that how the same can be done using Powershell. It is very sweet and quick solution. Here is the powershell script. Run the same in your powershell windows. Get-WmiObject -Class Win32_LogicalDisk | Select -Property DeviceID, @{Name=’FreeSpaceMB’;Expression={$_.FreeSpace/1MB} } | Format-Table -AutoSize Well, I ran this script in my powershell window, it gave me following result – very accurately and easily. Thanks Ravikanth one more time for excellent tip. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, PostADay, SQL, SQL Authority, SQL Query, SQL Scripts, SQL Server, SQL Stored Procedure, SQL Tips and Tricks, T SQL, Technology Tagged: Powershell

    Read the article

  • Error in Using Gparted to add Unallocated space

    - by Ba7a7chy
    I'm trying to Add Unallocated space to my Ubuntu partition which is inside an extended partition. i get this error: GParted 0.12.1 --enable-libparted-dmraid Libparted 2.3 Move /dev/sda3 to the left and grow it from 93.31 GiB to 219.75 GiB 00:00:00 ( ERROR ) calibrate /dev/sda3 00:00:00 ( SUCCESS ) path: /dev/sda3 start: 292,704,254 end: 488,394,751 size: 195,690,498 (93.31 GiB) move partition to the left and grow it from 93.31 GiB to 219.75 GiB 00:00:00 ( ERROR ) old start: 292,704,254 old end: 488,394,751 old size: 195,690,498 (93.31 GiB) requested start: 27,545,600 requested end: 488,392,703 requested size: 460,847,104 (219.75 GiB) libparted messages ( INFO ) Unable to satisfy all constraints on the partition. Can't have overlapping partitions. ========================================

    Read the article

  • expand window to free space on screen in kde

    - by Pascal Rosin
    I am using Kubuntu and I want to expand the current window to the free space on the screen or to say it more precisely: I want to make the current window as big as possible without overlapping new windows (windows already overlapped should be ignored). Is there a keyboard shortcut or an extension to the KDE Window management, that realizes such a shortcut or a window button? I would also appreciate a hint, how to write a script that could do this window thing on keyboard shortcut invocation. I am a programmer but don't know what the best way is to control KDE Windows via script.

    Read the article

  • Star Trail Photos Taken from the International Space Station

    - by Jason Fitzpatrick
    While most people have seen a star trail photo or two, seeing a set of star trail photos taken from over 300 miles above the Earth’s surface is a treat. Courtesy of Astronaut and Expedition 31 Flight Engineer Don Pettit, the photos capture star trails from the vantage point of the International Space Station. He explains his technique: My star trail images are made by taking a time exposure of about 10 to 15 minutes. However, with modern digital cameras, 30 seconds is about the longest exposure possible, due to electronic detector noise effectively snowing out the image. To achieve the longer exposures I do what many amateur astronomers do. I take multiple 30-second exposures, then ‘stack’ them using imaging software, thus producing the longer exposure. Hit up the link below for the full Flickr set of the star trails. ISS Star Trails [via Smithsonian Magazine] HTG Explains: What Is RSS and How Can I Benefit From Using It? HTG Explains: Why You Only Have to Wipe a Disk Once to Erase It HTG Explains: Learn How Websites Are Tracking You Online

    Read the article

  • Why aren't tangent space normal maps completely blue?

    - by seahorse
    Why aren't normal maps just blue? I would think that normal maps should be predominantly blue in color because the Z component of the normal is represented by blue. Normals point out of the surface in the Z direction so we should see blue as the predominant colour since the Z component is dominant. By definition tangent space is perpendicular to the surface. At any point we should have the normal always pointing in the Z (blue direction) with no X (red direction) or Y (green direction). Thus the normal map (since it is a "normal map") should have the colour of the normals which is just blue (R = x = 0, G = y = 0, B = z = 1) with no shades in between. But normal maps are not so, and they have gradients of shades in them. Why is this so?

    Read the article

  • My Ubuntu drive is running out of space, how to fix, something is wrong

    - by Jamie Flores
    I'm moving from windows and am having trouble figuring this out: I'm getting a message that pops up saying disk space is low. It says I have 800MB free. I click on the disk usage analyzer and it shows 24.6 total capacity and 22.5 used. When I look in GParted it shows a partition at 72.6GB where I have Ubuntu installed. It also shows that 70.65GB used and 1.94 free in that partition. How do I figure out what else is in that partition? It's the only ext4 format. What am I missing?

    Read the article

  • Unallocated space with important data

    - by Chethan S.
    I used GParted to convert a primary partition to extended one after copying the data to another partition. After having the extended partition I moved the data back. To my utter shock after a restart I found out that the new extended partition did convert into "unallocated space". I tried installing testdisk. Testdisk could identify the partition as a primary partition and not the newly created extended partition. So what should I do now? I badly want the data back.

    Read the article

  • Detect Open Space in Farseer

    - by Tom G
    I'm working on a 2D platformer using XNA and Farseer. I would like the player's character to be able to grab and climb up ledges. Detecting a collision between the player and the side of a wall is simple enough with the OnCollision delegate, but I have to admit I'm a bit stumped on how to detect that there's enough clearance for the PC to mount the ledge. Essentially, I want to ensure there's an appropriately sized rectangle above and to the left or right of the PC (depending on their direction) and I'm not sure how I would check for such a space. Any suggestions on how to determine there is nothing in the simulated world within some bounding rectangle?

    Read the article

  • Meaning of the free space indication in Deluge

    - by Tjae Beamon
    Recently I installed Ubuntu 12.0.4 using Wubi with my current Windows Vista. I have already installed all the 265 updates from the Ubuntu software center and downloaded Deluge from there. My hardrive is 80GB according to the disc usage analyzer. It also says 31.2 GB used and 47.8GB free. The confusion comes when I run Deluge. At the bottom it says 2.0GB free space. Is that 2.0GB just a size set from the torrent client and can be changed or am I limited to just that 2.0GB?

    Read the article

  • matrix to transform unit cube to space defined by 8 arbitrary points

    - by aadster
    I asked a question relating to similar to this already, but I think this is a clearer objective of what Im trying to achieve.. or whether its possible at all! Im trying to find a transformation (matrix ideally) which would transform the 8 points of a 3d unit cube to 8 arbitrary points in space. The 8 target points have no known structure. e.g: My gut feeling is that a matrix is unable to provide this xform since the cube faces vertices can be concave.. but are there any other methods of transformation? Thanks!

    Read the article

  • Why does Bing not show my adcenter ads though there is enough space

    - by gamma
    I created several campaigns using the MS adcenter. I'm targeting the whole world at any time with 2-3 placement texts per keyword group. The bids I placed are sometimes quite high, so they should get displayed. When I try to search for my keywords in Bing nothing gets displayed, though there is plenty of space for it. Bing mostly displays 2-3 ads, but the ones at the right side rather seldom. I'd like to know, how I can improve the fact that my ads are not being displayed - without increasing the bids any further.

    Read the article

  • Update failing. Not enough space on /tmp

    - by KodeSeeker
    Im not being able to run update manager as I get the error saying that there is not enough free space in the /tmp directory. I've practically cleaned out the tmp directory but the error persists. Any help would be appreciated. here's df-h /dev/loop0 13G 11G 952M 92% / udev 2.0G 4.0K 2.0G 1% /dev tmpfs 785M 920K 784M 1% /run none 5.0M 0 5.0M 0% /run/lock none 2.0G 584K 2.0G 1% /run/shm /dev/sda6 20G 14G 6.4G 68% /host overflow 1.0M 16K 1008K 2% /tmp Thanks.

    Read the article

  • SQL Outer Join on a bunch of Inner Joined results

    - by Matthew Frederick
    I received some great help on joining a table to itself and am trying to take it to the next level. The SQL below is from the help but with my addition of the select line beginning with COUNT, the inner join to the Recipient table, and the Group By. SELECT Event.EventID AS EventID, Event.EventDate AS EventDateUTC, Participant2.ParticipantID AS AwayID, Participant1.ParticipantID AS HostID, COUNT(Recipient.ChallengeID) AS AllChallenges FROM Event INNER JOIN Matchup Matchup1 ON (Event.EventID = Matchup1.EventID) INNER JOIN Matchup Matchup2 ON (Event.EventID = Matchup2.EventID) INNER JOIN Participant Participant1 ON (Matchup1.Host = 1 AND Matchup1.ParticipantID = Participant1.ParticipantID) INNER JOIN Participant Participant2 ON (Matchup2.Host != 1 AND Matchup2.ParticipantID = Participant2.ParticipantID) INNER JOIN Recipient ON (Event.EventID = Recipient.EventID) WHERE Event.CategoryID = 1 AND Event.Resolved = 0 AND Event.Type = 1 GROUP BY Recipient.ChallengeID ORDER BY EventDateUTC ASC My goal is to get a count of how many rows in the Recipient table match the EventID in Event. This code works fine except that I also want to get results where there are 0 matching rows in Recipient. I want 15 rows (= the number of events) but I get 2 rows, one with a count of 1 and one with a count of 2 (which is appropriate for an inner join as there are 3 rows in the sample Recipient table, one for one EventID and two for another EventID). I thought that either a LEFT join or an OUTER join was what I was looking for, but I know that I'm not quite getting how the tables are actually joined. A LEFT join there gives me one more row with 0, which happens to be EventID 1 (first thing in the table), but that's all. Errors advise me that I can't just change that INNER join to an OUTER. I tried some parenthesizing and some subselects and such but can't seem to make it work.

    Read the article

  • Using outer query result in a subquery in postgresql

    - by brad
    I have two tables points and contacts and I'm trying to get the average points.score per contact grouped on a monthly basis. Note that points and contacts aren't related, I just want the sum of points created in a month divided by the number of contacts that existed in that month. So, I need to sum points grouped by the created_at month, and I need to take the count of contacts FOR THAT MONTH ONLY. It's that last part that's tricking me up. I'm not sure how I can use a column from an outer query in the subquery. I tried something like this: SELECT SUM(score) AS points_sum, EXTRACT(month FROM created_at) AS month, date_trunc('MONTH', created_at) + INTERVAL '1 month' AS next_month, (SELECT COUNT(id) FROM contacts WHERE contacts.created_at <= next_month) as contact_count FROM points GROUP BY month, next_month ORDER BY month So, I'm extracting the actual month that my points are being summed, and at the same time, getting the beginning of the next_month so that I can say "Get me the count of contacts where their created at is < next_month" But it complains that column next_month doesn't exist This is understandable as the subquery knows nothing about the outer query. Qualifying with points.next_month doesn't work either. So can someone point me in the right direction of how to achieve this? Tables: Points score | created_at 10 | "2011-11-15 21:44:00.363423" 11 | "2011-10-15 21:44:00.69667" 12 | "2011-09-15 21:44:00.773289" 13 | "2011-08-15 21:44:00.848838" 14 | "2011-07-15 21:44:00.924152" Contacts id | created_at 6 | "2011-07-15 21:43:17.534777" 5 | "2011-08-15 21:43:17.520828" 4 | "2011-09-15 21:43:17.506452" 3 | "2011-10-15 21:43:17.491848" 1 | "2011-11-15 21:42:54.759225" sum, month and next_month (without the subselect) sum | month | next_month 14 | 7 | "2011-08-01 00:00:00" 13 | 8 | "2011-09-01 00:00:00" 12 | 9 | "2011-10-01 00:00:00" 11 | 10 | "2011-11-01 00:00:00" 10 | 11 | "2011-12-01 00:00:00"

    Read the article

  • Convert SQL with Inner AND Outer Join to L2S

    - by Refracted Paladin
    I need to convert the below Sproc to a Linq query. At the very bottom is what I have so far. For reference the fields behind the "splat"(not my sproc) are ImmunizationID int, HAReviewID int, ImmunizationMaintID int, ImmunizationOther varchar(50), ImmunizationDate smalldatetime, ImmunizationReasonID int The first two are PK and FK, respectively. The other two ints are linke to the Maint Table where there description is stored. That is what I am stuck on, the INNER JOIN AND the LEFT OUTER JOIN Thanks, SELECT tblHAReviewImmunizations.*, tblMaintItem.ItemDescription, tblMaintItem2.ItemDescription as Reason FROM dbo.tblHAReviewImmunizations INNER JOIN dbo.tblMaintItem ON dbo.tblHAReviewImmunizations.ImmunizationMaintID = dbo.tblMaintItem.ItemID LEFT OUTER JOIN dbo.tblMaintItem as tblMaintItem2 ON dbo.tblHAReviewImmunizations.ImmunizationReasonID = tblMaintItem2.ItemID WHERE HAReviewID = @haReviewID My attempt so far -- public static DataTable GetImmunizations(int haReviewID) { using (var context = McpDataContext.Create()) { var currentImmunizations = from haReviewImmunization in context.tblHAReviewImmunizations where haReviewImmunization.HAReviewID == haReviewID join maintItem in context.tblMaintItems on haReviewImmunization.ImmunizationReasonID equals maintItem.ItemID into g from maintItem in g.DefaultIfEmpty() let Immunization = GetImmunizationNameByID( haReviewImmunization.ImmunizationMaintID) select new { haReviewImmunization.ImmunizationDate, haReviewImmunization.ImmunizationOther, Immunization, Reason = maintItem == null ? " " : maintItem.ItemDescription }; return currentImmunizations.CopyLinqToDataTable(); } } private static string GetImmunizationNameByID(int? immunizationID) { using (var context = McpDataContext.Create()) { var domainName = from maintItem in context.tblMaintItems where maintItem.ItemID == immunizationID select maintItem.ItemDescription; return domainName.SingleOrDefault(); } }

    Read the article

  • SQL Standard Regarding Left Outer Join and Where Conditions

    - by Ryan
    I am getting different results based on a filter condition in a query based on where I place the filter condition. My questions are: Is there a technical difference between these queries? Is there anything in the SQL standard that explains the different resultsets in the queries? Given the simplified scenario: --Table: Parent Columns: ID, Name, Description --Table: Child Columns: ID, ParentID, Name, Description --Query 1 SELECT p.ID, p.Name, p.Description, c.ID, c.Name, c.Description FROM Parent p LEFT OUTER JOIN Child c ON (p.ID = c.ParentID) WHERE c.ID IS NULL OR c.Description = 'FilterCondition' --Query 2 SELECT p.ID, p.Name, p.Description, c.ID, c.Name, c.Description FROM Parent p LEFT OUTER JOIN Child c ON (p.ID = c.ParentID AND c.Description = 'FilterCondition') I assumed the queries would return the same resultsets and I was surprised when they didn't. I am using MS SQL2005 and in the actual queries, query 1 returned ~700 rows and query 2 returned ~1100 rows and I couldn't detect a pattern on which rows were returned and which rows were excluded. There were still many rows in query 1 with child rows with data and NULL data. I prefer the style of query 2 (and I think it is more optimal), but I thought the queries would return the same results.

    Read the article

  • Inner or Outer left Join

    - by user1557856
    I'm having difficulty modifying a script for this situation and wondering if someone maybe able to help: I have an address table and a phone table both sharing the same column called id_number. So id_number = 2 on both tables refers to the same entity. Address and phone information used to be stored in one table (the address table) but it is now split into address and phone tables since we moved to Oracle 11g. There is a 3rd table called both_ids. This table also has an id_number column in addition to an other_ids column storing SSN and some other ids. Before the table was split into address and phone tables, I had this script: (Written in Sybase) INSERT INTO sometable_3 ( SELECT a.id_number, a.other_id, NVL(a1.addr_type_code,0) home_addr_type_code, NVL(a1.addr_status_code,0) home_addr_status_code, NVL(a1.addr_pref_ind,0) home_addr_pref_ind, NVL(a1.street1,0) home_street1, NVL(a1.street2,0) home_street2, NVL(a1.street3,0) home_street3, NVL(a1.city,0) home_city, NVL(a1.state_code,0) home_state_code, NVL(a1.zipcode,0) home_zipcode, NVL(a1.zip_suffix,0) home_zip_suffix, NVL(a1.telephone_status_code,0) home_phone_status, NVL(a1.area_code,0) home_area_code, NVL(a1.telephone_number,0) home_phone_number, NVL(a1.extension,0) home_phone_extension, NVL(a1.date_modified,'') home_date_modified FROM both_ids a, address a1 WHERE a.id_number = a1.id_number(+) AND a1.addr_type_code = 'H'); Now that we moved to Oracle 11g, the address and phone information are split. How can I modify the above script to generate the same result in Oracle 11g? Do I have to first do INNER JOIN between address and phone tables and then do a LEFT OUTER JOIN to both_ids? I tried the following and it did not work: Insert Into.. select ... FROM a1. address INNER JOIN t.Phone ON a1.id_number = t.id_number LEFT OUTER JOIN both_ids a ON a.id_number = a1.id_number WHERE a1.adrr_type_code = 'H'

    Read the article

  • Outer product using CBLAS

    - by The Dude
    I am having trouble utilizing CBLAS to perform an Outer Product. My code is as follows: //===SET UP===// double x1[] = {1,2,3,4}; double x2[] = {1,2,3}; int dx1 = 4; int dx2 = 3; double X[dx1 * dx2]; for (int i = 0; i < (dx1*dx2); i++) {X[i] = 0.0;} //===DO THE OUTER PRODUCT===// cblas_dgemm(CblasRowMajor, CblasNoTrans, CblasTrans, dx1, dx2, 1, 1.0, x1, dx1, x2, 1, 0.0, X, dx1); //===PRINT THE RESULTS===// printf("\nMatrix X (%d x %d) = x1 (*) x2 is:\n", dx1, dx2); for (i=0; i<4; i++) { for (j=0; j<3; j++) { printf ("%lf ", X[j+i*3]); } printf ("\n"); } I get: Matrix X (4 x 3) = x1 (*) x2 is: 1.000000 2.000000 3.000000 0.000000 -1.000000 -2.000000 -3.000000 0.000000 7.000000 14.000000 21.000000 0.000000 But the correct answer is found here: https://www.sharcnet.ca/help/index.php/BLAS_and_CBLAS_Usage_and_Examples I have seen: Efficient computation of kronecker products in C But, it doesn't help me because they don't actually say how to utilize dgemm to actually do this... Any help? What am I doing wrong here?

    Read the article

  • LVM mirror attempt results in "Insufficient free space"

    - by MattK
    Attempting to add a disk to mirror an LVM volume on CentOS 7 always fails with "Insufficient free space: 1 extents needed, but only 0 available". Having searched for a solution, I have tried specifying disks, multiple logging options, adding 3rd log partition, but have not found a solution Not sure if I am making a rookie mistake, or there is something more subtle wrong (I am more familiar with ZFS, new to using LVM): # lvconvert -m1 centos_bi/home Insufficient free space: 1 extents needed, but only 0 available # lvconvert -m1 --corelog centos_bi/home Insufficient free space: 1 extents needed, but only 0 available # lvconvert -m1 --corelog --alloc anywhere centos_bi/home Insufficient free space: 1 extents needed, but only 0 available # lvconvert -m1 --mirrorlog mirrored --alloc anywhere centos_bi/home /dev/sda2 Insufficient free space: 1 extents needed, but only 0 available # lvconvert -m1 --corelog --alloc anywhere centos_bi/home /dev/sdi2 /dev/sda2 Insufficient free space: 1 extents needed, but only 0 available The two disks are of the same size, and have identical partition layouts via "sfdisk -d /dev/sdi part_table; sfdisk /dev/sda < part_table". The current configuration is detailed below. # pvs PV VG Fmt Attr PSize PFree /dev/sda1 centos_bi lvm2 a-- 496.00m 496.00m /dev/sda2 centos_bi lvm2 a-- 465.27g 465.27g /dev/sdi2 centos_bi lvm2 a-- 465.27g 0 # vgs VG #PV #LV #SN Attr VSize VFree centos_bi 3 3 0 wz--n- 931.02g 465.75g # lvs -a -o +devices LV VG Attr LSize Pool Origin Data% Move Log Cpy%Sync Convert Devices home centos_bi -wi-ao---- 391.64g /dev/sdi2(6050) root centos_bi -wi-ao---- 50.00g /dev/sdi2(106309) swap centos_bi -wi-ao---- 23.63g /dev/sdi2(0) # pvdisplay --- Physical volume --- PV Name /dev/sdi2 VG Name centos_bi PV Size 465.27 GiB / not usable 3.00 MiB Allocatable yes (but full) PE Size 4.00 MiB Total PE 119109 Free PE 0 Allocated PE 119109 --- Physical volume --- PV Name /dev/sda2 VG Name centos_bi PV Size 465.27 GiB / not usable 3.00 MiB Allocatable yes PE Size 4.00 MiB Total PE 119109 Free PE 119109 Allocated PE 0 --- Physical volume --- PV Name /dev/sda1 VG Name centos_bi PV Size 500.00 MiB / not usable 4.00 MiB Allocatable yes PE Size 4.00 MiB Total PE 124 Free PE 124 Allocated PE 0 # vgdisplay --- Volume group --- VG Name centos_bi System ID Format lvm2 Metadata Areas 3 Metadata Sequence No 10 VG Access read/write VG Status resizable MAX LV 0 Cur LV 3 Open LV 3 Max PV 0 Cur PV 3 Act PV 3 VG Size 931.02 GiB PE Size 4.00 MiB Total PE 238342 Alloc PE / Size 119109 / 465.27 GiB Free PE / Size 119233 / 465.75 GiB # lvdisplay --- Logical volume --- LV Path /dev/centos_bi/swap LV Name swap VG Name centos_bi LV Write Access read/write LV Creation host, time localhost, 2014-08-07 16:34:34 -0400 LV Status available # open 2 LV Size 23.63 GiB Current LE 6050 Segments 1 Allocation inherit Read ahead sectors auto - currently set to 256 Block device 253:1 --- Logical volume --- LV Path /dev/centos_bi/home LV Name home VG Name centos_bi LV Write Access read/write LV Creation host, time localhost, 2014-08-07 16:34:35 -0400 LV Status available # open 1 LV Size 391.64 GiB Current LE 100259 Segments 1 Allocation inherit Read ahead sectors auto - currently set to 256 Block device 253:2 --- Logical volume --- LV Path /dev/centos_bi/root LV Name root VG Name centos_bi LV Write Access read/write LV Creation host, time localhost, 2014-08-07 16:34:37 -0400 LV Status available # open 1 LV Size 50.00 GiB Current LE 12800 Segments 1 Allocation inherit Read ahead sectors auto - currently set to 256 Block device 253:0

    Read the article

< Previous Page | 6 7 8 9 10 11 12 13 14 15 16 17  | Next Page >