Search Results

Search found 13669 results on 547 pages for 'left handed'.

Page 3/547 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • Customizing Hibernate Criteria - Adding conditions to a left join

    - by Douglas Ferguson
    I need to be able to do the following: Select * from Table1 left join Table2 on id1 = id2 AND i1 = ? Hibernate criteria doesn't allow be to specify the i1 = ? part. The existing code is using hibernate criteria and it would be a huge refactor to swap out for HQL Does anybody have any tips how I could implement this differently or any way to override the Hibernate Criteria? I'm not opposed to cracking open hibernate and modifying, but when I began to dig it, there seems to be layers upon layers of abstractions. I never found the the point where SQL is actually generated...

    Read the article

  • Left outer join null using VB.NET and LINQ

    - by jvcoach23
    I've got what I think is a working left outer join LINQ query, but I'm having problems with the select because of null values in the right hand side of the join. Here is what I have so far Dim Os = From e In oExcel Group Join c In oClassIndexS On c.tClassCode Equals Mid(e.ClassCode, 1, 4) Into right1 = Group _ From c In right1.DefaultIfEmpty I want to return all of e and one column from c called tClassCode. I was wondering what the syntax would be. As you can see, I'm using VB.NET. Update... Here is the query doing join where I get the error: _message = "Object reference not set to an instance of an object." Dim Os = From e In oExcel Group Join c In oClassIndexS On c.tClassCode Equals Mid(e.ClassCode, 1, 4) Into right1 = Group _ From c In right1.DefaultIfEmpty Select e, c.tClassCode If I remove the c.tClassCode from the select, the query runs without error. So I thought perhaps I needed to do a select new, but I don't think I was doing that correctly either.

    Read the article

  • LINQ, Left Join, Only Get where null in join table

    - by kmehta
    Hi. I am trying to do a left outer join on two tables, but I only want to return the results from the first table where the second table does not have a record (null). var agencies = from a in agencyList join aa in joinTable on a.AgencyId equals aa.AgencyId into joined from aa in joined.DefaultIfEmpty() where aa == null) select a; But this does not exclude the non null values of aa, and returns all the records just the same as if the 'where aa == null' was not there. Any help is appreciated. Thanks.

    Read the article

  • mySQL Left Join on multiple tables

    - by Jarrod
    Hi I'm really struggling with this query. I have 4 tables (http://oberto.co.nz/db-sql.png): Invoice_Payement, Invoice, Client and Calendar. I'm trying to create a report by summing up the 'paid_amount' col, in Invoice_Payment, by month/year. The query needs to include all months, even those with no data There query needs the condition (Invoice table): registered_id = [id] I have tried with the below query, which works, but falls short when 'paid_date' does not have any records for a month. The outcome is that month does not show in the results I added a Calendar table to resolved this but not sure how to left join to it. SELECT MONTHNAME(Invoice_Payments.date_paid) as month, SUM(Invoice_Payments.paid_amount) AS total FROM Invoice, Client, Invoice_Payments WHERE Client.registered_id = 1 AND Client.id = Invoice.client_id And Invoice.id = Invoice_Payments.invoice_id AND date_paid IS NOT NULL GROUP BY YEAR(Invoice_Payments.date_paid), MONTH(Invoice_Payments.date_paid) Please see the above link for a basic ERD diagram of my scenario. Thanks for reading. I've posted this Q before but I think I worded it badly.

    Read the article

  • SQL query for getting count on same table using left outer join

    - by Sasi
    Hi all, I have a table from which i need to get the count grouped on two columns. the table has two columns one datetime column and another one is success value(-1,1,0) What i am looking for is something like this... count of success value for each month month----success-----count 11------- -1 ------- 50 11------- 1 --------- 50 11------- 0 ------- 50 12------- -1 ------- 50 12------- 1 ------- 50 12------- 0 ------- 50 if there is no success value for a month then the count should be null or zero. I have tried with left outer join as well but of no use it gives the count incorrectly. Thanks in advance Sasi

    Read the article

  • mysql filtering result using left outer join

    - by user288178
    my query: SELECT content.*, activity_log.content_id FROM content LEFT JOIN activity_log ON content.id = activity_log.content_id AND sess_id = '$sess_id' WHERE activity_log.content_id IS NULL AND visibility = $visibility AND content.reported < ".REPORTED_LIMIT." AND content.file_ready = 1 LIMIT 1 The purpose of that query is to get 1 row from the content table that has not been viewed by the user (identified by session_id), but it still returns contents that have been viewed. What is wrong? ( I have checked the table making sure that the content_ids are there) Note: I think this is more efficient than using subqueries, thoughts?

    Read the article

  • Left Join not returning all rows

    - by DisgruntledGoat
    I have this query in MySQL: SELECT pr.*, pr7.value AS `room_price_high` FROM `jos_hp_properties` pr LEFT OUTER JOIN `jos_hp_properties2` pr7 ON pr7.property=pr.id WHERE pr7.field=23 The jos_hp_properties table has 27 rows but the query only returns one. Based on this question I think it may be because of the WHERE clause. The jos_hp_properties2 table has fields id, property, field, value, where field is a foreign key to a third table (which I don't need to get data from). Is there a way to select all the rows from the first table, including the value from table #2 where the field is 23 (or NULL if there is no field 23)?

    Read the article

  • SQL LEFT JOIN help

    - by Stolz
    My scenario: There are 3 tables for storing tv show information; season, episode and episode_translation. My data: There are 3 seasons, with 3 episodes each one, but there is only translation for one episode. My objetive: I want to get a list of all the seasons and episodes for a show. If there is a translation available in a specified language, show it, otherwise show null. My attempt to get serie 1 information in language 1: SELECT season_number AS season,number AS episode,name FROM season NATURAL JOIN episode NATURAL LEFT JOIN episode_trans WHERE id_serie=1 AND id_lang=1 ORDER BY season_number,number result: +--------+---------+--------------------------------+ | season | episode | name | +--------+---------+--------------------------------+ | 3 | 3 | Episode translated into lang 1 | +--------+---------+--------------------------------+ expected result +-----------------+--------------------------------+ | season | episode| name | +-----------------+--------------------------------+ | 1 | 1 | NULL | | 1 | 2 | NULL | | 1 | 3 | NULL | | 2 | 1 | NULL | | 2 | 2 | NULL | | 2 | 3 | NULL | | 3 | 1 | NULL | | 3 | 2 | NULL | | 3 | 3 | Episode translated into lang 1 | +--------+--------+--------------------------------+ Full DB dump http://pastebin.com/Y8yXNHrH

    Read the article

  • Using ANTLR with Left-Recursive Rules

    - by CNevin561
    Basically Ive written a Parse for a language with just basic arithmetic operators ( +, -, * / ) etc, but for the minus and plus cases, the Abstract Syntax Tree which is generated has parsed them as right associative when they need to be left associative. Having a googled for a solution, i found a tutorial that suggests rewriting the rule from: Expression ::= Expression <operator> Term | Term as Expression ::= Term <operator> Expression*. However in my head this seems to generate the tree the wrong way round. Any pointers on a way to resolve this issue?

    Read the article

  • Left Join only returning one row

    - by Adam
    I am trying to join two tables. I would like all the columns from the product_category table (there are a total of 6 now) and count the number of products, CatCount, that are in each category from the products_has_product_category table. My query result is 1 row with the first category and a total count of 68, when I am looking for 6 rows with each individual category's count. <?php $result = mysql_query(" SELECT a.*, COUNT(b.category_id) AS CatCount FROM `product_category` a LEFT JOIN `products_has_product_category` b ON a.product_category_id = b.category_id "); while($row = mysql_fetch_array($result)) { echo ' <li class="ui-shadow" data-count-theme="d"> <a href="' . $row['product_category_ref_page'] . '.php" data-icon="arrow-r" data-iconpos="right">' . $row['product_category_name'] . '</a><span class="ui-li-count">' . $row['CatCount'] . '</span></li>'; } ?> I have been working on this for a couple of hours and would really appreciate any help on what I am doing wrong.

    Read the article

  • HQL : LEFT OUTER JOIN

    - by Parama
    Hi all, I have two tables and respective classes in java.The mapping in the HBM.xml is as follows : The query in the HBM.xml is as follows : from Reports as rep left join rep.parts as parts I am getting the following exception while executing the code : May 19, 2010 10:47:04 AM org.hibernate.util.JDBCExceptionReporter logExceptions WARNING: SQL Error: 904, SQLState: 42000 May 19, 2010 10:47:04 AM org.hibernate.util.JDBCExceptionReporter logExceptions SEVERE: ORA-00904: "REPORTS0_"."PARTNO": invalid identifier org.hibernate.exception.SQLGrammarException: could not execute query at org.hibernate.exception.SQLStateConverter.convert(SQLStateConverter.java:67) at org.hibernate.exception.JDBCExceptionHelper.convert(JDBCExceptionHelper.java:43) at org.hibernate.loader.Loader.doList(Loader.java:2223) at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2104) at org.hibernate.loader.Loader.list(Loader.java:2099) at org.hibernate.loader.hql.QueryLoader.list(QueryLoader.java:378) at org.hibernate.hql.ast.QueryTranslatorImpl.list(QueryTranslatorImpl.java:338) at org.hibernate.engine.query.HQLQueryPlan.performList(HQLQueryPlan.java:172) at org.hibernate.impl.SessionImpl.list(SessionImpl.java:1121) at org.hibernate.impl.QueryImpl.list(QueryImpl.java:79) at com.hcl.spring.db.sample.dao.ItemDAOImpl.loadItems(ItemDAOImpl.java:43) at com.hcl.spring.db.sample.service.ItemServiceImpl.loadItems(ItemServiceImpl.java:20) at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) at sun.reflect.NativeMethodAccessorImpl.invoke(Unknown Source) at sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source) at java.lang.reflect.Method.invoke(Unknown Source) at org.springframework.aop.support.AopUtils.invokeJoinpointUsingReflection(AopUtils.java:304) at org.springframework.aop.framework.ReflectiveMethodInvocation.invokeJoinpoint(ReflectiveMethodInvocation.java:182) at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:149) at org.springframework.transaction.interceptor.TransactionInterceptor.invoke(TransactionInterceptor.java:106) at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:171) at org.springframework.aop.framework.JdkDynamicAopProxy.invoke(JdkDynamicAopProxy.java:204) at $Proxy3.loadItems(Unknown Source) at com.hcl.spring.db.sample.Main.loadItems(Main.java:40) at com.hcl.spring.db.sample.Main.main(Main.java:19) Caused by: java.sql.SQLException: ORA-00904: "REPORTS0_"."PARTNO": invalid identifier at oracle.jdbc.driver.DatabaseError.throwSqlException(DatabaseError.java:112) at oracle.jdbc.driver.T4CTTIoer.processError(T4CTTIoer.java:331) at oracle.jdbc.driver.T4CTTIoer.processError(T4CTTIoer.java:288) at oracle.jdbc.driver.T4C8Oall.receive(T4C8Oall.java:743) at oracle.jdbc.driver.T4CPreparedStatement.doOall8(T4CPreparedStatement.java:216) at oracle.jdbc.driver.T4CPreparedStatement.executeForDescribe(T4CPreparedStatement.java:799) at oracle.jdbc.driver.OracleStatement.executeMaybeDescribe(OracleStatement.java:1038) at oracle.jdbc.driver.T4CPreparedStatement.executeMaybeDescribe(T4CPreparedStatement.java:839) at oracle.jdbc.driver.OracleStatement.doExecuteWithTimeout(OracleStatement.java:1133) at oracle.jdbc.driver.OraclePreparedStatement.executeInternal(OraclePreparedStatement.java:3285) at oracle.jdbc.driver.OraclePreparedStatement.executeQuery(OraclePreparedStatement.java:3329) at org.hibernate.jdbc.AbstractBatcher.getResultSet(AbstractBatcher.java:186) at org.hibernate.loader.Loader.getResultSet(Loader.java:1787) at org.hibernate.loader.Loader.doQuery(Loader.java:674) at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:236) at org.hibernate.loader.Loader.doList(Loader.java:2220) ... 22 more Do request your help on the same.

    Read the article

  • Linq to Entity Left Outer Join

    - by radman
    Hi All, I have an Entity model with Invoices, AffiliateCommissions and AffiliateCommissionPayments. Invoice to AffiliateCommission is a one to many, AffiliateCommission to AffiliateCommissionPayment is also a one to many I am trying to make a query that will return All Invoices that HAVE a commission but not necessarily have a related commissionPayment. I want to show the invoices with commissions whether they have a commission payment or not. Query looks something like: using (var context = new MyEntitities()) { var invoices = from i in context.Invoices from ac in i.AffiliateCommissions join acp in context.AffiliateCommissionPayments on ac.affiliateCommissionID equals acp.AffiliateCommission.affiliateCommissionID where ac.Affiliate.affiliateID == affiliateID select new { companyName = i.User.companyName, userName = i.User.fullName, email = i.User.emailAddress, invoiceEndDate = i.invoicedUntilDate, invoiceNumber = i.invoiceNumber, invoiceAmount = i.netAmount, commissionAmount = ac.amount, datePaid = acp.paymentDate, checkNumber = acp.checkNumber }; return invoices.ToList(); } This query above only returns items with an AffiliateCommissionPayment.

    Read the article

  • How to get Word 2003 to make my print layout go from left to right?

    - by Shaul
    My copy of MS Word 2003 was installed on my computer with the locale set to Israel, so among other things my Normal.dot template was set up for right-to-left. I managed to fix most of the Hebrew support things so that I am working in English by default now. The only thing I haven't found a cure for is how to make the "print layout" view also go from left to right; as things are, the page flow always appears from right to left, even in English documents - IOW, page 1 appears on the right of page 2, as shown below. I can't see any obvious option to change this. How do I do it?

    Read the article

  • T-SQL - Left Outer Joins - Filters in the where clause versus the on clause.

    - by Greg Potter
    I am trying to compare two tables to find rows in each table that is not in the other. Table 1 has a groupby column to create 2 sets of data within table one. groupby number ----------- ----------- 1 1 1 2 2 1 2 2 2 4 Table 2 has only one column. number ----------- 1 3 4 So Table 1 has the values 1,2,4 in group 2 and Table 2 has the values 1,3,4. I expect the following result when joining for Group 2: `Table 1 LEFT OUTER Join Table 2` T1_Groupby T1_Number T2_Number ----------- ----------- ----------- 2 2 NULL `Table 2 LEFT OUTER Join Table 1` T1_Groupby T1_Number T2_Number ----------- ----------- ----------- NULL NULL 3 The only way I can get this to work is if I put a where clause for the first join: PRINT 'Table 1 LEFT OUTER Join Table 2, with WHERE clause' select table1.groupby as [T1_Groupby], table1.number as [T1_Number], table2.number as [T2_Number] from table1 LEFT OUTER join table2 --****************************** on table1.number = table2.number --****************************** WHERE table1.groupby = 2 AND table2.number IS NULL and a filter in the ON for the second: PRINT 'Table 2 LEFT OUTER Join Table 1, with ON clause' select table1.groupby as [T1_Groupby], table1.number as [T1_Number], table2.number as [T2_Number] from table2 LEFT OUTER join table1 --****************************** on table2.number = table1.number AND table1.groupby = 2 --****************************** WHERE table1.number IS NULL Can anyone come up with a way of not using the filter in the on clause but in the where clause? The context of this is I have a staging area in a database and I want to identify new records and records that have been deleted. The groupby field is the equivalent of a batchid for an extract and I am comparing the latest extract in a temp table to a the batch from yesterday stored in a partioneds table, which also has all the previously extracted batches as well. Code to create table 1 and 2: create table table1 (number int, groupby int) create table table2 (number int) insert into table1 (number, groupby) values (1, 1) insert into table1 (number, groupby) values (2, 1) insert into table1 (number, groupby) values (1, 2) insert into table2 (number) values (1) insert into table1 (number, groupby) values (2, 2) insert into table2 (number) values (3) insert into table1 (number, groupby) values (4, 2) insert into table2 (number) values (4) EDIT: A bit more context - depending on where I put the filter I different results. As stated above the where clause gives me the correct result in one state and the ON in the other. I am looking for a consistent way of doing this. Where - select table1.groupby as [T1_Groupby], table1.number as [T1_Number], table2.number as [T2_Number] from table1 LEFT OUTER join table2 --****************************** on table1.number = table2.number --****************************** WHERE table1.groupby = 2 AND table2.number IS NULL Result: T1_Groupby T1_Number T2_Number ----------- ----------- ----------- 2 2 NULL On - select table1.groupby as [T1_Groupby], table1.number as [T1_Number], table2.number as [T2_Number] from table1 LEFT OUTER join table2 --****************************** on table1.number = table2.number AND table1.groupby = 2 --****************************** WHERE table2.number IS NULL Result: T1_Groupby T1_Number T2_Number ----------- ----------- ----------- 1 1 NULL 2 2 NULL 1 2 NULL Where (table 2 this time) - select table1.groupby as [T1_Groupby], table1.number as [T1_Number], table2.number as [T2_Number] from table2 LEFT OUTER join table1 --****************************** on table2.number = table1.number AND table1.groupby = 2 --****************************** WHERE table1.number IS NULL Result: T1_Groupby T1_Number T2_Number ----------- ----------- ----------- NULL NULL 3 On - select table1.groupby as [T1_Groupby], table1.number as [T1_Number], table2.number as [T2_Number] from table2 LEFT OUTER join table1 --****************************** on table2.number = table1.number --****************************** WHERE table1.number IS NULL AND table1.groupby = 2 Result: T1_Groupby T1_Number T2_Number ----------- ----------- ----------- (0) rows returned

    Read the article

  • Photoshop: left-handed copy/paste?

    - by Diodeus
    Being left-handed, I use the mouse in my left hand. In most applications I use CTRL+INSERT/SHIFT+INSERT to copy and paste (with my right hand). For some bone-headed reason, this is not supported in Photoshop, so I have to use right-click COPY sub-menus, which is a lot slower. Is there a way to configure Photoshop to use CTRL+INSERT to copy and SHIFT+INSERT to paste?

    Read the article

  • My Right-to-Left Foot (T-SQL Tuesday #13)

    - by smisner
    As a business intelligence consultant, I often encounter the situation described in this month's T-SQL Tuesday, hosted by Steve Jones ( Blog | Twitter) – “What the Business Says Is Not What the  Business Wants.” Steve posed the question, “What issues have you had in interacting with the business to get your job done?” My profession requires me to have one foot firmly planted in the technology world and the other foot planted in the business world. I learned long ago that the business never says exactly what the business wants because the business doesn't have the words to describe what the business wants accurately enough for IT. Not only do technological-savvy barriers exist, but there are also linguistic barriers between the two worlds. So how do I cope? The adage "a picture is worth a thousand words" is particularly helpful when I'm called in to help design a new business intelligence solution. Many of my students in BI classes have heard me explain ("rant") about left-to-right versus right-to-left design. To understand what I mean about these two design options, let's start with a picture: When we design a business intelligence solution that includes some sort of traditional data warehouse or data mart design, we typically place the data sources on the left, the new solution in the middle, and the users on the right. When I've been called in to help course-correct a failing BI project, I often find that IT has taken a left-to-right approach. They look at the data sources, decide how to model the BI solution as a _______ (fill in the blank with data warehouse, data mart, cube, etc.), and then build the new data structures and supporting infrastructure. (Sometimes, they actually do this without ever having talked to the business first.) Then, when they show what they've built to the business, the business says that is not what we want. Uh-oh. I prefer to take a right-to-left approach. Preferably at the beginning of a project. But even if the project starts left-to-right, I'll do my best to swing it around so that we’re back to a right-to-left approach. (When circumstances are beyond my control, I carry on, but it’s a painful project for everyone – not because of me, but because the approach just doesn’t get to what the business wants in the most effective way.) By using a right to left approach, I try to understand what it is the business is trying to accomplish. I do this by having them explain reports to me, and explaining the decision-making process that relates to these reports. Sometimes I have them explain to me their business processes, or better yet show me their business processes in action because I need pictures, too. I (unofficially) call this part of the project "getting inside the business's head." This is starting at the right side of the diagram above. My next step is to start moving leftward. I do this by preparing some type of prototype. Depending on the nature of the project, this might mean that I simply mock up some data in a relational database and build a prototype report in Reporting Services. If I'm lucky, I might be able to use real data in a relational database. I'll either use a subset of the data in the prototype report by creating a prototype database to hold the sample data, or select data directly from the source. It all depends on how much data there is, how complex the queries are, and how fast I need to get the prototype completed. If the solution will include Analysis Services, then I'll build a prototype cube. Analysis Services makes it incredibly easy to prototype. You can sit down with the business, show them the prototype, and have a meaningful conversation about what the BI solution should look like. I know I've done a good job on the prototype when I get knocked out of my chair so that the business user can explore the solution further independently. (That's really happened to me!) We can talk about dimensions, hierarchies, levels, members, measures, and so on with something tangible to look at and without using those terms. It's not helpful to use sample data like Adventure Works or to use BI terms that they don't really understand. But when I show them their data using the BI technology and talk to them in their language, then they truly have a picture worth a thousand words. From that, we can fine tune the prototype to move it closer to what they want. They have a better idea of what they're getting, and I have a better idea of what to build. So right to left design is not truly moving from the right to the left. But it starts from the right and moves towards the middle, and once I know what the middle needs to look like, I can then build from the left to meet in the middle. And that’s how I get past what the business says to what the business wants.

    Read the article

  • Why / how does XNA's right-handed coordinate system effect anything if you can specify near/far Z values?

    - by vargonian
    I am told repeatedly that XNA Game Studio uses a right-handed coordinate system, and I understand the difference between a right-handed and left-handed coordinate system. But given that you can use a method like Matrix.CreateOrthographicOffCenter to create your own custom projection matrix, specifying the left, right, top, bottom, zNear and zFar values, when does XNA's coordinate system come into play? For example, I'm told that in a right-handed coordinate system, increasingly negative Z values go "into" the screen. But I can easily create my projection matrix like this: Matrix.CreateOrthographicOffCenter(left, right, bottom, top, 0.1f, 10000f); I've now specified a lower value for the near Z than the far Z, which, as I understand it, means that positive Z now goes into the screen. I can similarly tweak the values of left/right/top/bottom to achieve similar results. If specifying a lower zNear than zFar value doesn't affect the Z direction of the coordinate system, what does it do? And when is the right-handed coordinate system enforced? The reason I ask is that I'm trying to implement a 2.5D camera that supports zooming and rotation, and I've spent two full days encountering one unexpected result after another.

    Read the article

  • Android edittext right align [closed]

    - by Yoav
    I am using my application with hebrew (right-to-left) layout. I have a feature where I open an activity with EditText in it - where I put some text (previously entered by the user) to be edited by him. However, when I do setText I find out that the text is aligned to the left of the edittext instead of of the right. (If I start with empty edittext then it is automatically right aligned when user starts inputting hebrew, but cursor is positioned to the left) (android:gravity="right" does not work)

    Read the article

  • Why Swipe left doesn't work? [on hold]

    - by Hitesh
    I wrote the below code to detect and perform a sprite action on the single tap and swipe right event. @Override public boolean onSceneTouchEvent(Scene pScene, TouchEvent pSceneTouchEvent) { float x = 0F; int tapCount = 0; boolean playermoving = false; // TODO Auto-generated method stub if (pSceneTouchEvent.getAction() == MotionEvent.ACTION_MOVE) { if (pSceneTouchEvent.getX() > x) { playermoving = true; players.runRight(); } if (pSceneTouchEvent.getX() < x) { Log.i("Run Left", "SPRITE Left"); } /* * if (pSceneTouchEvent.getX() < x) { System.exit(0); * Log.i("SWIPE left", "SPRITE LEFT"); } */ } if (pSceneTouchEvent.getAction() == MotionEvent.ACTION_DOWN) { playermoving = false; x = pSceneTouchEvent.getX(); tapCount++; Log.i("X CORD", String.valueOf(x)); } if (pSceneTouchEvent.isActionDown()) { if (tapCount == 1 && playermoving != true) { tapCount = 0; players.jumpRight(); } } return true; } The code works fine. The only problem is that the swipe left event is not being detected due to some reasons. What can i do to make the swipe left action work? Please help

    Read the article

  • How can one-handed work in Ubuntu be eased?

    - by N.N.
    My right hand is temporarily immobilized and I would like to do some minor general work on my computer. Mostly web browsing, mailing and file and directory browsing and editing. For this I currently use Firefox, Thunderbird, Nautilus and the GNOME terminal (I have already asked a specific question about Emacs). Are there ways to ease such, or any other general, one-handed work in Ubuntu? I have found http://stackoverflow.com/questions/2391805/how-can-i-remain-productive-with-one-hand-completely-immobilized but that is not exactly what I am asking for. I want to ease whatever little time spent one-handed in Ubuntu and this is also interesting for situations where there is no injury involved, such as when one hand is occupied. I do realize I should avoid unnecessary strain. The main thing that is much slower one-handed is writing. Since I am only temporarily immobilized it seems to make no sense learn a new keyboard layout. I would be surprised if I managed to learn and become more effective with a new keyboard layout (than one-handed QWERTY) before I can use my other hand again. What I have already found: Sticky keys for making it easier to enter keyboard commands. When writing one-handed there are more cases of where it is useful to paste in phrases rather than to reenter them. It is easier to use Super+S rather than CtrlAlt+arrow keys to switch work space.

    Read the article

  • T-SQL - Left Outer Joins - Fileters in the where clause versus the on clause.

    - by Greg Potter
    I am trying to compare two tables to find rows in each table that is not in the other. Table 1 has a groupby column to create 2 sets of data within table one. groupby number ----------- ----------- 1 1 1 2 2 1 2 2 2 4 Table 2 has only one column. number ----------- 1 3 4 So Table 1 has the values 1,2,4 in group 2 and Table 2 has the values 1,3,4. I expect the following result when joining for Group 2: `Table 1 LEFT OUTER Join Table 2` T1_Groupby T1_Number T2_Number ----------- ----------- ----------- 2 2 NULL `Table 2 LEFT OUTER Join Table 1` T1_Groupby T1_Number T2_Number ----------- ----------- ----------- NULL NULL 3 The only way I can get this to work is if I put a where clause for the first join: PRINT 'Table 1 LEFT OUTER Join Table 2, with WHERE clause' select table1.groupby as [T1_Groupby], table1.number as [T1_Number], table2.number as [T2_Number] from table1 LEFT OUTER join table2 --****************************** on table1.number = table2.number --****************************** WHERE table1.groupby = 2 AND table2.number IS NULL and a filter in the ON for the second: PRINT 'Table 2 LEFT OUTER Join Table 1, with ON clause' select table1.groupby as [T1_Groupby], table1.number as [T1_Number], table2.number as [T2_Number] from table2 LEFT OUTER join table1 --****************************** on table2.number = table1.number AND table1.groupby = 2 --****************************** WHERE table1.number IS NULL Can anyone come up with a way of not using the filter in the on clause but in the where clause? The context of this is I have a staging area in a database and I want to identify new records and records that have been deleted. The groupby field is the equivalent of a batchid for an extract and I am comparing the latest extract in a temp table to a the batch from yesterday stored in a partioneds table, which also has all the previously extracted batches as well. Code to create table 1 and 2: create table table1 (number int, groupby int) create table table2 (number int) insert into table1 (number, groupby) values (1, 1) insert into table1 (number, groupby) values (2, 1) insert into table1 (number, groupby) values (1, 2) insert into table2 (number) values (1) insert into table1 (number, groupby) values (2, 2) insert into table2 (number) values (3) insert into table1 (number, groupby) values (4, 2) insert into table2 (number) values (4)

    Read the article

  • What's the different between these 2 mysql queries? one using left join

    - by Lyon
    Hi, I see people using LEFT JOIN in their mysql queries to fetch data from two tables. But I normally do it without left join. Is there any differences besides the syntax, e.g. performance? Here's my normal query style: SELECT * FROM table1 as tbl1, table2 as tbl2 WHERE tbl1.id=tbl2.table_id as compared to SELECT * FROM table1 as tbl1 LEFT JOIN table2 as tbl2 on tbl1.id=tbl2.id Personally I prefer the first style...hmm..

    Read the article

  • Difference left/right super button

    - by Erik Keemink
    When I press my left super key the gnome shell appears and when I press the right super key it does not. Moreover pressing right super + T does open a terminal at once, but when using left super I have to press the t twice, when I press the t only once it is similar to just pressing the t without holding super left. This last point also occurs with other shortcuts that I defined (like super+L, super+E), but not with super+up/down/left/right. What I want is to press either super key to get the gnome shell and to use either super key in combination with T to open a terminal immediately (and similar with other shortcuts). I use Ubuntu 12.04 LTS and the gnome 3 shell.

    Read the article

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >