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  • wide and long table in latex

    - by Tim
    Hi, I have a wide and long table. I am not sure how to put it in my thesis. Since it is wide, sidewaystable may be the choice. Meanwhile it is also to long to fit in one page, so longtable comes into my mind. However, I cannot make sideaystable and longtable working together for one table, e.g. \begin{sidewaystable} \begin{longtable}{| c ||c| c| c |c| c|| c |c| c|c|c| } \caption{A glance of images.} \centering % table content \end{longtable} \end{sidewaystable} What shall I do? Thanks and regards!

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  • SQL query to return data from two separate rows in a table joined to a master table

    - by Ali
    I have a TWO tables of data with following fields table1=(ITTAG,ITCODE,ITDESC,SUPcode) table2=(ACCODE,ACNAME,ROUTE,SALMAN) This is my customer master table that contains my customer data such as customer code, customer name and so on... Every Route has a supervisor (table1=supcode) and I need to know the supervisor name in my table which both supervisor name and code exist in one table. table1 has contain all names separated by ITTAG. For example, supervisor's name has ITTAG='K'; also salesman's name has ITTAG='S'. ITTAG ITCODE ITDESC SUPCODE ------ ------ ------ ------- S JT JOHN TOMAS TF K WK VIKI KOO NULL Now this is the result which I want ACCODE ACNAME ROUTE SALEMANNAME SUPERVISORNAME ------- ------ ------ ------------ --------------- IMC1010 ABC HOTEL 01 JOHN TOMAS VIKI KOO I hope this this information is sufficient to get the query..

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  • SQLite table creation date

    - by legr3c
    Is there a way to query the creation date of a table in SQLite? I am new to SQL, overall. I just found this http://stackoverflow.com/questions/1171019/sql-server-table-creation-date-query. I am assuming that sqlite_master is the equivalent to sys.tables in SQLite. Is that correct? But then my sqlite_master table only has the columns "type", "name", "tbl_name", "rootpage" and "sql". If this is not possible in SQLite, what would be the best way to implement this functionality by myself?

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  • jQuery add border to table.

    - by Kyle Sevenoaks
    Hi, I'm a jQuery noob, I tried this: <input value="1" type="checkbox" name="mytable" id="checkbox2" style="float:left;" /> {literal} <script src="http://code.jquery.com/jquery-latest.js"></script> <script type="text/javascript"> $(function() { //checkbox $(".mytable").click(function(){ $(".mytable").toggleClass('mytableborders'); }); }); </script> {/literal} <table class="mytable" id="cart">....</table> But it doesn't work, I want the checkbox to change the class of the table from .mytable to .mytableborders. Thanks :)

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  • Resetting AUTO_INCREMENT on myISAM without rebuilding the table

    - by Artem
    Please help I am in major trouble with our production database. I had accidentally inserted a key with a very large value into an autoincrement column, and now I can't seem to change this value without a huge rebuild time. "ALTER TABLE tracks_copy AUTO_INCREMENT = 661482981" Is super-slow. How can I fix this in production? I can't get this to work either (has no effect): myisamchk tracks.MYI --set-auto-increment=661482982 Any ideas? Basically, no matter what I do I get an overflow: SHOW CREATE TABLE tracks CREATE TABLE tracks ( ... ) ENGINE=MYISAM AUTO_INCREMENT=2147483648 DEFAULT CHARSET=latin1

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  • MySQL multi CREATE TABLE syntax help?

    - by rlb.usa
    Hi guys, I'm trying to write a MySQL script that creates several tables. I have: CREATE TABLE `DataBase1`.`tbl_this`( ... ); CREATE TABLE `DataBase1`.`tbl_that`( ... ); ... (14 more) ... BUT, only the first CREATE TABLE statement is executed. I get no syntax errors. Erm, am I missing the MSSQL equivalent of GO ? What am I doing wrong here; how do I get this baby to run all the tables?

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  • How to handle large table in MySQL ?

    - by Frantz Miccoli
    I've a database used to store items and properties about these items. The number of properties is extensible, thus there is a join table to store each property associated to an item value. CREATE TABLE `item_property` ( `property_id` int(11) NOT NULL, `item_id` int(11) NOT NULL, `value` double NOT NULL, PRIMARY KEY (`property_id`,`item_id`), KEY `item_id` (`item_id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci; This database has two goals : storing (which has first priority and has to be very quick, I would like to perform many inserts (hundreds) in few seconds), retrieving data (selects using item_id and property_id) (this is a second priority, it can be slower but not too much because this would ruin my usage of the DB). Currently this table hosts 1.6 billions entries and a simple count can take up to 2 minutes... Inserting isn't fast enough to be usable. I'm using Zend_Db to access my data and would really be happy if you don't suggest me to develop any php side part. Thanks for your advices !

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  • PHP Doctrine 1.2 table names

    - by Ofir
    I'm trying to upgrade my doctrine ORM from 1.1.6 to 1.2.1 but i've enountered a BC issue with table names. Some of my table names have several words (e.g. t_foo_bar for class FooBar) where the t_ prefix is generated automatically with: $manager->setAttribute(Doctrine_Core::ATTR_TBLNAME_FORMAT, 't_%s'); This worked well in previous versions. In 1.2.1 however, it looks like doctrine is looking for t_foobar (instead of t_foo_bar with an underscore). Do you know how to solve this without changing the table names?

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  • Sqlite + Java: table not updating

    - by Ben L
    I'm using a Java wrapper for SQLite called SQLiteJDBC - this may have some bearing on any answers... I have a table which I've displayed in a GUI, within that UI I have a button for the fields for an individual row of that table. When I save my changes I do this... Statement stmt = connection.createStatement(); stmt.execute("update 'tableName' set 'fieldName'=1 where userid=1"); int updateCount = stmt.getUpdateCount(); My connection is valid, I get no exceptions thrown and getUpdateCount() returns '1' indicating that one row has been updated. However my table is not updated. I've spent the last few hours trying to work out what's going on but I'm not having any luck. Help!!

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  • Trying to use table aliases in SQL...

    - by user366685
    I'm a graphic designer trying my best to understand table aliases, but it's not working. Here's what I have so far: SELECT colours.colourid AS colourid1, combinations.manufacturercolourid AS colourmanid1, colours.colourname AS colourname1, colours.colourhex AS colourhex1, combinations.qecolourid2 AS colouridqe2, colours.colourid AS colourid2, colours.colourname AS colourname2, colours.colourhex AS colourhex2, colours.colourid AS colourid3, combinations.qecolourid3 AS colouridqe3, colours.colourname AS colourname3, colours.colourhex AS colourhex3, colours.colourid AS colourid4, combinations.qecolourid4 AS colouridqe4, colours.colourname AS colourname4, colours.colourhex AS colourhex4, combinations.coloursupplierid FROM combinations INNER JOIN colours ON colours.colourid = combinations.manufacturercolourid; Now, the idea is that in the colours lookup table, the id will pull the colour code, hex and name from the lookup table so that I can pull the colour code, hex and name for the 4 colours that I'm looking for. I can get this to work, but it only pulls up the first name, code and hex and I'm just not seeing what I'm doing wrong.

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  • One table for uploads for multiple resources in CakePHP

    - by Mef
    Hi, I have CakePHP app in which I'd like to attach gallery to multiple resources. Let's say I've got artists, each one has own gallery. I've got articles, every article has some images attached to it and so on. Now I set up tables like this: Artists hasMany Artistimages, fields in artistimages table are: id, artist_id, filename, filetype, filesize etc. Articles hasMany Articleimages, fields in articleimages table are: id, article_id, filename, filetype, filesize etc. ...but this is not how it should be, I think. Is there possibility to have one table called for example uploads which will contain all images with foreign key pointing to resource its reffering to? How to tell CakePHP which image is coming from which resource?

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  • Drupal: How to render a form and table on same page

    - by Aaron
    Can someone help me render a form and table on the same page? I'm sure it's easy, but can't think of how to do it. Here's hook_menu: function ncbi_subsites_menu() { $items = array(); $items['admin/content/ncbi_subsites'] = array( 'title' => 'NCBI Subsites Module', 'description' => 'Informs Drupal about NCBI subsites as defined by the Content Inventory database', 'page callback' => 'ncbi_subsites_show_main_page', 'access arguments' => array( 'administer site configuration' ), 'type' => MENU_NORMAL_ITEM, ); return $items; } Here's the callback: function ncbi_subsites_show_main_page() { $subsites = ncbi_subsites_get_subsites_from_inventory(); // fnc returns associative array from inventory, defined in include return ncbi_subsites_make_table( $subsites ); } In the callback I call some helper functions that return a themed, paged table. What I want is a small form above the table. How would I that?

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  • Find mininum unused value in mysql table

    - by splattru
    I have a MySQL table like this: +--------+-------+ + Server + Port + +--------+-------+ + 1 + 27001 + +--------+-------+ + 2 + 27002 + +--------+-------+ + 4 + 27004 + +--------+-------+ How you can see - here was 27003 port - but it was deleted some time ago (just for example). Is there some way to know the minimum value that is not used (higher than 27000 for example) in mysql table? Or there is no way and I need to make an separate table with all ports and column with "used" or "not used" ? Situation explanation: I am creating game hosting website, and I can use auto-increment for ports, but I will have a lot of deleted/unused ports after some time, and I want to use all of them before increasing port range. Thanks for your answers!

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  • how to sort an existing table in greasemonkey ?

    - by user570512
    i'm writing a greasemonkey user.js for a page with a table in it. (table is 100 rows by 18 columns.) now what i want to do is to make it sortable on column. and also have it run in both chrome and firefox. all searches for answers sofar resulted in suggestions to use jquery/dojo or something alike. can i be done without any external code? after all it's just a matter of replacing the row's in a different order, right? or is that a silly thing to say? the thing is that i'm already using dojo for some query needs but since i want it to run in both firefox and chrome, i just copy paste the whole dojo thing in my script.. also, most of the solutions i found sofar seem to be more for use when building a table. not for altering an existing one. any help is appreciated.

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  • Operations in table data with javascript

    - by Zangrandi
    I'm working with rails and I don't know javascript. I have a table with a select_tag field and I want to have another field that capture the option selected, multiply for the price captured in another field and display the total. Like this <table> <tr> <td>(select_tag)</td> <td>price</td> <td>total</td> </tr> </table

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  • SQL Server dynamic pivot table

    - by user972255
    In SQL Server, I have two tables TableA and TableB, based on these I need to generate a report which is kind of very complex and after doing some research I come to a conclusion that I have to go with SQL Pivot table but I dont have any idea about the SQL Pivot feature so, can anyone please help me on this. Please see the details below: Create table TableA ( ProjectID INT NOT NULL, ControlID INT NOT NULL, ControlCode Varchar(2) NOT NULL, ControlPoint Decimal NULL, ControlScore Decimal NULL, ControlValue Varchar(50) ) Sample Data ------------- ProjectID | ControlID | ControlCode | ControlPoint | ControlScore | ControlValue P001 1 A 30.44 65 Invalid P001 2 C 45.30 85 Valid Create table TableB ( ControlID INT NOT NULL, ControlChildID INT NOT NULL, ControlChildValue Varchar(200) NULL ) Sample Data ------------ ControlID | ControlChildID | ControlChildValue 1 100 Yes 1 101 No 1 102 NA 1 103 Others 2 104 Yes 2 105 SomeValue Output should be in a single row for a given ProjectID with all its Control values first & followed by child control values (based on the ControlCode (i.e.) ControlCode_Child (1, 2, 3...) and it should look like this

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  • Tool to automatically combine many large pivot table in many large Excel sheet ?

    - by Sim
    Hi all, We have several Excel files that contains large pivot data table with the same structure For example File A Pivot table (Field A, B, C) File B Pivot table (Field A, B, C) We want to combine them into 1 pivot table (A, B, C). Just want to know which ways we can do it ? Manual way : open a new empty sheet, copy and paste the pivot there and create the pivot again Automatic way : is there some tool that did this ? Thanks

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  • ASP.NET MVC 3: Layouts and Sections with Razor

    - by ScottGu
    This is another in a series of posts I’m doing that cover some of the new ASP.NET MVC 3 features: Introducing Razor (July 2nd) New @model keyword in Razor (Oct 19th) Layouts with Razor (Oct 22nd) Server-Side Comments with Razor (Nov 12th) Razor’s @: and <text> syntax (Dec 15th) Implicit and Explicit code nuggets with Razor (Dec 16th) Layouts and Sections with Razor (Today) In today’s post I’m going to go into more details about how Layout pages work with Razor.  In particular, I’m going to cover how you can have multiple, non-contiguous, replaceable “sections” within a layout file – and enable views based on layouts to optionally “fill in” these different sections at runtime.  The Razor syntax for doing this is clean and concise. I’ll also show how you can dynamically check at runtime whether a particular layout section has been defined, and how you can provide alternate content (or even an alternate layout) in the event that a section isn’t specified within a view template.  This provides a powerful and easy way to customize the UI of your site and make it clean and DRY from an implementation perspective. What are Layouts? You typically want to maintain a consistent look and feel across all of the pages within your web-site/application.  ASP.NET 2.0 introduced the concept of “master pages” which helps enable this when using .aspx based pages or templates.  Razor also supports this concept with a feature called “layouts” – which allow you to define a common site template, and then inherit its look and feel across all the views/pages on your site. I previously discussed the basics of how layout files work with Razor in my ASP.NET MVC 3: Layouts with Razor blog post.  Today’s post will go deeper and discuss how you can define multiple, non-contiguous, replaceable regions within a layout file that you can then optionally “fill in” at runtime. Site Layout Scenario Let’s look at how we can implement a common site layout scenario with ASP.NET MVC 3 and Razor.  Specifically, we’ll implement some site UI where we have a common header and footer on all of our pages.  We’ll also add a “sidebar” section to the right of our common site layout.  On some pages we’ll customize the SideBar to contain content specific to the page it is included on: And on other pages (that do not have custom sidebar content) we will fall back and provide some “default content” to the sidebar: We’ll use ASP.NET MVC 3 and Razor to enable this customization in a nice, clean way.  Below are some step-by-step tutorial instructions on how to build the above site with ASP.NET MVC 3 and Razor. Part 1: Create a New Project with a Layout for the “Body” section We’ll begin by using the “File->New Project” menu command within Visual Studio to create a new ASP.NET MVC 3 Project.  We’ll create the new project using the “Empty” template option: This will create a new project that has no default controllers in it: Creating a HomeController We will then right-click on the “Controllers” folder of our newly created project and choose the “Add->Controller” context menu command.  This will bring up the “Add Controller” dialog: We’ll name the new controller we create “HomeController”.  When we click the “Add” button Visual Studio will add a HomeController class to our project with a default “Index” action method that returns a view: We won’t need to write any Controller logic to implement this sample – so we’ll leave the default code as-is.  Creating a View Template Our next step will be to implement the view template associated with the HomeController’s Index action method.  To implement the view template, we will right-click within the “HomeController.Index()” method and select the “Add View” command to create a view template for our home page: This will bring up the “Add View” dialog within Visual Studio.  We do not need to change any of the default settings within the above dialog (the name of the template was auto-populated to Index because we invoked the “Add View” context menu command within the Index method).  When we click the “Add” Button within the dialog, a Razor-based “Index.cshtml” view template will be added to the \Views\Home\ folder within our project.  Let’s add some simple default static content to it: Notice above how we don’t have an <html> or <body> section defined within our view template.  This is because we are going to rely on a layout template to supply these elements and use it to define the common site layout and structure for our site (ensuring that it is consistent across all pages and URLs within the site).  Customizing our Layout File Let’s open and customize the default “_Layout.cshtml” file that was automatically added to the \Views\Shared folder when we created our new project: The default layout file (shown above) is pretty basic and simply outputs a title (if specified in either the Controller or the View template) and adds links to a stylesheet and jQuery.  The call to “RenderBody()” indicates where the main body content of our Index.cshtml file will merged into the output sent back to the browser. Let’s modify the Layout template to add a common header, footer and sidebar to the site: We’ll then edit the “Site.css” file within the \Content folder of our project and add 4 CSS rules to it: And now when we run the project and browse to the home “/” URL of our project we’ll see a page like below: Notice how the content of the HomeController’s Index view template and the site’s Shared Layout template have been merged together into a single HTML response.  Below is what the HTML sent back from the server looks like: Part 2: Adding a “SideBar” Section Our site so far has a layout template that has only one “section” in it – what we call the main “body” section of the response.  Razor also supports the ability to add additional "named sections” to layout templates as well.  These sections can be defined anywhere in the layout file (including within the <head> section of the HTML), and allow you to output dynamic content to multiple, non-contiguous, regions of the final response. Defining the “SideBar” section in our Layout Let’s update our Layout template to define an additional “SideBar” section of content that will be rendered within the <div id=”sidebar”> region of our HTML.  We can do this by calling the RenderSection(string sectionName, bool required) helper method within our Layout.cshtml file like below:   The first parameter to the “RenderSection()” helper method specifies the name of the section we want to render at that location in the layout template.  The second parameter is optional, and allows us to define whether the section we are rendering is required or not.  If a section is “required”, then Razor will throw an error at runtime if that section is not implemented within a view template that is based on the layout file (which can make it easier to track down content errors).  If a section is not required, then its presence within a view template is optional, and the above RenderSection() code will render nothing at runtime if it isn’t defined. Now that we’ve made the above change to our layout file, let’s hit refresh in our browser and see what our Home page now looks like: Notice how we currently have no content within our SideBar <div> – that is because the Index.cshtml view template doesn’t implement our new “SideBar” section yet. Implementing the “SideBar” Section in our View Template Let’s change our home-page so that it has a SideBar section that outputs some custom content.  We can do that by opening up the Index.cshtml view template, and by adding a new “SiderBar” section to it.  We’ll do this using Razor’s @section SectionName { } syntax: We could have put our SideBar @section declaration anywhere within the view template.  I think it looks cleaner when defined at the top or bottom of the file – but that is simply personal preference.  You can include any content or code you want within @section declarations.  Notice above how I have a C# code nugget that outputs the current time at the bottom of the SideBar section.  I could have also written code that used ASP.NET MVC’s HTML/AJAX helper methods and/or accessed any strongly-typed model objects passed to the Index.cshtml view template. Now that we’ve made the above template changes, when we hit refresh in our browser again we’ll see that our SideBar content – that is specific to the Home Page of our site – is now included in the page response sent back from the server: The SideBar section content has been merged into the proper location of the HTML response : Part 3: Conditionally Detecting if a Layout Section Has Been Implemented Razor provides the ability for you to conditionally check (from within a layout file) whether a section has been defined within a view template, and enables you to output an alternative response in the event that the section has not been defined.  This provides a convenient way to specify default UI for optional layout sections.  Let’s modify our Layout file to take advantage of this capability.  Below we are conditionally checking whether the “SideBar” section has been defined without the view template being rendered (using the IsSectionDefined() method), and if so we render the section.  If the section has not been defined, then we now instead render some default content for the SideBar:  Note: You want to make sure you prefix calls to the RenderSection() helper method with a @ character – which will tell Razor to execute the HelperResult it returns and merge in the section content in the appropriate place of the output.  Notice how we wrote @RenderSection(“SideBar”) above instead of just RenderSection(“SideBar”).  Otherwise you’ll get an error. Above we are simply rendering an inline static string (<p>Default SideBar Content</p>) if the section is not defined.  A real-world site would more likely refactor this default content to be stored within a separate partial template (which we’d render using the Html.RenderPartial() helper method within the else block) or alternatively use the Html.Action() helper method within the else block to encapsulate both the logic and rendering of the default sidebar. When we hit refresh on our home-page, we will still see the same custom SideBar content we had before.  This is because we implemented the SideBar section within our Index.cshtml view template (and so our Layout rendered it): Let’s now implement a “/Home/About” URL for our site by adding a new “About” action method to our HomeController: The About() action method above simply renders a view back to the client when invoked.  We can implement the corresponding view template for this action by right-clicking within the “About()” method and using the “Add View” menu command (like before) to create a new About.cshtml view template.  We’ll implement the About.cshtml view template like below. Notice that we are not defining a “SideBar” section within it: When we browse the /Home/About URL we’ll see the content we supplied above in the main body section of our response, and the default SideBar content will rendered: The layout file determined at runtime that a custom SideBar section wasn’t present in the About.cshtml view template, and instead rendered the default sidebar content. One Last Tweak… Let’s suppose that at a later point we decide that instead of rendering default side-bar content, we just want to hide the side-bar entirely from pages that don’t have any custom sidebar content defined.  We could implement this change simply by making a small modification to our layout so that the sidebar content (and its surrounding HTML chrome) is only rendered if the SideBar section is defined.  The code to do this is below: Razor is flexible enough so that we can make changes like this and not have to modify any of our view templates (nor make change any Controller logic changes) to accommodate this.  We can instead make just this one modification to our Layout file and the rest happens cleanly.  This type of flexibility makes Razor incredibly powerful and productive. Summary Razor’s layout capability enables you to define a common site template, and then inherit its look and feel across all the views/pages on your site. Razor enables you to define multiple, non-contiguous, “sections” within layout templates that can be “filled-in” by view templates.  The @section {} syntax for doing this is clean and concise.  Razor also supports the ability to dynamically check at runtime whether a particular section has been defined, and to provide alternate content (or even an alternate layout) in the event that it isn’t specified.  This provides a powerful and easy way to customize the UI of your site - and make it clean and DRY from an implementation perspective. Hope this helps, Scott P.S. In addition to blogging, I am also now using Twitter for quick updates and to share links. Follow me at: twitter.com/scottgu

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  • SQL SERVER – Finding Different ColumnName From Almost Identitical Tables

    - by pinaldave
    I have mentioned earlier on this blog that I love social media – Facebook and Twitter. I receive so many interesting questions that sometimes I wonder how come I never faced them in my real life scenario. Well, let us see one of the similar situation. Here is one of the questions which I received on my social media handle. “Pinal, I have a large database. I did not develop this database but I have inherited this database. In our database we have many tables but all the tables are in pairs. We have one archive table and one current table. Now here is interesting situation. For a while due to some reason our organization has stopped paying attention to archive data. We did not archive anything for a while. If this was not enough we  even changed the schema of current table but did not change the corresponding archive table. This is now becoming a huge huge problem. We know for sure that in current table we have added few column but we do not know which ones. Is there any way we can figure out what are the new column added in the current table and does not exist in the archive tables? We cannot use any third party tool. Would you please guide us?” Well here is the interesting example of how we can use sys.column catalogue views and get the details of the newly added column. I have previously written about EXCEPT over here which is very similar to MINUS of Oracle. In following example we are going to create two tables. One of the tables has extra column. In our resultset we will get the name of the extra column as we are comparing the catalogue view of the column name. USE AdventureWorks2012 GO CREATE TABLE ArchiveTable (ID INT, Col1 VARCHAR(10), Col2 VARCHAR(100), Col3 VARCHAR(100)); CREATE TABLE CurrentTable (ID INT, Col1 VARCHAR(10), Col2 VARCHAR(100), Col3 VARCHAR(100), ExtraCol INT); GO -- Columns in ArchiveTable but not in CurrentTable SELECT name ColumnName FROM sys.columns WHERE OBJECT_NAME(OBJECT_ID) = 'ArchiveTable' EXCEPT SELECT name ColumnName FROM sys.columns WHERE OBJECT_NAME(OBJECT_ID) = 'CurrentTable' GO -- Columns in CurrentTable but not in ArchiveTable SELECT name ColumnName FROM sys.columns WHERE OBJECT_NAME(OBJECT_ID) = 'CurrentTable' EXCEPT SELECT name ColumnName FROM sys.columns WHERE OBJECT_NAME(OBJECT_ID) = 'ArchiveTable' GO DROP TABLE ArchiveTable; DROP TABLE CurrentTable; GO The above query will return us following result. I hope this solves the problems. It is not the most elegant solution ever possible but it works. Here is the puzzle back to you – what native T-SQL solution would you have provided in this situation? Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL System Table, SQL Tips and Tricks, T SQL, Technology

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  • MySQL "ERROR 1005 (HY000): Can't create table 'foo.#sql-12c_4' (errno: 150)"

    - by Ankur Banerjee
    Hi, I was working on creating some tables in database foo, but every time I end up with errno 150 regarding the foreign key. Firstly, here's my code for creating tables: CREATE TABLE Clients ( client_id CHAR(10) NOT NULL , client_name CHAR(50) NOT NULL , provisional_license_num CHAR(50) NOT NULL , client_address CHAR(50) NULL , client_city CHAR(50) NULL , client_county CHAR(50) NULL , client_zip CHAR(10) NULL , client_phone INT NULL , client_email CHAR(255) NULL , client_dob DATETIME NULL , test_attempts INT NULL ); CREATE TABLE Applications ( application_id CHAR(10) NOT NULL , office_id INT NOT NULL , client_id CHAR(10) NOT NULL , instructor_id CHAR(10) NOT NULL , car_id CHAR(10) NOT NULL , application_date DATETIME NULL ); CREATE TABLE Instructors ( instructor_id CHAR(10) NOT NULL , office_id INT NOT NULL , instructor_name CHAR(50) NOT NULL , instructor_address CHAR(50) NULL , instructor_city CHAR(50) NULL , instructor_county CHAR(50) NULL , instructor_zip CHAR(10) NULL , instructor_phone INT NULL , instructor_email CHAR(255) NULL , instructor_dob DATETIME NULL , lessons_given INT NULL ); CREATE TABLE Cars ( car_id CHAR(10) NOT NULL , office_id INT NOT NULL , engine_serial_num CHAR(10) NULL , registration_num CHAR(10) NULL , car_make CHAR(50) NULL , car_model CHAR(50) NULL ); CREATE TABLE Offices ( office_id INT NOT NULL , office_address CHAR(50) NULL , office_city CHAR(50) NULL , office_County CHAR(50) NULL , office_zip CHAR(10) NULL , office_phone INT NULL , office_email CHAR(255) NULL ); CREATE TABLE Lessons ( lesson_num INT NOT NULL , client_id CHAR(10) NOT NULL , date DATETIME NOT NULL , time DATETIME NOT NULL , milegage_used DECIMAL(5, 2) NULL , progress CHAR(50) NULL ); CREATE TABLE DrivingTests ( test_num INT NOT NULL , client_id CHAR(10) NOT NULL , test_date DATETIME NOT NULL , seat_num INT NOT NULL , score INT NULL , test_notes CHAR(255) NULL ); ALTER TABLE Clients ADD PRIMARY KEY (client_id); ALTER TABLE Applications ADD PRIMARY KEY (application_id); ALTER TABLE Instructors ADD PRIMARY KEY (instructor_id); ALTER TABLE Offices ADD PRIMARY KEY (office_id); ALTER TABLE Lessons ADD PRIMARY KEY (lesson_num); ALTER TABLE DrivingTests ADD PRIMARY KEY (test_num); ALTER TABLE Applications ADD CONSTRAINT FK_Applications_Offices FOREIGN KEY (office_id) REFERENCES Offices (office_id); ALTER TABLE Applications ADD CONSTRAINT FK_Applications_Clients FOREIGN KEY (client_id) REFERENCES Clients (client_id); ALTER TABLE Applications ADD CONSTRAINT FK_Applications_Instructors FOREIGN KEY (instructor_id) REFERENCES Instructors (instructor_id); ALTER TABLE Applications ADD CONSTRAINT FK_Applications_Cars FOREIGN KEY (car_id) REFERENCES Cars (car_id); ALTER TABLE Lessons ADD CONSTRAINT FK_Lessons_Clients FOREIGN KEY (client_id) REFERENCES Clients (client_id); ALTER TABLE Cars ADD CONSTRAINT FK_Cars_Offices FOREIGN KEY (office_id) REFERENCES Offices (office_id); ALTER TABLE Clients ADD CONSTRAINT FK_DrivingTests_Clients FOREIGN KEY (client_id) REFERENCES Clients (client_id); These are the errors that I get: mysql> ALTER TABLE Applications ADD CONSTRAINT FK_Applications_Cars FOREIGN KEY (car_id) REFERENCES Cars (car_id); ERROR 1005 (HY000): Can't create table 'foo.#sql-12c_4' (errno: 150) I ran SHOW ENGINE INNODB STATUS which gives a more detailed error description: ------------------------ LATEST FOREIGN KEY ERROR ------------------------ 100509 20:59:49 Error in foreign key constraint of table practice9/#sql-12c_4: FOREIGN KEY (car_id) REFERENCES Cars (car_id): Cannot find an index in the referenced table where the referenced columns appear as the first columns, or column types in the table and the referenced table do not match for constraint. Note that the internal storage type of ENUM and SET changed in tables created with >= InnoDB-4.1.12, and such columns in old tables cannot be referenced by such columns in new tables. See http://dev.mysql.com/doc/refman/5.1/en/innodb-foreign-key-constraints.html for correct foreign key definition. ------------ I searched around on StackOverflow and elsewhere online - came across a helpful blog post here with pointers on how to resolve this error - but I can't figure out what's going wrong. Any help would be appreciated!

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  • CSS: Labels in table columns

    - by hello
    Hello. BACKGROUND: I would like to have small labels in columns of a table. I'm using some implemented parts of HTML5/CSS3 in my project, and this section specifically is for mobile devices. While both facts are not necessarily relevant, the bottom line is that I don't have to support Internet Explorer or even Firefox for that matter (just WebKit). THE PROBLEM With my current CSS approach, the vertical padding of the cell comes from the <span element (set to display: block with top/bottom margins), which contains the "value" of the column. As a result there's no padding when the <span> is empty or missing (no value) and the label is not in place. The "full" coulmns should give you the idea of where I want the labels to be, even if there's no value, and the <span> is not there. I realize that I could use "non-breaking-space", but I would really like to avoid it. I wonder if any of you have a fix / better way to do this? current code is below. Thank you for any help. <!DOCTYPE html> <html lang="en"> <head> <title>ah</title> <style> body { width: 320px; } /* TABLE */ table { width: 100%; border-collapse: collapse; font-family: arial; } th, td { border: 1px solid #ccc; border-width: 0px 0px 1px 1px; } th:last-child, td:last-child { border-right-width: 1px; } tr:first-child th { border-top-width: 1px; background: #efefef; } /* RELEVANT STUFF */ td { padding: 3px; } td sup { display: block; } td span { display: block; margin: 3px 0px; text-align: center; } </style> </head> <body> <table> <tr> <th colspan="3">something</th> </tr> <tr> <td><sup>some label</sup><span>any content</span></td> <td><sup>some label</sup><span>any content</span></td> <td><sup>some label</sup><span></span></td><!-- No content, just a label --> </tr> </table> </body> </html>

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  • PHP: table structure

    - by A3efan
    I'm developing a website that has some audio courses, each course can have multiple lessons. I want to display each course in its own table with its different lessons. This is my sql statement: Table: courses id, title Table: lessons id, cid (course id), title, date, file $sql = "SELECT lessons.*, courses.title AS course FROM lessons INNER JOIN courses ON courses.id = lessons.cid GROUP BY lessons.id ORDER BY lessons.id" ; Can someone help me with the PHP code? This is the I code I have written: mysql_select_db($database_config, $config); mysql_query("set names utf8"); $sql = "SELECT lessons.*, courses.title AS course FROM lessons INNER JOIN courses ON courses.id = lessons.cid GROUP BY lessons.id ORDER BY lessons.id" ; $result = mysql_query($sql) or die(mysql_error()); while ($row = mysql_fetch_assoc($result)) { echo "<p><span class='heading1'>" . $row['course'] . "</span> </p> "; echo "<p class='datum'>Posted onder <a href='*'>*</a>, latest update on " . strftime("%A %d %B %Y %H:%M", strtotime($row['date'])); } echo "</p>"; echo "<class id='text'>"; echo "<p>...</p>"; echo "<table border: none cellpadding='1' cellspacing='1'>"; echo "<tr>"; echo "<th>Nr.</th>"; echo "<th width='450'>Lesso</th>"; echo "<th>Date</th>"; echo "<th>Download</th>"; echo "</tr>"; echo "<tr>"; echo "<td>" . $row['nr'] . "</td>"; echo "<td>" . $row['title'] . "</td>"; echo "<td>" . strftime("%d/%m/%Y", strtotime($row['date'])) . "</td>"; echo "<td><a href='audio/" . rawurlencode($row['file']) . "'>MP3</a></td>"; echo "</tr>"; echo "</table>"; echo "<br>"; } ?>

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  • CSS menu broken in Firefox (display:table-cell;)

    - by roman m
    HTML: <td align="center" width="100%"> <a class="Forum_ib_moderate" href="Default.aspx" title="Moderate"></a> <a class="Forum_ib_admin" href="Default.aspx" title="Admin"></a> ... CSS: A.Forum_ib_moderate:link, A.Forum_ib_moderate:visited, A.Forum_ib_moderate:active, A.Forum_ib_moderate:hover { background-image: url(images/ib_moderate.png); background-repeat: no-repeat; background-position: center; padding-left: 2px; padding-right: 2px; padding-top: 8px; padding-bottom: 0px; height: 35px; width: 35px; display:table-cell; } A.Forum_ib_admin:hover { background-image: url(images/ib_admin_hover.png); } the menu looks just fine in IE, shows up vertical in Firefox. If i turn off "display:table-cell;" style in Firebug and then turn it back on, it fixes that menu node. any ideas? p.s.: i don't want to mess with the menu itself, since it's a part of a DNN Forum 4.4.3. I'd rather fix the CSS to make it show correctly.

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