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  • Can I delete EFI System Partition without harming other data on drive?

    - by Andy
    I have three external HDD's in a USB enclosure. After a recent upgrade to Win7, during which these three drives were actually installed inside the PC tower, two of the three drives now have a 200MB EFI partition, and the two drives do not show up as usable drives under either Win7 or Snow Leopard. One of the drives is empty; the other one, however, has a bunch of stuff on it that I want to save if possible. My question is, how can I get back to this data? Can I simply delete the EFI partition, and all will be well? Or do I have to do something trickier? Or am I just hosed?

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  • Partition and mount my secondary hard drive on CentOS 5.5 64bit?

    - by Andrew Fashion
    I am trying to prepare my second hard drive for user image uploads. Here is the current layout: # sudo parted /dev/sda print Model: ATA WDC WD2500KS-00M (scsi) Disk /dev/sda: 250GB Sector size (logical/physical): 512B/512B Partition Table: msdos Number Start End Size Type File system Flags 1 32.3kB 107MB 107MB primary ext3 boot 2 107MB 8595MB 8488MB primary linux-swap 3 8595MB 10.7GB 2147MB primary ext3 4 10.7GB 250GB 239GB extended 5 10.7GB 250GB 239GB logical ext3 Information: Don't forget to update /etc/fstab, if necessary. I am assuming #4 is my secondary drive? How do I partition and mount it so I can begin using it? And how do I add to fstab? I understand if it's to many questions in one, just help me with whatever you can I guess :) Thank you for any help!

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  • SQL update table from another table

    - by LtDan
    Using SQL in Access, trying to "Update" a table, with the user name, from another table. The 3rd line below (SQLnm2...) says error-2465 cant find field '|'. I've tried changing the expression many ways but no success. Any assistance would be greatly appreciated. Dim SQLnm As String Dim SQLnm2 As String SQLnm2 = SQLnm2 & "', '" & [Employees]![NBK] & "');" SQLnm = " Update tbl_DateTracking SET NBK = " SQLnm = SQLnm & "'" & SQLnm2 & "' WHERE " SQLnm = SQLnm & "CaseId = '" & CaseId & "' AND OCC_Scenario = '" & OCC_Scenario & "';" DoCmd.RunSQL SQLnm

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  • Drop down table and jquery

    - by Marcelo
    Hi, I'm trying to make a drop down table using jQuery, with a similar code like here: (from the topic: Conditional simple drop down list?) <body> <div id="myQuestions"> <select id="QuestionOptions"> <option value="A">Question A</option> <option value="B">Question B</option> </select> </div> <div id="myAnswers"> <div id="A" style="display: none;"> <div id="QuestionC"> <p>Here is an example question C.</p> </div> <div id="QuestionD"> <select id="QuestionOptionsD"> <option value="G">Question G</option> <option value="H">Question H</option> </select> </div> </div> <div id="B" style="display: none;"> <div id="QuestionE"> <p>Here is an example question E.</p> </div> <div id="QuestionF"> <select id="QuestionOptionsF"> <option value="I">Question I</option> <option value="J">Question J</option> </select> </div> </div> </div> And the jQuery part $(function () { $('#QuestionOptions').change(function () { $('#myAnswers > div').hide(); $('#myAnswers').find('#' + $(this).val()).show(); }); }); My problem is, when I finish to table the part of "myQuestions", and start to table the part of "myAnswers", the dynamic part of the table doesn't work. In this case, the myAnswers part won't be hidden, it'll be shown since the beginning. I tried to put everything in one table, then I tried to create a different table for myQuestions, then another table for myAnswers and it didn't work. Does anyone know where am I mistaking ? Sorry for any mistake in English, I'm not a native speaker. Thanks in advance.

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  • Javascript table construction bug using JQuery in Firefox on Mac OSX

    - by Poita_
    I'm using some code to build up tables using JQuery, but in Firefox 3.5.3 on Mac OSX, the table cells all appear on separate lines by themselves, instead of in their respective rows. Chrome 5.0.342.7 beta on OSX correctly produces the table, as does Safari 4.0.5. Here is a minimal reproduction case: <html> <body> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script> <script type="text/javascript"> $(document).ready(function() { var b = $('body'); b.append("<table>"); for (var i = 0; i < 3; ++i) { b.append("<tr>"); for (var j = 0; j < 3; ++j) b.append("<td>x</td>"); b.append("</tr>"); } b.append("</table>"); }); </script> </body> </html> In Chrome and Safari, I get this correct output: x x x x x x x x x but Firefox produces: x x x x x x x x x Note that if I manually create that exact table without using Javascript (i.e. direct into the HTML) then the table appears correctly in Firefox. Also, if I change the JS to append then entire table in one call then it also works -- the only time it doesn't work is if you append it part-by-part as I have done before. My question is: is this to be expected, or should I report this as a bug to Firefox? I'm pretty sure this is a Firefox bug, but I'm a bit of a newbie to JS and web development in general, so perhaps there's something I'm missing? P.S. obviously there are easy ways to get around this -- that's not my concern. See above.

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  • Make a table start on the same line as header

    - by ripper234
    I am trying to get a table of icons appear on the same line as the header. In the HTML below, the icons appear on a separate line. I tried using 'top' attribute to move the table, but this is not a good solution because then there's an ugly space between the icons table and the rest of the document. How can I fix this? <html> <head> <style type="text/css"> #action-icons { float:right; position:relative; border:0; } </style> </head> <body> <h1 class="edit">Bla bla</h1> <table id="action-icons"> <tbody> <tr> <td><img width="64" height="64"/></td> <td><img width="60" height="60"/></td> </tr> <tr> <td><img width="36" height="36"/></td> <td><img width="36" height="36"/></td> </tr> </tbody> </table> <table width="100%" class="tasksgrid"> <tbody> <tr> <th class='taskcell'>One</th> <th class='taskcell'>Two</th> </tr> </tbody> </table> </body> </html>

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  • Create table with PHP and populate from MySQL

    - by typoknig
    Hi all, I am creating a table to display on a web page and that table is populated from data in a MySQL database. I am trying to do a couple of things that are making it difficult for me. First I am trying to have call the PHP code that exists in a separate file in HTML via JavaScript. I think I have that working right but I am not 100% sure (because the table will not display). I think it is working right because some of the code for the table (which is in the PHP file) displays in FireBug. Second I am trying to make it so the rows alternate colors for easy viewing too. My PHP code so far is below. The table does not display at all in any browser. $query = "SELECT * FROM employees"; $result = mysql_query($query); $num = mysql_num_rows($result); echo '<table>'; for ($i = 0; $i < $num; $i++){ $row = mysql_fetch_array($result); $id = $row['id']; $l_name = $row['l_name']; $f_name = $row['f_name']; $ssn = $row['ssn']; $class = (($i % 2) == 0) ? "table_odd_row" : "table_even_row"; echo "<tr>"; echo "<td class=" . $class . ">$wrap_id</td>"; echo "<td class=" . $class . ">$wrap_l_name</td>"; echo "<td class=" . $class . ">$wrap_f_name</td>"; echo "<td class=" . $class . ">$wrap_ssn</td>"; echo "</tr>"; } echo '</table>'; mysql_close($link); }

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  • Join and sum not compatible matrices through data.table

    - by leodido
    My goal is to "sum" two not compatible matrices (matrices with different dimensions) using (and preserving) row and column names. I've figured this approach: convert the matrices to data.table objects, join them and then sum columns vectors. An example: > M1 1 3 4 5 7 8 1 0 0 1 0 0 0 3 0 0 0 0 0 0 4 1 0 0 0 0 0 5 0 0 0 0 0 0 7 0 0 0 0 1 0 8 0 0 0 0 0 0 > M2 1 3 4 5 8 1 0 0 1 0 0 3 0 0 0 0 0 4 1 0 0 0 0 5 0 0 0 0 0 8 0 0 0 0 0 > M1 %ms% M2 1 3 4 5 7 8 1 0 0 2 0 0 0 3 0 0 0 0 0 0 4 2 0 0 0 0 0 5 0 0 0 0 0 0 7 0 0 0 0 1 0 8 0 0 0 0 0 0 This is my code: M1 <- matrix(c(0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0), byrow = TRUE, ncol = 6) colnames(M1) <- c(1,3,4,5,7,8) M2 <- matrix(c(0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0), byrow = TRUE, ncol = 5) colnames(M2) <- c(1,3,4,5,8) # to data.table objects DT1 <- data.table(M1, keep.rownames = TRUE, key = "rn") DT2 <- data.table(M2, keep.rownames = TRUE, key = "rn") # join and sum of common columns if (nrow(DT1) > nrow(DT2)) { A <- DT2[DT1, roll = TRUE] A[, list(X1 = X1 + X1.1, X3 = X3 + X3.1, X4 = X4 + X4.1, X5 = X5 + X5.1, X7, X8 = X8 + X8.1), by = rn] } That outputs: rn X1 X3 X4 X5 X7 X8 1: 1 0 0 2 0 0 0 2: 3 0 0 0 0 0 0 3: 4 2 0 0 0 0 0 4: 5 0 0 0 0 0 0 5: 7 0 0 0 0 1 0 6: 8 0 0 0 0 0 0 Then I can convert back this data.table to a matrix and fix row and column names. The questions are: how to generalize this procedure? I need a way to automatically create list(X1 = X1 + X1.1, X3 = X3 + X3.1, X4 = X4 + X4.1, X5 = X5 + X5.1, X7, X8 = X8 + X8.1) because i want to apply this function to matrices which dimensions (and row/columns names) are not known in advance. In summary I need a merge procedure that behaves as described. there are other strategies/implementations that achieve the same goal that are, at the same time, faster and generalized? (hoping that some data.table monster help me) to what kind of join (inner, outer, etc. etc.) is assimilable this procedure? Thanks in advance. p.s.: I'm using data.table version 1.8.2 EDIT - SOLUTIONS @Aaron solution. No external libraries, only base R. It works also on list of matrices. add_matrices_1 <- function(...) { a <- list(...) cols <- sort(unique(unlist(lapply(a, colnames)))) rows <- sort(unique(unlist(lapply(a, rownames)))) out <- array(0, dim = c(length(rows), length(cols)), dimnames = list(rows,cols)) for (m in a) out[rownames(m), colnames(m)] <- out[rownames(m), colnames(m)] + m out } @MadScone solution. Used reshape2 package. It works only on two matrices per call. add_matrices_2 <- function(m1, m2) { m <- acast(rbind(melt(M1), melt(M2)), Var1~Var2, fun.aggregate = sum) mn <- unique(colnames(m1), colnames(m2)) rownames(m) <- mn colnames(m) <- mn m } BENCHMARK (100 runs with microbenchmark package) Unit: microseconds expr min lq median uq max 1 add_matrices_1 196.009 257.5865 282.027 291.2735 549.397 2 add_matrices_2 13737.851 14697.9790 14864.778 16285.7650 25567.448 No need to comment the benchmark: @Aaron solution wins. I'll continue to investigate a similar solution for data.table objects. I'll add other solutions eventually reported or discovered.

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  • SQLite join selection from the same table using reference from another table

    - by daikini
    I have two tables: table: points |key_id | name | x | y | ------------------------ |1 | A |10 |20 | |2 | A_1 |11 |21 | |3 | B |30 |40 | |4 | B_1 |31 |42 | table: pairs |f_key_p1 | f_key_p2 | ---------------------- |1 | 2 | |3 | 4 | Table 'pairs' defines which rows in table 'points' should be paired. How can I query database to select paired rows? My desired query result would be like this: |name_1|x_1|x_2|name_2|x_2|y_2| ------------------------------- |A |10 |20 |A_1 |11 |21 | |B |30 |40 |B_1 |31 |41 |

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  • <asp:Table> Vs html <table>

    - by keith
    What are the pros and cons between using the ASP.Net control compared to the old reliable table html implementation. I know that the asp:Table will end up on the returned page as a html table, and from looking into it so far people are saying its easier to work with the asp:Table in the server side code, but I'd love to hear what the stackoverflow community has to say about the matter.

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  • Free space not reclaimed after online resizing ext4 in Ubuntu 9.10

    - by TiansHUo
    My root partition was filling up, with only 500 mbs left, I wanted to resize my root partition from 20 Gb to 40Gb So I resized my partition by using these steps: Using Gparted to resize another partition to give space for the EXT4 Using fdisk, deleting the root partition (on /dev/sda2), and creating it again using the new size resize2fs /dev/sda2 Updating grub2 But now the problem is that although I can boot in my new partition and the new partition shows it is 40Gb, but the free size was still 500mb. So I booted from a LiveCD and checked with e2fsck -p /dev/sda2, it reported clean. So I added the -f flag (force check), still, the drive is full.

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  • Cannot resize OS X partition

    - by David Pearce
    I am trying to resize my existing Mac OS Extended partition on my Macbook to install Windows 7 (using steps similar to these), but when ever I go to apply the changes, I get this error: Partition failed Partition failed with the error: The partition cannot be resized. Try reducing the amount of change in the size of the partition. The total capacity of the hard drive in question is 260GB, with the entirety being taken up by the OS X boot partition. There is I am aiming to shrink that partition down to 60GB. How can I fix this problem? I have been reducing the amount of change by 10GB each attempt, but it still is not working. I assume the problem is that there is not a large amount of continuous space on the device. Is there some way to can do a manual defrag that would rectify this problem?

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  • Cannot resize OS X partition

    - by joshhunt
    I am trying to resize my existing Mac OS Extended partition on my Macbook to install Windows 7 (using steps similar to these), but when ever I go to apply the changes, I get this error: Partition failed Partition failed with the error: The partition cannot be resized. Try reducing the amount of change in the size of the partition. The total capacity of the hard drive in question is 260GB, with the entirety being taken up by the OS X boot partition. There is I am aiming to shrink that partition down to 60GB. How can I fix this problem? I have been reducing the amount of change by 10GB each attempt, but it still is not working. I assume the problem is that there is not a large amount of continuous space on the device. Is there some way to can do a manual defrag that would rectify this problem?

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  • Partitioning a bootable Flash drive

    - by mmc
    Is it possible to have a 2 partition Flash drive that looks like the following: A partition that is bootable to OS X (this will require a GUID partition table) A second partition formatted either FAT32 or NTFS that is readable on both OS X and various flavors of Windows I have set up a disk using Disk Utility on the Mac, and it boots fine with a second FAT32 partition... but Windows does not see it. Any flavor of Windows wants to format the entire drive. Has anyone done this, and if so, can you explain the steps you followed? EDIT: Making it bootable is no problem. I have that. I'm wondering how to make the second partition on a Flash drive visible to Windows. It's possible that the "second partition" is the problem, and I need Windows to be first, and HFS to be second. I'll try that tonight.

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  • Can a virus on a windows 7 partition make its way into the OS X partition?

    - by hatorade
    I have a Windows 7 partition on my MBP that I installed with Boot Camp. I have reason to believe that there was a virus on my Windows 7 partition (did some scans, got some sketchy results from Avira). I decided to just wipe the entire partition using Boot Camp Restore to reformat the old partition and add it back to my OS X partition. I'm wondering however if in the time period I had the two partitions up a virus could have jumped from the Windows 7 partition onto the OS X partition, in which case I now need to worry about a virus on my OS X installation?

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  • HELP! Free space not reclaimed after online resizing ext4 in Ubuntu 9.10

    - by TiansHUo
    My root partition was filling up, with only 500 mbs left, I wanted to resize my root partition from 20 Gb to 40Gb So I resized my partition by using these steps: Using Gparted to resize another partition to give space for the EXT4 Using fdisk, deleting the root partition (on /dev/sda2), and creating it again using the new size resize2fs /dev/sda2 Updating grub2 But now the problem is that although I can boot in my new partition and the new partition shows it is 40Gb, but the free size was still 500mb. So I booted from a LiveCD and checked with e2fsck -p /dev/sda2, it reported clean. So I added the -f flag (force check), still, the drive is full.

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  • Create new partition on ssd hard (remove hibernate)

    - by Aleks
    I bought Dell Vostro 3360 notebook with Windows 7 It has 128 GB SSD hard disk, it is already has 4 partitions: Dell partition, Recovery partition, OS partition, Hibernate partition. Here is screenshot (Russian language, but I made some marks): I want to split OS partition, because I need c:\ and d:\ . Reason is that I have a lot of difficulties with administration mode on c:. So I tried to split OS partition but I already have 4 partitions. Can I remove hibernate partition without consequences? I have Hibernate disabled in settings, but I have hiberfil.sys file on c:\ If I can remove it, how can I do this, I can't do this with standard GUI disk managment tool

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  • Combine OS partion with data partition on NAS4Free/FreeNAS

    - by Pak
    I recently built a NAS4Free (formerly FreeNAS) machine using a 256MB (yes, MB) USB drive for the OS. When I did the original install, I had the bright idea of making the OS partition just big enough for the OS and a then creating a second partition using the remainder of the drive to store stuff pertaining to the OS. I never really found a use for the data partition and I ended up running out of space on the OS partition, so now I'd like to combine the partitions into a single partition. Is this something that is possible to do while everything is up and running? If it comes down to it, I can take down the machine and do a fresh install of the OS using the entire space of the USB drive, but I'd like to use this as an opportunity to better familiarize myself with FreeBSD/UNIX type systems. If this is possible, will it interfere with the NAS4Free things? The data partition shows up in the web interface under the disks section. If I end up manually changing the partitions, I'd be concerned with NAS4Free getting confused by the missing partition.

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  • Partition Wise Joins

    - by jean-pierre.dijcks
    Some say they are the holy grail of parallel computing and PWJ is the basis for a shared nothing system and the only join method that is available on a shared nothing system (yes this is oversimplified!). The magic in Oracle is of course that is one of many ways to join data. And yes, this is the old flexibility vs. simplicity discussion all over, so I won't go there... the point is that what you must do in a shared nothing system, you can do in Oracle with the same speed and methods. The Theory A partition wise join is a join between (for simplicity) two tables that are partitioned on the same column with the same partitioning scheme. In shared nothing this is effectively hard partitioning locating data on a specific node / storage combo. In Oracle is is logical partitioning. If you now join the two tables on that partitioned column you can break up the join in smaller joins exactly along the partitions in the data. Since they are partitioned (grouped) into the same buckets, all values required to do the join live in the equivalent bucket on either sides. No need to talk to anyone else, no need to redistribute data to anyone else... in short, the optimal join method for parallel processing of two large data sets. PWJ's in Oracle Since we do not hard partition the data across nodes in Oracle we use the Partitioning option to the database to create the buckets, then set the Degree of Parallelism (or run Auto DOP - see here) and get our PWJs. The main questions always asked are: How many partitions should I create? What should my DOP be? In a shared nothing system the answer is of course, as many partitions as there are nodes which will be your DOP. In Oracle we do want you to look at the workload and concurrency, and once you know that to understand the following rules of thumb. Within Oracle we have more ways of joining of data, so it is important to understand some of the PWJ ideas and what it means if you have an uneven distribution across processes. Assume we have a simple scenario where we partition the data on a hash key resulting in 4 hash partitions (H1 -H4). We have 2 parallel processes that have been tasked with reading these partitions (P1 - P2). The work is evenly divided assuming the partitions are the same size and we can scan this in time t1 as shown below. Now assume that we have changed the system and have a 5th partition but still have our 2 workers P1 and P2. The time it takes is actually 50% more assuming the 5th partition has the same size as the original H1 - H4 partitions. In other words to scan these 5 partitions, the time t2 it takes is not 1/5th more expensive, it is a lot more expensive and some other join plans may now start to look exciting to the optimizer. Just to post the disclaimer, it is not as simple as I state it here, but you get the idea on how much more expensive this plan may now look... Based on this little example there are a few rules of thumb to follow to get the partition wise joins. First, choose a DOP that is a factor of two (2). So always choose something like 2, 4, 8, 16, 32 and so on... Second, choose a number of partitions that is larger or equal to 2* DOP. Third, make sure the number of partitions is divisible through 2 without orphans. This is also known as an even number... Fourth, choose a stable partition count strategy, which is typically hash, which can be a sub partitioning strategy rather than the main strategy (range - hash is a popular one). Fifth, make sure you do this on the join key between the two large tables you want to join (and this should be the obvious one...). Translating this into an example: DOP = 8 (determined based on concurrency or by using Auto DOP with a cap due to concurrency) says that the number of partitions >= 16. Number of hash (sub) partitions = 32, which gives each process four partitions to work on. This number is somewhat arbitrary and depends on your data and system. In this case my main reasoning is that if you get more room on the box you can easily move the DOP for the query to 16 without repartitioning... and of course it makes for no leftovers on the table... And yes, we recommend up-to-date statistics. And before you start complaining, do read this post on a cool way to do stats in 11.

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  • Back up a single table in SQL Server

    - by BuckWoody
    SQL Server doesn’t have an easy way to take a table backup, so I often use the bcp (Bulk Copy Program) to accomplish the same goal. I’ve mentioned this before, and someone told me when they tried it they couldn’t restore the table – ah the dangers of telling people half the information! I should have mentioned that you need to have a “format file” ready if the table does not exist at the destination. In my case I already had the table, in this person’s case they did not. The format file can be used to rebuild that table structure before the data is bcp’d in, and you can read more about it here: http://msdn.microsoft.com/en-us/library/ms191516.aspx There’s another way to back up a table, and that’s to create a Filegroup and place the table there. Then you can take a Filegroup backup to back up a single table. Of course, there are other methods of moving a single table’s data in an out, including SQL Server Integration Services and even the older Data Transformation Services, or simply by using hte SQLCMD or PowerShell utilities to run a query and just save the output to a file. In fact, these days I’m using a PowerShell script to build INSERT statements from that query. That could also easily be modified to create the table structure (or modify one if needed) quite easily. Share this post: email it! | bookmark it! | digg it! | reddit! | kick it! | live it!

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  • Adding Column to a SQL Server Table

    - by Dinesh Asanka
    Adding a column to a table is  common task for  DBAs. You can add a column to a table which is a nullable column or which has default values. But are these two operations are similar internally and which method is optimal? Let us start this with an example. I created a database and a table using following script: USE master Go --Drop Database if exists IF EXISTS (SELECT 1 FROM SYS.databases WHERE name = 'AddColumn') DROP DATABASE AddColumn --Create the database CREATE DATABASE AddColumn GO USE AddColumn GO --Drop the table if exists IF EXISTS ( SELECT 1 FROM sys.tables WHERE Name = 'ExistingTable') DROP TABLE ExistingTable GO --Create the table CREATE TABLE ExistingTable (ID BIGINT IDENTITY(1,1) PRIMARY KEY CLUSTERED, DateTime1 DATETIME DEFAULT GETDATE(), DateTime2 DATETIME DEFAULT GETDATE(), DateTime3 DATETIME DEFAULT GETDATE(), DateTime4 DATETIME DEFAULT GETDATE(), Gendar CHAR(1) DEFAULT 'M', STATUS1 CHAR(1) DEFAULT 'Y' ) GO -- Insert 100,000 records with defaults records INSERT INTO ExistingTable DEFAULT VALUES GO 100000 Before adding a Column Before adding a column let us look at some of the details of the database. DBCC IND (AddColumn,ExistingTable,1) By running the above query, you will see 637 pages for the created table. Adding a Column You can add a column to the table with following statement. ALTER TABLE ExistingTable Add NewColumn INT NULL Above will add a column with a null value for the existing records. Alternatively you could add a column with default values. ALTER TABLE ExistingTable Add NewColumn INT NOT NULL DEFAULT 1 The above statement will add a column with a 1 value to the existing records. In the below table I measured the performance difference between above two statements. Parameter Nullable Column Default Value CPU 31 702 Duration 129 ms 6653 ms Reads 38 116,397 Writes 6 1329 Row Count 0 100000 If you look at the RowCount parameter, you can clearly see the difference. Though column is added in the first case, none of the rows are affected while in the second case all the rows are updated. That is the reason, why it has taken more duration and CPU to add column with Default value. We can verify this by several methods. Number of Pages The number of data pages can be obtained by using DBCC IND command. Though, this an undocumented dbcc command, many experts are ok to use this command in production. However, since there is no official word from Microsoft, use this “at your own risk”. DBCC IND (AddColumn,ExistingTable,1) Before Adding the Columns 637 Adding a Column with NULL 637 Adding a column with DEFAULT value 1270 This clearly shows that pages are physically modified. Please note, a high value indicated in the Adding a column with DEFAULT value  column is also a result of page splits. Continues…

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  • How-to populate different select list content per table row

    - by frank.nimphius
    Normal 0 false false false EN-US X-NONE X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} A frequent requirement posted on the OTN forum is to render cells of a table column using instances of af:selectOneChoices with each af:selectOneChoice instance showing different list values. To implement this use case, the select list of the table column is populated dynamically from a managed bean for each row. The table's current rendered row object is accessible in the managed bean using the #{row} expression, where "row" is the value added to the table's var property. <af:table var="row">   ...   <af:column ...>     <af:selectOneChoice ...>         <f:selectItems value="#{browseBean.items}"/>     </af:selectOneChoice>   </af:column </af:table> The browseBean managed bean referenced in the code snippet above has a setItems and getItems method defined that is accessible from EL using the #{browseBean.items} expression. When the table renders, then the var property variable - the #{row} reference - is filled with the data object displayed in the current rendered table row. The managed bean getItems method returns a List<SelectItem>, which is the model format expected by the f:selectItems tag to populate the af:selectOneChoice list. public void setItems(ArrayList<SelectItem> items) {} //this method is executed for each table row public ArrayList<SelectItem> getItems() {   FacesContext fctx = FacesContext.getCurrentInstance();   ELContext elctx = fctx.getELContext();   ExpressionFactory efactory =          fctx.getApplication().getExpressionFactory();          ValueExpression ve =          efactory.createValueExpression(elctx, "#{row}", Object.class);      Row rw = (Row) ve.getValue(elctx);         //use one of the row attributes to determine which list to query and   //show in the current af:selectOneChoice list  // ...  ArrayList<SelectItem> alsi = new ArrayList<SelectItem>();  for( ... ){      SelectItem item = new SelectItem();        item.setLabel(...);        item.setValue(...);        alsi.add(item);   }   return alsi;} For better performance, the ADF Faces table stamps it data rows. Stamping means that the cell renderer component - af:selectOneChoice in this example - is instantiated once for the column and then repeatedly used to display the cell data for individual table rows. This however means that you cannot refresh a single select one choice component in a table to change its list values. Instead the whole table needs to be refreshed, rerunning the managed bean list query. Be aware that having individual list values per table row is an expensive operation that should be used only on small tables for Business Services with low latency data fetching (e.g. ADF Business Components and EJB) and with server side caching strategies for the queried data (e.g. storing queried list data in a managed bean in session scope).

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  • DB Schema for ACL involving 3 subdomains

    - by blacktie24
    Hi, I am trying to design a database schema for a web app which has 3 subdomains: a) internal employees b) clients c) contractors. The users will be able to communicate with each other to some degree, and there may be some resources that overlap between them. Any thoughts about this schema? Really appreciate your time and thoughts on this. Cheers! -- -- Table structure for table locations CREATE TABLE IF NOT EXISTS locations ( id bigint(20) NOT NULL, name varchar(250) NOT NULL ) ENGINE=InnoDB DEFAULT CHARSET=latin1; -- -- Table structure for table privileges CREATE TABLE IF NOT EXISTS privileges ( id int(11) NOT NULL AUTO_INCREMENT, name varchar(255) NOT NULL, resource_id int(11) NOT NULL, PRIMARY KEY (id) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=10 ; -- -- Table structure for table resources CREATE TABLE IF NOT EXISTS resources ( id int(11) NOT NULL AUTO_INCREMENT, name varchar(255) NOT NULL, user_type enum('internal','client','expert') NOT NULL, PRIMARY KEY (id) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ; -- -- Table structure for table roles CREATE TABLE IF NOT EXISTS roles ( id int(11) NOT NULL AUTO_INCREMENT, name varchar(255) NOT NULL, type enum('position','department') NOT NULL, parent_id int(11) DEFAULT NULL, user_type enum('internal','client','expert') NOT NULL, PRIMARY KEY (id) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ; -- -- Table structure for table role_perms CREATE TABLE IF NOT EXISTS role_perms ( id int(11) NOT NULL AUTO_INCREMENT, role_id int(11) NOT NULL, privilege_id int(11) NOT NULL, mode varchar(250) NOT NULL, PRIMARY KEY (id) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ; -- -- Table structure for table users CREATE TABLE IF NOT EXISTS users ( id int(10) unsigned NOT NULL AUTO_INCREMENT, email varchar(255) NOT NULL, password varchar(255) NOT NULL, salt varchar(255) NOT NULL, type enum('internal','client','expert') NOT NULL, first_name varchar(255) NOT NULL, last_name varchar(255) NOT NULL, location_id int(11) NOT NULL, phone varchar(255) NOT NULL, status enum('active','inactive') NOT NULL DEFAULT 'active', PRIMARY KEY (id) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=4 ; -- -- Table structure for table user_perms CREATE TABLE IF NOT EXISTS user_perms ( id int(11) NOT NULL AUTO_INCREMENT, user_id int(11) NOT NULL, privilege_id int(11) NOT NULL, mode varchar(250) NOT NULL, PRIMARY KEY (id) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ; -- -- Table structure for table user_roles CREATE TABLE IF NOT EXISTS user_roles ( id int(11) NOT NULL, user_id int(11) NOT NULL, role_id int(11) NOT NULL ) ENGINE=InnoDB DEFAULT CHARSET=latin1;

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