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  • problem when trying to empty a stack in c

    - by frx08
    Hi all, (probably it's a stupid thing but) I have a problem with a stack implementation in C language, when I try to empty it, the function to empty the stack does an infinite loop.. the top of the stack is never null. where I commit an error? thanks bye! #include <stdio.h> #include <stdlib.h> typedef struct stack{ size_t a; struct stack *next; } stackPos; typedef stackPos *ptr; void push(ptr *top, size_t a){ ptr temp; temp = malloc(sizeof(stackPos)); temp->a = a; temp->next = *top; *top = temp; } void freeStack(ptr *top){ ptr temp = *top; while(*top!=NULL){ //the program does an infinite loop *top = temp->next; free(temp); } } int main(){ ptr top = NULL; push(&top, 4); push(&top, 8); //down here the problem freeStack(&top); return 0; }

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  • Compiling scipy on Windows 32-bit

    - by Sridhar Ratnakumar
    Has anyone tried compiling SciPy on Windows using numpy-1.3.0 that was built with the pre-built ATLAS libraries (atlas3.6.0_WinNT_P4SSE2.zip) linked in the installation document. I get the following linker error, and have no ideas as to how to fix this issue. $ python setup.py config --compiler=mingw32 build --compiler=mingw32 install --root=i [...] creating build\temp.win32-2.6\Release creating build\temp.win32-2.6\Release\scipy creating build\temp.win32-2.6\Release\scipy\integrate compile options: '-DNO_ATLAS_INFO=2 -I"C:\Documents and Settings\apy\Application Data\Python\Python26\site-packages\numpy\core\inc lude" -IC:\Python26\include -IC:\Python26\PC -c' gcc -mno-cygwin -O2 -Wall -Wstrict-prototypes -DNO_ATLAS_INFO=2 -I"C:\Documents and Settings\apy\Application Data\Python\Python26\ site-packages\numpy\core\include" -IC:\Python26\include -IC:\Python26\PC -c scipy\integrate\_odepackmo dule.c -o build\temp.win32-2.6\Release\scipy\integrate\_odepackmodule.o C:\MinGW\bin\g77.exe -g -Wall -mno-cygwin -g -Wall -mno-cygwin -shared build\temp.win32-2.6\Release\scipy\integrate\_odepackmodule .o -LC:\atlas3.6.0_WinNT_P4SSE2 -LC:\MinGW\lib -LC:\MinGW\lib\gcc\mingw32\3.4.5 -LC:\Python26\libs -LC:\Act ivePython32Python26\PCbuild -Lbuild\temp.win32-2.6 -lodepack -llinpack_lite -lmach -latlas -lcblas -lf77blas -llapack -lpython26 - lg2c -o build\lib.win32-2.6\scipy\integrate\_odepack.pyd C:\atlas3.6.0_WinNT_P4SSE2/libf77blas.a(ATL_F77wrap_daxpy.o):ATL_F77wrap_axpy.c:(.text+0x3c): undefined reference to `ATL _daxpy' C:\atlas3.6.0_WinNT_P4SSE2/libf77blas.a(ATL_F77wrap_dscal.o):ATL_F77wrap_scal.c:(.text+0x26): undefined reference to `ATL _dscal' C:\atlas3.6.0_WinNT_P4SSE2/libf77blas.a(ATL_F77wrap_dcopy.o):ATL_F77wrap_copy.c:(.text+0x3d): undefined reference to `ATL _dcopy' C:\atlas3.6.0_WinNT_P4SSE2/libf77blas.a(ATL_F77wrap_idamax.o):ATL_F77wrap_amax.c:(.text+0x1e): undefined reference to `AT L_idamax' C:\atlas3.6.0_WinNT_P4SSE2/libf77blas.a(ATL_F77wrap_ddot.o):ATL_F77wrap_dot.c:(.text+0x36): undefined reference to `ATL_d dot' collect2: ld returned 1 exit status error: Command "C:\MinGW\bin\g77.exe -g -Wall -mno-cygwin -g -Wall -mno-cygwin -shared build\temp.win32-2.6\Release\scipy\integrat e\_odepackmodule.o -LC:\atlas3.6.0_WinNT_P4SSE2 -LC:\MinGW\lib -LC:\MinGW\lib\gcc\mingw32\3.4.5 -LC:\Python 26\libs -LC:\Python26\PCbuild -Lbuild\temp.win32-2.6 -lodepack -llinpack_lite -lmach -latlas -lcblas -lf77blas -llap ack -lpython26 -lg2c -o build\lib.win32-2.6\scipy\integrate\_odepack.pyd" failed with exit status 1 Does anyone know what could have gone wrong here?

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  • Unreasonable errors in merge sort

    - by Alexxx
    i have the following errors - please help me to find the error: 9 IntelliSense: expected a '}' 70 4 it points on the end of the code - but there are no open { anywhere!! so why?? 8 IntelliSense: expected a ';' 57 1 it points on the { after the void main but why to put ; after the { of the void main?? Error 7 error C1075: end of file found before the left brace '{' at 70 1 points to the beginig of the code - why??? #include <stdio.h> #include <stdlib.h> void merge(int *a,int p,int q,int r) { int i=p,j=q+1,k=0; int* temp=(int*)calloc(r-p+1, sizeof(int)); while ((i<=q)&& (j<=r)) if(a[i]<a[j]) temp[k++]=a[i++]; else temp[k++]=a[j++]; while(j<=r) // if( i>q ) temp[k++]=a[j++]; while(i<=q) // j>r temp[k++]=a[i++]; for(i=p,k=0;i<=r;i++,k++) // copy temp[] to a[] a[i]=temp[k]; free(temp); } void merge_sort(int *a,int first, int last) { int middle; if(first < last) { middle=(first+last)/2; merge_sort(a,first,middle); merge_sort(a,middle+1,last); merge(a,first,middle,last); { } void main() { int a[] = {9, 7, 2, 3, 5, 4, 1, 8, 6, 10}; int i; merge_sort(a, 0, 9); for (i = 0; i < 10; i++) printf ("%d ", a[i]);

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  • isalpha(<mychar>) == true evaluates to false??

    - by Buttink
    string temp is equal to "ZERO:\t.WORD\t1" from my debugger. (the first line of my file) string temp = RemoveWhiteSpace(data); int i = 0; if ( temp.length() > 0 && isalpha(temp[0]) ) cout << "without true worked" << endl; if ( temp.length() > 0 && isalpha(temp[0]) == true ) cout << "with true worked" << endl; This is my code to check if first character of temp is a a-z,A-Z. The first if statement will evaluate to true and the 2nd to false. WHY?!?!?! I have tried this even without the "temp.length() 0 &&" and it still evaluates false. It just hates the "== true". The only thing I can think of is that isalpha() returns != 0 and true == 1. Then, you could get isalpha() == 2 != 1. But, I have no idea if C++ is that ... weird. BTW, I dont need to know that the "== true" is logically pointless. I know. output was without true worked Compiled with CodeBlock using GNU GCC on Ubuntu 9.10 (if this matters any)

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  • How to set incremental CSS classes in each Table Cell with jQuery?

    - by Mark Rapp
    I have a table populated via a DB and it renders like so (it could have any number of columns referring to "time", 5 columns, 8 columns, 2 columns, etc): <table id="eventInfo"> <tr> <td class="name">John</td> <td class="date">Dec 20</td> <td class="**time**">2pm</td> <td class="**time**">3pm</td> <td class="**time**">4pm</td> <td class="event">Birthday</td> </tr> <tr> <td class="name">Billy</td> <td class="date">Dec 19</td> <td class="**time**">6pm</td> <td class="**time**">7pm</td> <td class="**time**">8pm</td> <td class="event">Birthday</td> </tr> With jQuery, I'd like to go through each Table Row and incrementally set an additional class-name on only the Table Cells where "class='time'" so that the result would be: John Dec 20 2pm 3pm 4pm Birthday Billy Dec 19 6pm 7pm 8pm Birthday I've only been able to get it to count all of the Table Cells where "class='time'" and not each set within its own Table Row. This is what I've tried with jQuery: $(document).ready(function() { $("table#eventInfo tr").each(function() { var tcount = 0; $("td.time").attr("class", function() { return "timenum-" + tcount++; }) //writes out the results in each TD .each(function() { $("span", this).html("(class = '<b>" + this.className + "</b>')"); }); }); }); Unfortunately, this only results in: <table id="eventInfo"> <tr> <td class="name">John</td> <td class="date">Dec 20</td> <td class="**time** **timenum-1**">2pm</td> <td class="**time** **timenum-2**">3pm</td> <td class="**time** **timenum-3**">4pm</td> <td class="event">Birthday</td> </tr> <tr> <td class="name">Billy</td> <td class="date">Dec 19</td> <td class="**time** **timenum-4**">6pm</td> <td class="**time** **timenum-5**">7pm</td> <td class="**time** **timenum-6**">8pm</td> <td class="event">Birthday</td> </tr> Thanks for your help!

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  • BNF – how to read syntax?

    - by Piotr Rodak
    A few days ago I read post of Jen McCown (blog) about her idea of blogging about random articles from Books Online. I think this is a great idea, even if Jen says that it’s not exciting or sexy. I noticed that many of the questions that appear on forums and other media arise from pure fact that people asking questions didn’t bother to read and understand the manual – Books Online. Jen came up with a brilliant, concise acronym that describes very well the category of posts about Books Online – RTFM365. I take liberty of tagging this post with the same acronym. I often come across questions of type – ‘Hey, i am trying to create a table, but I am getting an error’. The error often says that the syntax is invalid. 1 CREATE TABLE dbo.Employees 2 (guid uniqueidentifier CONSTRAINT DEFAULT Guid_Default NEWSEQUENTIALID() ROWGUIDCOL, 3 Employee_Name varchar(60) 4 CONSTRAINT Guid_PK PRIMARY KEY (guid) ); 5 The answer is usually(1), ‘Ok, let me check it out.. Ah yes – you have to put name of the DEFAULT constraint before the type of constraint: 1 CREATE TABLE dbo.Employees 2 (guid uniqueidentifier CONSTRAINT Guid_Default DEFAULT NEWSEQUENTIALID() ROWGUIDCOL, 3 Employee_Name varchar(60) 4 CONSTRAINT Guid_PK PRIMARY KEY (guid) ); Why many people stumble on syntax errors? Is the syntax poorly documented? No, the issue is, that correct syntax of the CREATE TABLE statement is documented very well in Books Online and is.. intimidating. Many people can be taken aback by the rather complex block of code that describes all intricacies of the statement. However, I don’t know better way of defining syntax of the statement or command. The notation that is used to describe syntax in Books Online is a form of Backus-Naur notatiion, called BNF for short sometimes. This is a notation that was invented around 50 years ago, and some say that even earlier, around 400 BC – would you believe? Originally it was used to define syntax of, rather ancient now, ALGOL programming language (in 1950’s, not in ancient India). If you look closer at the definition of the BNF, it turns out that the principles of this syntax are pretty simple. Here are a few bullet points: italic_text is a placeholder for your identifier <italic_text_in_angle_brackets> is a definition which is described further. [everything in square brackets] is optional {everything in curly brackets} is obligatory everything | separated | by | operator is an alternative ::= “assigns” definition to an identifier Yes, it looks like these six simple points give you the key to understand even the most complicated syntax definitions in Books Online. Books Online contain an article about syntax conventions – have you ever read it? Let’s have a look at fragment of the CREATE TABLE statement: 1 CREATE TABLE 2 [ database_name . [ schema_name ] . | schema_name . ] table_name 3 ( { <column_definition> | <computed_column_definition> 4 | <column_set_definition> } 5 [ <table_constraint> ] [ ,...n ] ) 6 [ ON { partition_scheme_name ( partition_column_name ) | filegroup 7 | "default" } ] 8 [ { TEXTIMAGE_ON { filegroup | "default" } ] 9 [ FILESTREAM_ON { partition_scheme_name | filegroup 10 | "default" } ] 11 [ WITH ( <table_option> [ ,...n ] ) ] 12 [ ; ] Let’s look at line 2 of the above snippet: This line uses rules 3 and 5 from the list. So you know that you can create table which has specified one of the following. just name – table will be created in default user schema schema name and table name – table will be created in specified schema database name, schema name and table name – table will be created in specified database, in specified schema database name, .., table name – table will be created in specified database, in default schema of the user. Note that this single line of the notation describes each of the naming schemes in deterministic way. The ‘optionality’ of the schema_name element is nested within database_name.. section. You can use either database_name and optional schema name, or just schema name – this is specified by the pipe character ‘|’. The error that user gets with execution of the first script fragment in this post is as follows: Msg 156, Level 15, State 1, Line 2 Incorrect syntax near the keyword 'DEFAULT'. Ok, let’s have a look how to find out the correct syntax. Line number 3 of the BNF fragment above contains reference to <column_definition>. Since column_definition is in angle brackets, we know that this is a reference to notion described further in the code. And indeed, the very next fragment of BNF contains syntax of the column definition. 1 <column_definition> ::= 2 column_name <data_type> 3 [ FILESTREAM ] 4 [ COLLATE collation_name ] 5 [ NULL | NOT NULL ] 6 [ 7 [ CONSTRAINT constraint_name ] DEFAULT constant_expression ] 8 | [ IDENTITY [ ( seed ,increment ) ] [ NOT FOR REPLICATION ] 9 ] 10 [ ROWGUIDCOL ] [ <column_constraint> [ ...n ] ] 11 [ SPARSE ] Look at line 7 in the above fragment. It says, that the column can have a DEFAULT constraint which, if you want to name it, has to be prepended with [CONSTRAINT constraint_name] sequence. The name of the constraint is optional, but I strongly recommend you to make the effort of coming up with some meaningful name yourself. So the correct syntax of the CREATE TABLE statement from the beginning of the article is like this: 1 CREATE TABLE dbo.Employees 2 (guid uniqueidentifier CONSTRAINT Guid_Default DEFAULT NEWSEQUENTIALID() ROWGUIDCOL, 3 Employee_Name varchar(60) 4 CONSTRAINT Guid_PK PRIMARY KEY (guid) ); That is practically everything you should know about BNF. I encourage you to study the syntax definitions for various statements and commands in Books Online, you can find really interesting things hidden there. Technorati Tags: SQL Server,t-sql,BNF,syntax   (1) No, my answer usually is a question – ‘What error message? What does it say?’. You’d be surprised to know how many people think I can go through time and space and look at their screen at the moment they received the error.

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  • SQL SERVER – Shrinking Database is Bad – Increases Fragmentation – Reduces Performance

    - by pinaldave
    Earlier, I had written two articles related to Shrinking Database. I wrote about why Shrinking Database is not good. SQL SERVER – SHRINKDATABASE For Every Database in the SQL Server SQL SERVER – What the Business Says Is Not What the Business Wants I received many comments on Why Database Shrinking is bad. Today we will go over a very interesting example that I have created for the same. Here are the quick steps of the example. Create a test database Create two tables and populate with data Check the size of both the tables Size of database is very low Check the Fragmentation of one table Fragmentation will be very low Truncate another table Check the size of the table Check the fragmentation of the one table Fragmentation will be very low SHRINK Database Check the size of the table Check the fragmentation of the one table Fragmentation will be very HIGH REBUILD index on one table Check the size of the table Size of database is very HIGH Check the fragmentation of the one table Fragmentation will be very low Here is the script for the same. USE MASTER GO CREATE DATABASE ShrinkIsBed GO USE ShrinkIsBed GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Create FirstTable CREATE TABLE FirstTable (ID INT, FirstName VARCHAR(100), LastName VARCHAR(100), City VARCHAR(100)) GO -- Create Clustered Index on ID CREATE CLUSTERED INDEX [IX_FirstTable_ID] ON FirstTable ( [ID] ASC ) ON [PRIMARY] GO -- Create SecondTable CREATE TABLE SecondTable (ID INT, FirstName VARCHAR(100), LastName VARCHAR(100), City VARCHAR(100)) GO -- Create Clustered Index on ID CREATE CLUSTERED INDEX [IX_SecondTable_ID] ON SecondTable ( [ID] ASC ) ON [PRIMARY] GO -- Insert One Hundred Thousand Records INSERT INTO FirstTable (ID,FirstName,LastName,City) SELECT TOP 100000 ROW_NUMBER() OVER (ORDER BY a.name) RowID, 'Bob', CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%2 = 1 THEN 'Smith' ELSE 'Brown' END, CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 1 THEN 'New York' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 5 THEN 'San Marino' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 3 THEN 'Los Angeles' ELSE 'Houston' END FROM sys.all_objects a CROSS JOIN sys.all_objects b GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Insert One Hundred Thousand Records INSERT INTO SecondTable (ID,FirstName,LastName,City) SELECT TOP 100000 ROW_NUMBER() OVER (ORDER BY a.name) RowID, 'Bob', CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%2 = 1 THEN 'Smith' ELSE 'Brown' END, CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 1 THEN 'New York' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 5 THEN 'San Marino' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 3 THEN 'Los Angeles' ELSE 'Houston' END FROM sys.all_objects a CROSS JOIN sys.all_objects b GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Check Fragmentations in the database SELECT avg_fragmentation_in_percent, fragment_count FROM sys.dm_db_index_physical_stats (DB_ID(), OBJECT_ID('SecondTable'), NULL, NULL, 'LIMITED') GO Let us check the table size and fragmentation. Now let us TRUNCATE the table and check the size and Fragmentation. USE MASTER GO CREATE DATABASE ShrinkIsBed GO USE ShrinkIsBed GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Create FirstTable CREATE TABLE FirstTable (ID INT, FirstName VARCHAR(100), LastName VARCHAR(100), City VARCHAR(100)) GO -- Create Clustered Index on ID CREATE CLUSTERED INDEX [IX_FirstTable_ID] ON FirstTable ( [ID] ASC ) ON [PRIMARY] GO -- Create SecondTable CREATE TABLE SecondTable (ID INT, FirstName VARCHAR(100), LastName VARCHAR(100), City VARCHAR(100)) GO -- Create Clustered Index on ID CREATE CLUSTERED INDEX [IX_SecondTable_ID] ON SecondTable ( [ID] ASC ) ON [PRIMARY] GO -- Insert One Hundred Thousand Records INSERT INTO FirstTable (ID,FirstName,LastName,City) SELECT TOP 100000 ROW_NUMBER() OVER (ORDER BY a.name) RowID, 'Bob', CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%2 = 1 THEN 'Smith' ELSE 'Brown' END, CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 1 THEN 'New York' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 5 THEN 'San Marino' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 3 THEN 'Los Angeles' ELSE 'Houston' END FROM sys.all_objects a CROSS JOIN sys.all_objects b GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Insert One Hundred Thousand Records INSERT INTO SecondTable (ID,FirstName,LastName,City) SELECT TOP 100000 ROW_NUMBER() OVER (ORDER BY a.name) RowID, 'Bob', CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%2 = 1 THEN 'Smith' ELSE 'Brown' END, CASE WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 1 THEN 'New York' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 5 THEN 'San Marino' WHEN ROW_NUMBER() OVER (ORDER BY a.name)%10 = 3 THEN 'Los Angeles' ELSE 'Houston' END FROM sys.all_objects a CROSS JOIN sys.all_objects b GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Check Fragmentations in the database SELECT avg_fragmentation_in_percent, fragment_count FROM sys.dm_db_index_physical_stats (DB_ID(), OBJECT_ID('SecondTable'), NULL, NULL, 'LIMITED') GO You can clearly see that after TRUNCATE, the size of the database is not reduced and it is still the same as before TRUNCATE operation. After the Shrinking database operation, we were able to reduce the size of the database. If you notice the fragmentation, it is considerably high. The major problem with the Shrink operation is that it increases fragmentation of the database to very high value. Higher fragmentation reduces the performance of the database as reading from that particular table becomes very expensive. One of the ways to reduce the fragmentation is to rebuild index on the database. Let us rebuild the index and observe fragmentation and database size. -- Rebuild Index on FirstTable ALTER INDEX IX_SecondTable_ID ON SecondTable REBUILD GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Check Fragmentations in the database SELECT avg_fragmentation_in_percent, fragment_count FROM sys.dm_db_index_physical_stats (DB_ID(), OBJECT_ID('SecondTable'), NULL, NULL, 'LIMITED') GO You can notice that after rebuilding, Fragmentation reduces to a very low value (almost same to original value); however the database size increases way higher than the original. Before rebuilding, the size of the database was 5 MB, and after rebuilding, it is around 20 MB. Regular rebuilding the index is rebuild in the same user database where the index is placed. This usually increases the size of the database. Look at irony of the Shrinking database. One person shrinks the database to gain space (thinking it will help performance), which leads to increase in fragmentation (reducing performance). To reduce the fragmentation, one rebuilds index, which leads to size of the database to increase way more than the original size of the database (before shrinking). Well, by Shrinking, one did not gain what he was looking for usually. Rebuild indexing is not the best suggestion as that will create database grow again. I have always remembered the excellent post from Paul Randal regarding Shrinking the database is bad. I suggest every one to read that for accuracy and interesting conversation. Let us run following script where we Shrink the database and REORGANIZE. -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Check Fragmentations in the database SELECT avg_fragmentation_in_percent, fragment_count FROM sys.dm_db_index_physical_stats (DB_ID(), OBJECT_ID('SecondTable'), NULL, NULL, 'LIMITED') GO -- Shrink the Database DBCC SHRINKDATABASE (ShrinkIsBed); GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Check Fragmentations in the database SELECT avg_fragmentation_in_percent, fragment_count FROM sys.dm_db_index_physical_stats (DB_ID(), OBJECT_ID('SecondTable'), NULL, NULL, 'LIMITED') GO -- Rebuild Index on FirstTable ALTER INDEX IX_SecondTable_ID ON SecondTable REORGANIZE GO -- Name of the Database and Size SELECT name, (size*8) Size_KB FROM sys.database_files GO -- Check Fragmentations in the database SELECT avg_fragmentation_in_percent, fragment_count FROM sys.dm_db_index_physical_stats (DB_ID(), OBJECT_ID('SecondTable'), NULL, NULL, 'LIMITED') GO You can see that REORGANIZE does not increase the size of the database or remove the fragmentation. Again, I no way suggest that REORGANIZE is the solution over here. This is purely observation using demo. Read the blog post of Paul Randal. Following script will clean up the database -- Clean up USE MASTER GO ALTER DATABASE ShrinkIsBed SET SINGLE_USER WITH ROLLBACK IMMEDIATE GO DROP DATABASE ShrinkIsBed GO There are few valid cases of the Shrinking database as well, but that is not covered in this blog post. We will cover that area some other time in future. Additionally, one can rebuild index in the tempdb as well, and we will also talk about the same in future. Brent has written a good summary blog post as well. Are you Shrinking your database? Well, when are you going to stop Shrinking it? Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, PostADay, SQL, SQL Authority, SQL Index, SQL Performance, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, SQLServer, T SQL, Technology

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  • I thought the new AUTO_SAMPLE_SIZE in Oracle Database 11g looked at all the rows in a table so why do I see a very small sample size on some tables?

    - by Maria Colgan
    I recently got asked this question and thought it was worth a quick blog post to explain in a little more detail what is going on with the new AUTO_SAMPLE_SIZE in Oracle Database 11g and what you should expect to see in the dictionary views. Let’s take the SH.CUSTOMERS table as an example.  There are 55,500 rows in the SH.CUSTOMERS tables. If we gather statistics on the SH.CUSTOMERS using the new AUTO_SAMPLE_SIZE but without collecting histogram we can check what sample size was used by looking in the USER_TABLES and USER_TAB_COL_STATISTICS dictionary views. The sample sized shown in the USER_TABLES is 55,500 rows or the entire table as expected. In USER_TAB_COL_STATISTICS most columns show 55,500 rows as the sample size except for four columns (CUST_SRC_ID, CUST_EFF_TO, CUST_MARTIAL_STATUS, CUST_INCOME_LEVEL ). The CUST_SRC_ID and CUST_EFF_TO columns have no sample size listed because there are only NULL values in these columns and the statistics gathering procedure skips NULL values. The CUST_MARTIAL_STATUS (38,072) and the CUST_INCOME_LEVEL (55,459) columns show less than 55,500 rows as their sample size because of the presence of NULL values in these columns. In the SH.CUSTOMERS table 17,428 rows have a NULL as the value for CUST_MARTIAL_STATUS column (17428+38072 = 55500), while 41 rows have a NULL values for the CUST_INCOME_LEVEL column (41+55459 = 55500). So we can confirm that the new AUTO_SAMPLE_SIZE algorithm will use all non-NULL values when gathering basic table and column level statistics. Now we have clear understanding of what sample size to expect lets include histogram creation as part of the statistics gathering. Again we can look in the USER_TABLES and USER_TAB_COL_STATISTICS dictionary views to find the sample size used. The sample size seen in USER_TABLES is 55,500 rows but if we look at the column statistics we see that it is same as in previous case except  for columns  CUST_POSTAL_CODE and  CUST_CITY_ID. You will also notice that these columns now have histograms created on them. The sample size shown for these columns is not the sample size used to gather the basic column statistics. AUTO_SAMPLE_SIZE still uses all the rows in the table - the NULL rows to gather the basic column statistics (55,500 rows in this case). The size shown is the sample size used to create the histogram on the column. When we create a histogram we try to build it on a sample that has approximately 5,500 non-null values for the column.  Typically all of the histograms required for a table are built from the same sample. In our example the histograms created on CUST_POSTAL_CODE and the CUST_CITY_ID were built on a single sample of ~5,500 (5,450 rows) as these columns contained only non-null values. However, if one or more of the columns that requires a histogram has null values then the sample size maybe increased in order to achieve a sample of 5,500 non-null values for those columns. n addition, if the difference between the number of nulls in the columns varies greatly, we may create multiple samples, one for the columns that have a low number of null values and one for the columns with a high number of null values.  This scheme enables us to get close to 5,500 non-null values for each column. +Maria Colgan

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  • How do I keep a table up to date across 4 db's to be used in SQL Replication Filtering?

    - by Refracted Paladin
    I have a Win Form, Data Entry, application that uses 4 seperate Data Bases. This is an occasionally connected app that uses Merge Replication (SQL 2005) to stay in Sync. This is working just fine. The next hurdle I am trying to tackle is adding Filters to my Publications. Right now we are replicating 70mbs, compressed, to each of our 150 subscribers when, truthfully, they only need a tiny fraction of that. Using Filters I am able to accomplish this(see code below) but I had to make a mapping table in order to do so. This mapping table consists of 3 columns. A PrimaryID(Guid), WorkerName(varchar), and ClientID(int). The problem is I need this table present in all FOUR Databases in order to use it for the filter since, to my knowledge, views or cross-db query's are not allowed in a Filter Statement. What are my options? Seems like I would set it up to be maintained in 1 Database and then use Triggers to keep it updated in the other 3 Databases. In order to be a part of the Filter I have to include that table in the Replication Set so how do I flag it appropriately. Is there a better way, altogether? SELECT <published_columns> FROM [dbo].[tblPlan] WHERE [ClientID] IN (select ClientID from [dbo].[tblWorkerOwnership] where WorkerID = SUSER_SNAME()) Which allows you to chain together Filters, this next one is below the first one so it only pulls from the first's Filtered Set. SELECT <published_columns> FROM [dbo].[tblPlan] INNER JOIN [dbo].[tblHealthAssessmentReview] ON [tblPlan].[PlanID] = [tblHealthAssessmentReview].[PlanID] P.S. - I know how illogical the DB structure sounds. I didn't make it. I inherited it and was then told to make it a "disconnected app."

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  • How do I keep a table in sync across multiple SQL Databases?

    - by Refracted Paladin
    I have a Win Form, Data Entry, application that uses 4 seperate Data Bases. This is an occasionally connected app that uses Merge Replication (SQL 2005) to stay in Sync. This is working just fine. The next hurdle I am trying to tackle is adding Filters to my Publications. Right now we are replicating 70mbs, compressed, to each of our 150 subscribers when, truthfully, they only need a tiny fraction of that. Using Filters I am able to accomplish this(see code below) but I had to make a mapping table in order to do so. This mapping table consists of 3 columns. A PrimaryID(Guid), WorkerName(varchar), and ClientID(int). The problem is I need this table present in all FOUR Databases in order to use it for the filter since, to my knowledge, views or cross-db query's are not allowed in a Filter Statement. What are my options? Seems like I would set it up to be maintained in 1 Database and then use Triggers to keep it updated in the other 3 Databases. In order to be a part of the Filter I have to include that table in the Replication Set so how do I flag it appropriately. Is there a better way, altogether? SELECT <published_columns> FROM [dbo].[tblPlan] WHERE [ClientID] IN (select ClientID from [dbo].[tblWorkerOwnership] where WorkerID = SUSER_SNAME()) Which allows you to chain together Filters, this next one is below the first one so it only pulls from the first's Filtered Set. SELECT <published_columns> FROM [dbo].[tblPlan] INNER JOIN [dbo].[tblHealthAssessmentReview] ON [tblPlan].[PlanID] = [tblHealthAssessmentReview].[PlanID] P.S. - I know how illogical the DB structure sounds. I didn't make it. I inherited it and was then told to make it a "disconnected app." Go figure!

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  • Data from a table in 1 DB needed for filter in different DB...

    - by Refracted Paladin
    I have a Win Form, Data Entry, application that uses 4 seperate Data Bases. This is an occasionally connected app that uses Merge Replication (SQL 2005) to stay in Sync. This is working just fine. The next hurdle I am trying to tackle is adding Filters to my Publications. Right now we are replicating 70mbs, compressed, to each of our 150 subscribers when, truthfully, they only need a tiny fraction of that. Using Filters I am able to accomplish this(see code below) but I had to make a mapping table in order to do so. This mapping table consists of 3 columns. A PrimaryID(Guid), WorkerName(varchar), and ClientID(int). The problem is I need this table present in all FOUR Databases in order to use it for the filter since, to my knowledge, views or cross-db query's are not allowed in a Filter Statement. What are my options? Seems like I would set it up to be maintained in 1 Database and then use Triggers to keep it updated in the other 3 Databases. In order to be a part of the Filter I have to include that table in the Replication Set so how do I flag it appropriately. Is there a better way, altogether? SELECT <published_columns> FROM [dbo].[tblPlan] WHERE [ClientID] IN (select ClientID from [dbo].[tblWorkerOwnership] where WorkerID = SUSER_SNAME()) Which allows you to chain together Filters, this next one is below the first one so it only pulls from the first's Filtered Set. SELECT <published_columns> FROM [dbo].[tblPlan] INNER JOIN [dbo].[tblHealthAssessmentReview] ON [tblPlan].[PlanID] = [tblHealthAssessmentReview].[PlanID]

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  • How do I keep a table in Sync across 4 db's to be used in SQL Replication Filtering?

    - by Refracted Paladin
    I have a Win Form, Data Entry, application that uses 4 seperate Data Bases. This is an occasionally connected app that uses Merge Replication (SQL 2005) to stay in Sync. This is working just fine. The next hurdle I am trying to tackle is adding Filters to my Publications. Right now we are replicating 70mbs, compressed, to each of our 150 subscribers when, truthfully, they only need a tiny fraction of that. Using Filters I am able to accomplish this(see code below) but I had to make a mapping table in order to do so. This mapping table consists of 3 columns. A PrimaryID(Guid), WorkerName(varchar), and ClientID(int). The problem is I need this table present in all FOUR Databases in order to use it for the filter since, to my knowledge, views or cross-db query's are not allowed in a Filter Statement. What are my options? Seems like I would set it up to be maintained in 1 Database and then use Triggers to keep it updated in the other 3 Databases. In order to be a part of the Filter I have to include that table in the Replication Set so how do I flag it appropriately. Is there a better way, altogether? SELECT <published_columns> FROM [dbo].[tblPlan] WHERE [ClientID] IN (select ClientID from [dbo].[tblWorkerOwnership] where WorkerID = SUSER_SNAME()) Which allows you to chain together Filters, this next one is below the first one so it only pulls from the first's Filtered Set. SELECT <published_columns> FROM [dbo].[tblPlan] INNER JOIN [dbo].[tblHealthAssessmentReview] ON [tblPlan].[PlanID] = [tblHealthAssessmentReview].[PlanID] P.S. - I know how illogical the DB structure sounds. I didn't make it. I inherited it and was then told to make it a "disconnected app."

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  • Excel or Access: how to group several lines in a table and insert contents in columns? ("split column")

    - by Martin
    I have a table containing data of sold products (shown in the example on the left): Columns: Number of the order Product Name Attribute - specifies what is given in the following field "value", e. g. Customer Name or Product Variant Value - is the value of the Attribute Count - is the number of products of this variant sold in the order That means: Product B has 2 variants "c" and "d" Note that in Order 1 Product B was sold in Variant d only, because the letter "N" in field "D4" means "none". Note, that in OrdnerNo 3 Product B was sold only in Variant c, because for Variant d field "D9" is "N"!! This is confusing, but it is the structure of the original data (which I can not change). I need a way to convert the table on the left in a table like that on the right: one line for each product type Order Number Product Name Customer Name Count (number of products sold in this order) Variant - this is the problem, as it has to be filled with the So all rows with the same OrderNo and same product have to be grouped in to one, and I hope it is clear what I need. I tried to do it with Pivot Tables, but that fails, as the Count is always in each line, no matter if it has Value "N" or not and for the products without variants there is only one line for each order, however for products with variants there are several... So how could I create the right table with a VBA macro in MS Excel or maybe there is a trick in MS Access to do it directly or with an SQL query?

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  • Mysql: create index on 1.4 billion records

    - by SiLent SoNG
    I have a table with 1.4 billion records. The table structure is as follows: CREATE TABLE text_page ( text VARCHAR(255), page_id INT UNSIGNED ) ENGINE=MYISAM DEFAULT CHARSET=ascii The requirement is to create an index over the column text. The table size is about 34G. I have tried to create the index by the following statement: ALTER TABLE text_page ADD KEY ix_text (text) After 10 hours' waiting I finally give up this approach. Is there any workable solution on this problem? UPDATE: the table is unlikely to be updated or inserted or deleted. The reason why to create index on the column text is because this kind of sql query would be frequently executed: SELECT page_id FROM text_page WHERE text = ? UPDATE: I have solved the problem by partitioning the table. The table is partitioned into 40 pieces on column text. Then creating index on the table takes about 1 hours to complete. It seems that MySQL index creation becomes very slow when the table size becomes very big. And partitioning reduces the table into smaller trunks.

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  • bubble sort on array of c structures not sorting properly

    - by xmpirate
    I have the following program for books record and I want to sort the records on name of book. the code isn't showing any error but it's not sorting all the records. #include "stdio.h" #include "string.h" #define SIZE 5 struct books{ //define struct char name[100],author[100]; int year,copies; }; struct books book1[SIZE],book2[SIZE],*pointer; //define struct vars void sort(struct books *,int); //define sort func main() { int i; char c; for(i=0;i<SIZE;i++) //scanning values { gets(book1[i].name); gets(book1[i].author); scanf("%d%d",&book1[i].year,&book1[i].copies); while((c = getchar()) != '\n' && c != EOF); } pointer=book1; sort(pointer,SIZE); //sort call i=0; //printing values while(i<SIZE) { printf("##########################################################################\n"); printf("Book: %s\nAuthor: %s\nYear of Publication: %d\nNo of Copies: %d\n",book1[i].name,book1[i].author,book1[i].year,book1[i].copies); printf("##########################################################################\n"); i++; } } void sort(struct books *pointer,int n) { int i,j,sorted=0; struct books temp; for(i=0;(i<n-1)&&(sorted==0);i++) //bubble sort on the book name { sorted=1; for(j=0;j<n-i-1;j++) { if(strcmp((*pointer).name,(*(pointer+1)).name)>0) { //copy to temp val strcpy(temp.name,(*pointer).name); strcpy(temp.author,(*pointer).author); temp.year=(*pointer).year; temp.copies=(*pointer).copies; //copy next val strcpy((*pointer).name,(*(pointer+1)).name); strcpy((*pointer).author,(*(pointer+1)).author); (*pointer).year=(*(pointer+1)).year; (*pointer).copies=(*(pointer+1)).copies; //copy back temp val strcpy((*(pointer+1)).name,temp.name); strcpy((*(pointer+1)).author,temp.author); (*(pointer+1)).year=temp.year; (*(pointer+1)).copies=temp.copies; sorted=0; } *pointer++; } } } My Imput The C Programming Language X Y Z 1934 56 Inferno Dan Brown 1993 453 harry Potter and the soccers stone J K Rowling 2012 150 Ruby On Rails jim aurther nil 2004 130 Learn Python Easy Way gmaps4rails 1967 100 And the output ########################################################################## Book: Inferno Author: Dan Brown Year of Publication: 1993 No of Copies: 453 ########################################################################## ########################################################################## Book: The C Programming Language Author: X Y Z Year of Publication: 1934 No of Copies: 56 ########################################################################## ########################################################################## Book: Ruby On Rails Author: jim aurther nil Year of Publication: 2004 No of Copies: 130 ########################################################################## ########################################################################## Book: Learn Python Easy Way Author: gmaps4rails Year of Publication: 1967 No of Copies: 100 ########################################################################## ########################################################################## Book: Author: Year of Publication: 0 No of Copies: 0 ########################################################################## We can see the above sorting is wrong? What I'm I doing wrong?

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  • JQuery Tablesorter memorizeSortOrder widget

    - by echedey lorenzo
    Hi, I've found this code in the internet: $.tablesorter.addWidget({ id: "memorizeSortOrder", format: function(table) { if (!table.config.widgetMemorizeSortOrder.isBinded) { // only bind if not already binded table.config.widgetMemorizeSortOrder.isBinded = true; $("thead th:visible",table).click(function() { var i = $("thead th:visible",table).index(this); $.get(table.config.widgetMemorizeSortOrder.url+i+'|'+table.config.headerList[i].order); }); } // fi } }); Found in: http://www.adspeed.org/2008/10/jquery-extend-tablesorter-plugin.html I would like to memorize the sorting of my ajax tables so on each update (table changes completely so there is no append) it keeps sorted the as it was. Question is.. how can use this? $("#tablediv").load( "table.php", null, function (responseText, textStatus, req) { $("#table").trigger("update"); } ); What changes do I need?

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  • What is the easiest way to find a sql query returns a result or not?

    - by bala3569
    Consider the following sql server query , DECLARE @Table TABLE( Wages FLOAT ) INSERT INTO @Table SELECT 20000 INSERT INTO @Table SELECT 15000 INSERT INTO @Table SELECT 10000 INSERT INTO @Table SELECT 45000 INSERT INTO @Table SELECT 50000 SELECT * FROM ( SELECT *, ROW_NUMBER() OVER(ORDER BY Wages DESC) RowID FROM @Table ) sub WHERE RowID = 3 The result of the query would be 20000 ..... Thats fine as of now i have to find the result of this query, SELECT * FROM ( SELECT *, ROW_NUMBER() OVER(ORDER BY Wages DESC) RowID FROM @Table ) sub WHERE RowID = 6 It will not give any result because there are only 5 rows in the table..... so now my question is What is the easiest way to find a sql query returns a result or not?

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  • Disable Foreign key constraint on all tables didn't work ?

    - by Space Cracker
    i try a lot of commands to disable tables constraints in my database to make truncate to all tables but still now it give me the same error Cannot truncate table '' because it is being referenced by a FOREIGN KEY constraint. i try EXEC sp_msforeachtable "ALTER TABLE ? NOCHECK CONSTRAINT all" EXEC sp_MSforeachtable "TRUNCATE TABLE ?" and i tried this for each table ALTER TABLE [Table Name] NOCHECK CONSTRAINT ALL truncate table [Table Name] ALTER TABLE [Table Name] CHECK CONSTRAINT ALL and every time i have the previous error message .. could any please help me to solve sucha a problem ?

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  • Can you add identity to existing column in sql server 2008?

    - by bmutch
    In all my searching I see that you essentially have to copy the existing table to a new table to chance to identity column for pre-2008, does this apply to 2008 also? thanks. most concise solution I have found so far: CREATE TABLE Test ( id int identity(1,1), somecolumn varchar(10) ); INSERT INTO Test VALUES ('Hello'); INSERT INTO Test VALUES ('World'); -- copy the table. use same schema, but no identity CREATE TABLE Test2 ( id int NOT NULL, somecolumn varchar(10) ); ALTER TABLE Test SWITCH TO Test2; -- drop the original (now empty) table DROP TABLE Test; -- rename new table to old table's name EXEC sp_rename 'Test2','Test'; -- see same records SELECT * FROM Test;

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  • OBIA on Teradata - Part 1 Loader and Monitoring

    - by Mohan Ramanuja
    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;} The out-of-the-box (OOB) OBIA Informatica mappings come with TPump loader.   TPUMP  FASTLOAD TPump does not lock the table. FastLoad applies exclusive lock on the table. The table that TPump is loading can have data. The table that FastLoad is loading needs to be empty. 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;} TPump is not efficient with lookups. FastLoad is more efficient in the absence of lookups. 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;} The out-of the box Informatica mappings come with TPump loader. There is chance for bottleneck in writer thread The out-of the box tables in Teradata supplied with OBAW features all Dimension and Fact tables using ROW_WID as the key for primary index. Also, all staging tables use integration_id as the key for primary index. This reduces skewing of data across Teradata AMPs.You can use an SQL statement similar to the following to determine if data for a given table is distributed evenly across all AMP vprocs. The SQL statement displays the AMP with the most used through the AMP with the least-used space, investigating data distribution in the Message table in database RST.SELECT vproc,CurrentPermFROM DBC.TableSizeWHERE Databasename = ‘PRJ_CRM_STGC’AND Tablename = ‘w_party_per_d’ORDER BY 2 descIf you suspect distribution problems (skewing) among AMPS, the following is a sample of what you might enter for a three-column PI:SELECT HASHAMP (HASHBUCKET (HASHROW (col_x, col_y, col_z))), count (*)FROM hash15GROUP BY 1ORDER BY 2 desc; ETL Error Monitoring Error Table – These are tables that start with ET. Location and name can be specified in Informatica session as well as the loader connection.Loader Log – Loader log is available in the Informatica server under the session log folder. These give feedback on the loader parameters such as Packing Factor to use. These however need to be monitored in the production environment. The recommendations made in one environment may not be used in another environment.Log Table – These are tables that start with TL. These are sparse on information.Bad File – This is the Informatica file generated in case there is data quality issues

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  • Order of calls to set functions when invoking a flex component

    - by Jason
    I have a component called a TableDataViewer that contains the following pieces of data and their associated set functions: [Bindable] private var _dataSetLoader:DataSetLoader; public function get dataSetLoader():DataSetLoader {return _dataSetLoader;} public function set dataSetLoader(dataSetLoader:DataSetLoader):void { trace("setting dSL"); _dataSetLoader = dataSetLoader; } [Bindable] private var _table:Table = null; public function set table(table:Table):void { trace("setting table"); _table = table; _dataSetLoader.load(_table.definition.id, "viewData", _table.definition.id); } This component is nested in another component as follows: <ve:TableDataViewer width="100%" height="100%" paddingTop="10" dataSetLoader="{_openTable.dataSetLoader}" table="{_openTable.table}"/> Looking at the trace in the logs, the call to set table is coming before the call to set dataSetLoader. Which is a real shame because set table() needs dataSetLoader to already be set in order to call its load() function. So my question is, is there a way to enforce an order on the calls to the set functions when declaring a component?

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  • Is it Considered Good SQL practice to use GUID to link multiple tables to same Id field?

    - by Mallow
    I want to link several tables to a many-to-many(m2m) table. One table would be called location and this table would always be on one side of the m2m table. But I will have a list of several tables for example: Cards Photographs Illustrations Vectors Would using GUID's between these tables to link it to a single column in another table be considered 'Good Practice'? Will Mysql let me to have it automatically cascade updates and delete? If so, would multiple cascades lead to an issues? UPDATE I've read that GUID (a hex number) Generally takes up more space in a database and slows queries down. However I could still generate 'unique' ids by just having the table initial's as part of the id so that the table card's id would be c0001, and then Illustrations be I001. Regardless of this change, the questions still stands.

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  • Database design for a Fantasy league

    - by Samidh T
    Here's the basic schema for my database Table user{ userid numeber primary key, count number } Table player{ pid number primary key, } Table user-player{ userid number primary key foreign key(user), pid number primary key foreign key(player) } Table temp{ pid number primary key, points number } Here's what I intend to do... After every match the temp table is updated which holds the id of players that played the last match and the points they earned. Next run a procedure that will match the pid from temp table with every uid of user-player table having the same pid. add the points from temp table to the count of user table for every matching uid. empty temp table. My questions is considering 200 players and 10000 users,Will this method be efficient? I am going to be using mysql for this.

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  • Select records by comparing subsets

    - by devnull
    Given two tables (the rows in each table are distinct): 1) x | y z 2) x | y z ------- --- ------- --- 1 | a a 1 | a a 1 | b b 1 | b b 2 | a 1 | c 2 | b 2 | a 2 | c 2 | b 2 | c Is there a way to select the values in the x column of the first table for which all the values in the y column (for that x) are found in the z column of the second table? In case 1), expected result is 1. If c is added to the second table then the expected result is 2. In case 2), expected result is no record since neither of the subsets in the first table matches the subset in the second table. If c is added to the second table then the expected result is 1, 2. I've tried using except and intersect to compare subsets of first table with the second table, which works fine, but it takes too long on the intersect part and I can't figure out why (the first table has about 10.000 records and the second has around 10). EDIT: I've updated the question to provide an extra scenario.

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