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  • criticle join query mysql

    - by Bharanikumar
    i have table country and having country name and country code, i have customer table having countomer mobile number like customer_mobile field values like 0044-123456798 , 0024-582654753 , 012-52686145 , i want to populate my country oce into combo, that country name should in customer table which country code count is high that country name should come in the dropdown list top , How to write the join query... Thanks

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  • select query related problem

    - by user222585
    i have interest rate and amount in a table where interest rate are ranged in different values like 4.5,4.6,5.2,5.6 etc. i want to get sum of amounts classified by interest rate where interest rate will be separated by .25. for example all amount having interest rate 1.25,1.3,1.4 will be in one group and 1.5,1.67,1.9 will be in another group how can i write the query?

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  • Why does this properly escaped SQL query fail?

    - by Jason Rhodes
    Here's the query: INSERT INTO jobemails (jobid, to, subject, message, headers, datesent) VALUES ('340', '[email protected]', 'We\'ve received your request for a photo shoot called \'another\'.', 'message', 'headers', '2010-04-22 15:55:06') The datatypes are all correct, it always fails at the subject, so it must be how I'm escaping the values, I assume. I'm sure one of you will see my idiot mistake right away. A little help?

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  • Query a SharePoint List from InfoPath web form in code

    - by dawsonweb
    I'm writing an InfoPath web form, using C# code behind. The form is a holiday request, after the user inputs the start and end dates I want the code to query a bank-holiday sharepoint list and count occurrences between the dates. I've added the sharepoint list as a second datasource however I'm now stuck. Any ideas?

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  • SQL query joining rows with same value

    - by user1285737
    I need to write a query that creates a view that calculates the total cost of each sale, by considering quantity and price of each bought item. The view should return the debit and total cost. In the answer each debit-number should only occur once. Thanks in advance Table ITEM: ID NAME PRICE 118 Jeans 100 120 Towel 20 127 Shirt 55 Table DEBIT: DEBIT ITEM Quantity 100581 118 5 100581 120 1 100586 127 5

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  • Need some help with a NHibernate Query

    - by cwap
    Hi all Say I got 3 entities: Business, Employee and Payment. A payment has a foreign key to an Employee, while the Employee has an foreign key to a business. Now, I want to create a query which gives me all payments for a given business. I really don't have a clue about how to do this - I guess I want something like: mySession.CreateCriteria<Payment>() .Add(Criterion.Expression.Eq(/* Employee_FK => Employee.Business_FK == BusinessID */); Any help would be greatly appreciated :)

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  • how to do this in MySql Query??

    - by deep
    ID NAME AMT 1 Name1 1000 2 Name2 500 3 Name3 3000 4 Name1 5000 5 Name2 2000 6 Name1 3000 consider above table as sample. am having a problem in my sql query, Am using like this. Select name,amt from sample where amt between 1000 and 5000 it returns all the values in the table between 1000 and 5000, instead I want to get maximum amount record for each name i.e., 3 name3 3000 4 name1 5000 5 name2 2000

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  • SQL query construction - separate data in a column into two columns

    - by Tommy
    I have a column that contains links. The problem is that the titles of the links are in the same column, so it looks like this: linktitle|-|linkurl I want link title and linkurl in separate columns. I've created a new column for the urls, so I'm looking for a way to extract them and update the linkurl column with them. Is there any clever way to construct a query that does this?

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  • 4 table query / join. getting duplicate rows

    - by Horse
    So I have written a query that will grab an order (this is for an ecommerce type site), and from that order id it will get all order items (ecom_order_items), print options (c_print_options) and images (images). The eoi_p_id is currently a foreign key from the images table. This works fine and the query is: SELECT eoi_parentid, eoi_p_id, eoi_po_id, eoi_quantity, i_id, i_parentid, po_name, po_price FROM ecom_order_items, images, c_print_options WHERE eoi_parentid = '1' AND i_id = eoi_p_id AND po_id = eoi_po_id; The above would grab all the stuff I need for order #1 Now to complicate things I added an extra table (ecom_products), which needs to act in a similar way to the images table. The eoi_p_id can also point at a foreign key in this table too. I have added an extra field 'eoi_type' which will either have the value 'image', or 'product'. Now items in the order could be made up of a mix of items from images or ecom_products. Whatever I try it either ends up with too many records, wont actually output any with eoi_type = 'product', and just generally wont work. Any ideas on how to achieve what I am after? Can provide SQL samples if needed? SELECT eoi_id, eoi_parentid, eoi_p_id, eoi_po_id, eoi_po_id_2, eoi_quantity, eoi_type, i_id, i_parentid, po_name, po_price, po_id, ep_id FROM ecom_order_items, images, c_print_options, ecom_products WHERE eoi_parentid = '9' AND i_id = eoi_p_id AND po_id = eoi_po_id The above outputs duplicate rows and doesnt work as expected. Am I going about this the wrong way? Should I have seperate foreign key fields for the eoi_p_id depending it its an image or a product? Should I be using JOINs? Here is a mysql explain of the tables in question ecom_products +-------------+--------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +-------------+--------------+------+-----+---------+----------------+ | ep_id | int(8) | NO | PRI | NULL | auto_increment | | ep_title | varchar(255) | NO | | NULL | | | ep_link | text | NO | | NULL | | | ep_desc | text | NO | | NULL | | | ep_imgdrop | text | NO | | NULL | | | ep_price | decimal(6,2) | NO | | NULL | | | ep_category | varchar(255) | NO | | NULL | | | ep_hide | tinyint(1) | NO | | 0 | | | ep_featured | tinyint(1) | NO | | 0 | | +-------------+--------------+------+-----+---------+----------------+ ecom_order_items +--------------+-------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +--------------+-------------+------+-----+---------+----------------+ | eoi_id | int(8) | NO | PRI | NULL | auto_increment | | eoi_parentid | int(8) | NO | | NULL | | | eoi_type | varchar(32) | NO | | NULL | | | eoi_p_id | int(8) | NO | | NULL | | | eoi_po_id | int(8) | NO | | NULL | | | eoi_quantity | int(4) | NO | | NULL | | +--------------+-------------+------+-----+---------+----------------+ c_print_options +------------+--------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +------------+--------------+------+-----+---------+----------------+ | po_id | int(8) | NO | PRI | NULL | auto_increment | | po_name | varchar(255) | NO | | NULL | | | po_price | decimal(6,2) | NO | | NULL | | +------------+--------------+------+-----+---------+----------------+ images +--------------+--------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +--------------+--------------+------+-----+---------+----------------+ | i_id | int(8) | NO | PRI | NULL | auto_increment | | i_filename | varchar(255) | NO | | NULL | | | i_data | longtext | NO | | NULL | | | i_parentid | int(8) | NO | | NULL | | +--------------+--------------+------+-----+---------+----------------+

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  • a bugging query

    - by Sachindra
    i need to get a query where the elements are displayed in case the first letter is E (the word is electronics).. i have tried with the following : mysql_query("select * from nested_category where name like '[A-F]%'");

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  • help with sql-server query - dates

    - by Gold
    hi i have this table: id|date 1 | 10/11/2009 2 | 13/11/2009 1 | 20/12/2009 3 | 21/12/2009 1 | 30/12/2009 if i stand on the last record (id=1) and i need to see the last date where id=1 is appear -- will show me: 1 | 20/12/2009 what query will do it ? thank's in advance

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  • oracle query returns 4 duplicates of each row

    - by ajoe
    hello, I am Running a oracle query, it seems to work except that it returns 4 dupes of each result. here is the code: Select * from (Select a.*, rownum rnum From (SELECT NEW_USER.*, NEW_EHS_QUIZ_COMPLETE.datetime FROM NEW_USER, NEW_EHS_QUIZ_COMPLETE WHERE EXISTS(select * from NEW_EHS_QUIZ_COMPLETE where NEW_USER.id=NEW_EHS_QUIZ_COMPLETE.USER_ID) ORDER by last_name ASC ) a where rownum <= #pgtop# ) where rnum >= #pgbot# does anyone know why this isnt working properly? thanks in advance.

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  • Get insert statement query

    - by Karthick
    Hi, Is there a way to get the insert statements for a table via some query in MySql? for ex: if the table name is Cards,which has 5 rows, i need to get the insert statements for that particular table. Thanks.

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  • mysql query trouble

    - by Bharanikumar
    Hi , in my database i have phone numbers with country code , which look somthing like 0044-123456 0044-123456 0014-123456 0014-123456 0024-123456 0024-123456 0034-123456 0044-123456 0044-123456 0024-123456 0034-123456 084-123456 084-123456 i want to total up the numbers by country, something like this output 0044 (2) 0024 (2) 0034 (1) 084 (2) 064 (5) Is it possible to do this with a SQL query?

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  • Best way to use If() statement in My-Sql Query

    - by PHP-Prabhu
    Can any one please let me know the best way to use IF statement in mysql query to show if the "email" field is NULL then it should show as "no email"... Postcode Telephone Email ---------------------------------------------------------- BS20 0QN 1275373088 no email BS20 0QN 1275373088 no email PO9 4HG 023 92474208 [email protected] SO43 7DS 07801 715200 [email protected] ----------------------------------------------------------

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  • Slow query execution time

    - by rotor
    select p.id,p.title,p.slug,p.content, (select url from gallery where postid=p.id limit 1) as url, t.name from posts as p inner join termrel as tr on (tr.object = p.id) inner join termtax as tx on (tx.id = tr.termtax_id) inner join terms as t on (t.id = tx.term_id) where tx.taxonomy_id=3 and p.post_status is null order by t.name asc This query took about 0.2407s to execute. How to make it fast?

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  • sql query help join (i think)

    - by milan
    I am having trouble figuring our how I can get results only when products.published, product_types.published, and product_cats.published = 1 but my query isn't working. Please help: SELECT `products`.`title`, `products`.`menu_id`, `products`.`short_description`, `products`.`datasheet_icon`, `products`.`datasheet`, `products`.`ordering`, `products`.`product_type_id`, CASE WHEN CHAR_LENGTH(`products`.`alias`) THEN CONCAT_WS(':', `products`.`id`, `products`.`alias`) ELSE `products`.`id` END AS slug FROM `products`, `product_cats`, `product_types` WHERE `products`.published=1 AND `product_cats`.published=1 AND `product_types`.published=1 AND `products`.`product_cat_id`='42' AND `product_types`.`id` IN (1,40,48,49,50) GROUP BY `products`.`id` ORDER BY `product_types`.`ordering`, `products`.`ordering`

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  • Problem with PHP & SQL Query

    - by Shahd
    Hi .... i have a problem in php code (inserting values in database) i use PHPMyAdmin my DATABASE has 3 tables: 1) Member with this fields: MemberID, MemberName 2) Room with this fields: RoomID, RoomName 3) Join with this fields: MemberID, RoomID the idea is to join the member in the room. My query was mysql_query("INSERT INTO join (RoomID, MemberID) VALUES ('121', '131')"); but unfortunately it is not work

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  • SQL SERVER – Expanding Views – Contest Win Joes 2 Pros Combo (USD 198) – Day 4 of 5

    - by pinaldave
    August 2011 we ran a contest where every day we give away one book for an entire month. The contest had extreme success. Lots of people participated and lots of give away. I have received lots of questions if we are doing something similar this month. Absolutely, instead of running a contest a month long we are doing something more interesting. We are giving away USD 198 worth gift every day for this week. We are giving away Joes 2 Pros 5 Volumes (BOOK) SQL 2008 Development Certification Training Kit every day. One copy in India and One in USA. Total 2 of the giveaway (worth USD 198). All the gifts are sponsored from the Koenig Training Solution and Joes 2 Pros. The books are available here Amazon | Flipkart | Indiaplaza How to Win: Read the Question Read the Hints Answer the Quiz in Contact Form in following format Question Answer Name of the country (The contest is open for USA and India residents only) 2 Winners will be randomly selected announced on August 20th. Question of the Day: Which of the following key word will force the query to use indexes created on views? a) ENCRYPTION b) SCHEMABINDING c) NOEXPAND d) CHECK OPTION Query Hints: BIG HINT POST Usually, the assumption is that Index on the table will use Index on the table and Index on view will be used by view. However, that is the misconception. It does not happen this way. In fact, if you notice the image, you will find the both of them (table and view) use both the index created on the table. The index created on the view is not used. The reason for the same as listed in BOL. The cost of using the indexed view may exceed the cost of getting the data from the base tables, or the query is so simple that a query against the base tables is fast and easy to find. This often happens when the indexed view is defined on small tables. You can use the NOEXPAND hint if you want to force the query processor to use the indexed view. This may require you to rewrite your query if you don’t initially reference the view explicitly. You can get the actual cost of the query with NOEXPAND and compare it to the actual cost of the query plan that doesn’t reference the view. If they are close, this may give you the confidence that the decision of whether or not to use the indexed view doesn’t matter. Additional Hints: I have previously discussed various concepts from SQL Server Joes 2 Pros Volume 4. SQL Joes 2 Pros Development Series – Structured Error Handling SQL Joes 2 Pros Development Series – SQL Server Error Messages SQL Joes 2 Pros Development Series – Table-Valued Functions SQL Joes 2 Pros Development Series – Table-Valued Store Procedure Parameters SQL Joes 2 Pros Development Series – Easy Introduction to CHECK Options SQL Joes 2 Pros Development Series – Introduction to Views SQL Joes 2 Pros Development Series – All about SQL Constraints Next Step: Answer the Quiz in Contact Form in following format Question Answer Name of the country (The contest is open for USA and India) Bonus Winner Leave a comment with your favorite article from the “additional hints” section and you may be eligible for surprise gift. There is no country restriction for this Bonus Contest. Do mention why you liked it any particular blog post and I will announce the winner of the same along with the main contest. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: Joes 2 Pros, PostADay, SQL, SQL Authority, SQL Puzzle, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • SQL SERVER – Introduction to LEAD and LAG – Analytic Functions Introduced in SQL Server 2012

    - by pinaldave
    SQL Server 2012 introduces new analytical function LEAD() and LAG(). This functions accesses data from a subsequent row (for lead) and previous row (for lag) in the same result set without the use of a self-join . It will be very difficult to explain this in words so I will attempt small example to explain you this function. Instead of creating new table, I will be using AdventureWorks sample database as most of the developer uses that for experiment. Let us fun following query. USE AdventureWorks GO SELECT s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty, LEAD(SalesOrderDetailID) OVER (ORDER BY SalesOrderDetailID ) LeadValue, LAG(SalesOrderDetailID) OVER (ORDER BY SalesOrderDetailID ) LagValue FROM Sales.SalesOrderDetail s WHERE SalesOrderID IN (43670, 43669, 43667, 43663) ORDER BY s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty GO Above query will give us following result. When we look at above resultset it is very clear that LEAD function gives us value which is going to come in next line and LAG function gives us value which was encountered in previous line. If we have to generate the same result without using this function we will have to use self join. In future blog post we will see the same. Let us explore this function a bit more. This function not only provide previous or next line but it can also access any line before or after using offset. Let us fun following query, where LEAD and LAG function accesses the row with offset of 2. USE AdventureWorks GO SELECT s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty, LEAD(SalesOrderDetailID,2) OVER (ORDER BY SalesOrderDetailID ) LeadValue, LAG(SalesOrderDetailID,2) OVER (ORDER BY SalesOrderDetailID ) LagValue FROM Sales.SalesOrderDetail s WHERE SalesOrderID IN (43670, 43669, 43667, 43663) ORDER BY s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty GO Above query will give us following result. You can see the LEAD and LAG functions  now have interval of  rows when they are returning results. As there is interval of two rows the first two rows in LEAD function and last two rows in LAG function will return NULL value. You can easily replace this NULL Value with any other default value by passing third parameter in LEAD and LAG function. Let us fun following query. USE AdventureWorks GO SELECT s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty, LEAD(SalesOrderDetailID,2,0) OVER (ORDER BY SalesOrderDetailID ) LeadValue, LAG(SalesOrderDetailID,2,0) OVER (ORDER BY SalesOrderDetailID ) LagValue FROM Sales.SalesOrderDetail s WHERE SalesOrderID IN (43670, 43669, 43667, 43663) ORDER BY s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty GO Above query will give us following result, where NULL are now replaced with value 0. Just like any other analytic function we can easily partition this function as well. Let us see the use of PARTITION BY in this clause. USE AdventureWorks GO SELECT s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty, LEAD(SalesOrderDetailID) OVER (PARTITION BY SalesOrderID ORDER BY SalesOrderDetailID ) LeadValue, LAG(SalesOrderDetailID) OVER (PARTITION BY SalesOrderID ORDER BY SalesOrderDetailID ) LagValue FROM Sales.SalesOrderDetail s WHERE SalesOrderID IN (43670, 43669, 43667, 43663) ORDER BY s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty GO Above query will give us following result, where now the data is partitioned by SalesOrderID and LEAD and LAG functions are returning the appropriate result in that window. As now there are smaller partition in my query, you will see higher presence of NULL. In future blog post we will see how this functions are compared to SELF JOIN. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, PostADay, SQL, SQL Authority, SQL Function, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • SQL SERVER – OVER clause with FIRST _VALUE and LAST_VALUE – Analytic Functions Introduced in SQL Server 2012 – ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING

    - by pinaldave
    Yesterday I had discussed two analytical functions FIRST_VALUE and LAST_VALUE. After reading the blog post I received very interesting question. “Don’t you think there is bug in your first example where FIRST_VALUE is remain same but the LAST_VALUE is changing every line. I think the LAST_VALUE should be the highest value in the windows or set of result.” I find this question very interesting because this is very commonly made mistake. No there is no bug in the code. I think what we need is a bit more explanation. Let me attempt that first. Before you do that I suggest you read yesterday’s blog post as this question is related to that blog post. Now let’s have fun following query: USE AdventureWorks GO SELECT s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty, FIRST_VALUE(SalesOrderDetailID) OVER (ORDER BY SalesOrderDetailID) FstValue, LAST_VALUE(SalesOrderDetailID) OVER (ORDER BY SalesOrderDetailID) LstValue FROM Sales.SalesOrderDetail s WHERE SalesOrderID IN (43670, 43669, 43667, 43663) ORDER BY s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty GO The above query will give us the following result: As per the reader’s question the value of the LAST_VALUE function should be always 114 and not increasing as the rows are increased. Let me re-write the above code once again with bit extra T-SQL Syntax. Please pay special attention to the ROW clause which I have added in the above syntax. USE AdventureWorks GO SELECT s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty, FIRST_VALUE(SalesOrderDetailID) OVER (ORDER BY SalesOrderDetailID ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) FstValue, LAST_VALUE(SalesOrderDetailID) OVER (ORDER BY SalesOrderDetailID ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) LstValue FROM Sales.SalesOrderDetail s WHERE SalesOrderID IN (43670, 43669, 43667, 43663) ORDER BY s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty GO Now once again check the result of the above query. The result of both the query is same because in OVER clause the default ROWS selection is always UNBOUNDED PRECEDING AND CURRENT ROW. If you want the maximum value of the windows with OVER clause you need to change the syntax to UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING for ROW clause. Now run following query and pay special attention to ROW clause again. USE AdventureWorks GO SELECT s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty, FIRST_VALUE(SalesOrderDetailID) OVER (PARTITION BY SalesOrderID ORDER BY SalesOrderDetailID ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FstValue, LAST_VALUE(SalesOrderDetailID) OVER (PARTITION BY SalesOrderID ORDER BY SalesOrderDetailID ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) LstValue FROM Sales.SalesOrderDetail s WHERE SalesOrderID IN (43670, 43669, 43667, 43663) ORDER BY s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty GO Here is the resultset of the above query which is what questioner was asking. So in simple word, there is no bug but there is additional syntax needed to add to get your desired answer. The same logic also applies to PARTITION BY clause when used. Here is quick example of how we can further partition the query by SalesOrderDetailID with this new functions. USE AdventureWorks GO SELECT s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty, FIRST_VALUE(SalesOrderDetailID) OVER (PARTITION BY SalesOrderID ORDER BY SalesOrderDetailID ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FstValue, LAST_VALUE(SalesOrderDetailID) OVER (PARTITION BY SalesOrderID ORDER BY SalesOrderDetailID ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) LstValue FROM Sales.SalesOrderDetail s WHERE SalesOrderID IN (43670, 43669, 43667, 43663) ORDER BY s.SalesOrderID,s.SalesOrderDetailID,s.OrderQty GO Above query will give us windowed resultset on SalesOrderDetailsID as well give us FIRST and LAST value for the windowed resultset. There are lots to discuss for this two functions and we have just explored tip of the iceberg. In future post I will discover it further deep. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, PostADay, SQL, SQL Authority, SQL Function, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, T SQL, Technology

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