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  • MySQL: How to copy rows, but change a few fields?

    - by Andrew
    I have a large number of rows that I would like to copy, but I need to change one field. I can select the rows that I want to copy: select * from Table where Event_ID = "120" Now I want to copy all those rows and create new rows while setting the Event_ID to 155. How can I accomplish this?

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  • How do I test the speed of a mySQL query?

    - by Chris
    I have a select and query like below... $sql = "SELECT * FROM notifications WHERE to_id='".$userid."' AND (alert_read != '1' OR user_read != '1') ORDER BY alert_time DESC"; $result = mysql_query($sql); how do I test how long the query took to run?

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  • MySQL Query to receive random combinations from two tables.

    - by Michael
    Alright, here is my issue, I have two tables, one named firstnames and the other named lastnames. What I am trying to do here is to find 100 of the possible combinations from these names for test data. The firstnames table has 5494 entries in a single column, and the lastnames table has 88799 entries in a single column. The only query that I have been able to come up with that has some results is: select * from (select * from firstnames order by rand()) f LEFT JOIN (select * from lastnames order by rand()) l on 1=1 limit 10; The problem with this code is that it selects 1 firstname and gives every lastname that could go with it. While this is plausible, I will have to set the limit to 500000000 in order to get all the combinations possible without having only 20 first names(and I'd rather not kill my server). However, I only need 100 random generations of entries for test data, and I will not be able to get that with this code. Can anyone please give me any advice?

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  • Can anyone help me with a complex sum, 3 table join mysql query?

    - by Scarface
    Hey guys I have a query and it works fine, but I want to add another table to the mix. The invite table I want to add has two fields: username and user_invite. Much like this site, I am using a point system to encourage diligent users. The current query which is displayed below adds the up votes and down votes based on the user in question: $creator. I want to count the number of entries for that same user from the invite table, and add 50 for each row it finds to the current output/sum of my query. Is this possible with one query, or do I need two? "SELECT *, SUM(IF(points_id = \"1\", 1,0))-SUM(IF(points_id = \"2\", 1,0)) AS 'total' FROM points LEFT JOIN post ON post.post_id=points.points_id WHERE post.creator='$creator'"

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  • MySQL - accessing a table sum and compare to another table?

    - by assignment_operator
    This is for a homework assignment. I just plain don't understand how to do it. The instructions for this particular question is: List the branch name for all branches that have at least one book that has at least 4 copies on hand. Where the tables in question are: Branch: BranchName | BranchId Henry Downtown | 1 16 Riverview | 2 Henry On The Hill | 3 Inventory: BookId | BranchId | OnHand 1 | 1 | 2 2 | 3 | 4 3 | 1 | 8 4 | 3 | 1 5 | 1 | 2 6 | 2 | 3 From what I understand, I can get the number of OnHand per branch name with: SELECT BranchName, SUM(OnHand) FROM Branch B, Inventory I WHERE B.BranchId = I.BranchId GROUP BY BranchName; but I don't get how I'd do the comparison between the sum of OnHand per branch and 4. Any help would be appreciated, guys!

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  • MySQL - Combine two fields to create a unix timestamp?

    - by Dan
    Hi, I'm trying to retrieve a UNIX timestamp from a query by combining a date and a time field in the table, however it keeps returning as zero. SELECT *, UNIX_TIMESTAMP(startdate starttime) AS start, UNIX_TIMESTAMP(enddate endtime) AS end FROM mytable; Can anyone help me out? Thanks.

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  • How do I drop 'NOT NULL' from a column in MySQL?

    - by Will
    A show create table command shows the following: 'columnA' varchar(6) NOT NULL DEFAULT ''; How do I modify that column so that the not null is removed? I need it to be: 'columnA' varchar(6) DEFAULT NULL; I thought the following would work, but it has no effect: ALTER TABLE tbl_name MODIFY columnA varchar(6) DEFAULT NULL;

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  • MySQL: Query to obtain recipes using all given ingredients.

    - by John_A
    hi I have the following simplified tables: CREATE TABLE recipe(id int, name varchar(25)); CREATE TABLE ingredient(name varchar(25)); CREATE TABLE uses_ingredient(recipe_id int, name varchar(25)); I want to make a query that returns all id's of recipes that contain both Chicken and Cream. I have tried SELECT recipe_id FROM uses_ingredient INNER JOIN (SELECT * FROM ingredient WHERE name="Chicken" OR name="Cream") USING (name) GROUP BY recipe_id HAVING COUNT(recipe_id) >= (SELECT COUNT(*) FROM theme); which gives me :"ERROR 1248 (42000): Every derived table must have its own alias" and is probably wrong too. Next I tried SELECT recipe_id FROM (SELECT * FROM ingredient WHERE name="Chicken" OR name="Cream") AS t INNER JOIN uses_ingredient USING (name) GROUP BY recipe_id HAVING COUNT(recipe_id)>= (SELECT COUNT(*) FROM t); which gives "ERROR 1146 (42S02): Table 'recipedb.t' doesn't exist" I want to avoid creating temporary tables including using ENGINE=MEMORY.

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  • PHP, MySQL - would results-array shuffle be quicker than "select... order by rand()"?

    - by sombe
    I've been reading a lot about the disadvantages of using "order by rand" so I don't need update on that. I was thinking, since I only need a limited amount of rows retrieved from the db to be randomized, maybe I should do: $r = $db->query("select * from table limit 500"); for($i;$i<500;$i++) $arr[$i]=mysqli_fetch_assoc($r); shuffle($arr); (i know this only randomizes the 500 first rows, be it). would that be faster than $r = $db->("select * from table order by rand() limit 500"); let me just mention, say the db tables were packed with more than...10,000 rows. why don't you do it yourself?!? - well, i have, but i'm looking for your experienced opinion. thanks!

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  • MySQL: How to separate a name field in one table into firstname / lastname in two separate tables?

    - by Eileen
    I have a drupal database where the node table is full of profiles. The field node.title is "Firstname Lastname". I want to separate the names so that node.title = "Firstname", and over in another table entirely, content_type_profile.field_lastname_value = "Lastname". The entries in the two tables can be joined on the field nid. I'd love to run a SQL command to do this, and I am fine with taking the naive approach that the first word is the first name, and everything else in the field is last name -- it will mean a few manual corrections down the line, but that's much better than doing it all by hand in the first place. (I read this question and surely the answer lies in there but I am not that SQL-savvy and am not sure how to make it work for my database.) Thanks!

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  • Mysql: Order Results by number of matching rows in Second Table.

    - by KyleT
    I'm not sure the best way to word this question so bear with me. Table A has following columns: id name description Table B has the following columns: id a_id(foreign key to Table A) ip_address date Basically Table B contains a row for each time a user views a row from Table A. My question is how do I sort Table A results, based on the number of matching rows in Table B. i.e SELECT * FROM TableA ORDER BY (SELECT COUNT(*) FROM TableB where TableB.a_id = TableA.id) Thank you!

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  • MYSQL - Query to check against other table (hard to explain...)

    - by Sam
    I have a query that gets a list of emails who have subscribed for a newsletter trial which lasts 30 days.. $thirty = time() - 3024000; SELECT c.email FROM tbl_clients AS c JOIN tbl_clientoptions AS o ON o.client = c.id WHERE o.option = 'newsletter' AND c.datecreated $thirty What I want to do is do a check in that same query so it also returns clients OVER 30 days old if they have the tbl_clientoptions.option = 'trialoverride' (ie; a row in the client options table with the value "trialoverride") basic columns are: TBL_CLIENTS id,name,email,datecreated TBL_CLIENTOPTIONS id,client,option

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  • MySQL Order By Problem, Why is 1000 being seen as Smaller than 2?

    - by Jack
    I have a strange problem, I am trying to order the output of a set of records by a field called displayOrder. Now even though record A has a displayOrder of 2 and record B has a displayOrder of 1000, record B still shows up before record A. Here's my select statement: SELECT * FROM items ORDER BY displayOrder ASC It works fine until I have a record greater than 9, then 10, 11, 12, etc are seen as smaller than 2, 3, 4 because they start with the number 1. Any way to fox this?

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  • Adding A New Row Using SQL Server Management Studio

    - by Soo
    I'm learning how to use SQL Server Management Studio and can't figure out how to insert a new row into a table. Table Structure: ID, Field1, Field2 Query: INSERT INTO Table (Field1,Field2) VALUES(1,2) Error: Major Error 0x80040E14, Minor Error 25503 I'm probably missing something very noobie like. Any help would be appreciated.

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  • MySQL grouping by a previously declared alias, what do I wrap it in? ' OR `

    - by cgmojoco
    I have an SQL query that has an alias in the SELECT statement SELECT CONCAT(YEAR(r.Date),_utf8'-',_utf8'Q',QUARTER(r.Date)) AS 'QuarterYear' Later, I want to refer to this in my group by statement. I'm a little confused...should I wrap this with backticks, single quote or just leave it unwrapped int he group by GROUP BY `QuarterYear ` or should I do this?: GROUP BY 'QuarterYear' or just this?: GROUP BY QuarterYear

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  • MySql paging; "Showing result-set" of "total found" help

    - by Camran
    I need a formula for showing results on my classifieds website. I am now done with the paging of records, but this formula for showing results remains. I want it like this: Showing 1-50 of 123 found. Now what is the formula for this? I have these variables which should be enough I think: $results_per_page = 50; //results per page $page = 1; //current page Also a variable called $num_total contains the total nr of hits, in this case 123. Thanks

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  • MySQL is there a Single Select to Query Various Unrelated Values from a database?

    - by zzapper
    I saw somewhere what seemed to be nested selects, one "master" select on the "outside" and a series of selects inside- is this possible? I'm not talking about joins as there is particular relation between the selects. I seem not to be explaining myself very well. I want to do a single query which will pull out a series of stats from various tables latest order, latest customer, largest order. Obviously I can do that with a series of selects. The example I saw was something like select ( select ... from tbl_1 where .., select ... from tbl_2 where .., select ... from tbl_3 where .., ... )

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  • A way to search form table in MySQL database.

    - by Eugene
    I looked for a way to scan database for a specific table. For example i have: Database: system_ultimate Table: system_settings And let us say, that one doesn't know precise name of the table. He only knows, that it is some how connected to word settings. How could he search for that table name then? I understand, that usually people who develop know, what they develop, but I'm trying to get hang of MVC and I'm trying to stay as far away as possible from direct communication with table using the name. I know, that to see all tables I could use SHOW TABLES;

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  • mysql phpmyadmin, how to save date of birth that was attained through a form with the input type "se

    - by newbie-25
    Hello, I have a form which asks the user his or her date of birth. Its a drop down box since I use the select input type. The first drop down has Months, second has Days, the third has the Years. So the format should be Month(October), Day(25), Year(1990). In other words the months spell out the actual month and not simply the number representation of the month. However, I would like to save the Day and Year in their respective formats. How can I do this in phpMyAdmin? Should I use the DATE type? what should the value be? I just need to know how to correlate all of this in phpmyadmin, so I can save it properly. Thank you.

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