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  • checkbox, php and update mysql!

    - by Ronnie Chester Lynwood
    Hello now I got this form. i get values with "while": mysql_query("select * from mp3 where aktif = '0'"); <form name="form" method="post" action=""> <input type="text" size="10" name="id" value="<?=$haciosman['id']?>" /> <input type="text" name="baslik" value="<?=$haciosman['baslik']?> <textarea name="sarkisozu"><?=$haciosman['sarkisozu']?></textarea> <input type="text" name="a3" value="<?=$haciosman['ekleyen']?"> <input type="checkbox" name="onay[]" /> <input type="submit" name="0" id="0" value="Onayla" /> <form> and updating values with: <? if (isset($_POST['onay'])) { $cikti = mysql_query("update mp3 set aktif = '1', baslik = '$_POST[baslik]' where id = '$_POST[id]'"); if ($cikti) { echo "islem tamam"; exit; } } ?> but this code only updating only one value. how can i let it update multiple values?

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  • mysql data type confusion

    - by zen
    So this is more of a generalized question about MySQLs data types. I'd like to store a 5-digit US zip code (zip_code) properly in this example. A county has 10 different cities and 5 different zip codes. city | zip code -------+---------- city 0 | 33333 city 1 | 11111 city 2 | 22222 city 3 | 33333 city 4 | 44444 city 5 | 55555 city 6 | 33333 city 7 | 33333 city 8 | 44444 city 9 | 22222 I would typically structure a table like this as varchar(50), int(5) and not think twice about it. (1) If we wanted to ensure that this table had only one of 5 different zip codes we should use the enum data type, right? Now think of a similar scenario on a much larger scale. In a state, there are five-hundred cities with 418 different zip codes. (2) Should I store 418 zip codes as an enum data type OR as an int and create another table to reference?

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  • MySQL Subquery LIMIT

    - by atif089
    As the title says, I wanted a workaround for this... SELECT comments.comment_id, comments.content_id, comments.user_id, comments.`comment`, comments.comment_time, NULL FROM comments WHERE (comments.content_id IN (SELECT content.content_id FROM content WHERE content.user_id = 1 LIMIT 0, 10)) Cheers

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  • mysql select from multiple table

    - by Loon Yew
    I have 3 tables with values like below tbl_product recID pID price colour 1 BDPLA-0001 1.23 White 2 BDPLA-0002 2.23 Black 3 BDPLA-0003 2.28 Blue tbl_product_size recID pID size stock 1 1 2.0cm 10 2 1 3.0cm 20 3 2 2.5cm 30 4 3 3.6cm 40 5 3 3.8cm 50 tbl_order_details recID pID quantity size 201 BDPLA-0001 5 2.0cm 202 BDPLA-0002 10 2.5cm tbl_product.recID = tbl_product_size.pID tbl_product.pID = tbl_order_details.pID how can i combine the tables and produce result like this pID size stock quantity price BDPLA-0001 2.0cm 10 5 1.23 BDPLA-0001 3.0cm 20 null 1.23 BDPLA-0002 2.5cm 30 10 2.23 BDPLA-0003 3.6cm 40 null 2.28 BDPLA-0003 3.8cm 50 null 2.28

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  • Retrieve part of a MySQL column with PHP

    - by Gerardo Marset
    For instance, if I have the following table: +----+---+----------+ | id | a | position | +----+---+----------+ | 0 | 0 | 0 | | 1 | 0 | 1 | | 2 | 1 | 4 | | 3 | 1 | 9 | | 4 | 1 | 6 | | 5 | 1 | 1 | +----+---+----------+ and I want to get an array that contains the first 100 values from position where a is 1 in ascending order, what would I do? Im guessing something like this: $col = mysql_fetch_array( mysql_query(' SELECT `position` FROM `table` WHERE `a`="1" ORDER BY `position` ASC LIMIT 100 ')); I'd expect to get the following array: +-------+-------+ | index | value | +-------+-------+ | 0 | 1 | | 1 | 4 | | 2 | 6 | | 3 | 9 | +-------+-------+ but it doesn't work. ¿What should I do to make it work? Thanks

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  • Resuming MySQL indexing

    - by gmemon
    Hello All, I have been building index on a 200 million row table for almost 14 hours. Due to resource over-consumption on the machine (because of a separate incident), the machine cashed. Clearly, I want to avoid another 14 hours to re-construct the index. Is there a way that I can resume the construction of index from the point (or slightly back) where the machine crashed? I can see the temporary files created. Thanks

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  • Membership with Mysql, EF 1 and ASP.NET 3.5

    - by sanfra1983
    Hi, I created a web application with asp.net 3.5 and ado.net entity framework WebForms 1, but have not yet succeeded in creating a memebrship and roles. When I go on ASP.NET Configuration and click the Security Tab I get the following error: Keyword not supported. Parameter name: metadata Someone has already created an application with these same features to help me understand where is the problem? P.S.: I'm going crazy Thanks to all

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  • PHP, MySQL - My own version of SALT (I call salty) - Login Issue

    - by Fabio Anselmo
    Ok I wrote my own version of SALT I call it salty lol don't make fun of me.. Anyway the registration part of my script as follows is working 100% correctly. //generate SALTY my own version of SALT and I likes me salt.. lol function rand_string( $length ) { $chars = "ABCDEFGHIJKLMNOPQRSTUWXYZabcdefghijklmnopqrstuwxyz1234567890"; $size = strlen( $chars ); for( $i = 0; $i < $length; $i++ ) { $str .= $chars[ rand( 0, $size - 1 ) ]; } return $str; } $salty = rand_string( 256 ); //generate my extra salty pw $password = crypt('password'); $hash = $password . $salty; $newpass = $hash; //insert the data in the database include ('../../scripts/dbconnect.php'); //Update db record with my salty pw ;) // TESTED WITH AND WITHOUT SALTY //HENCE $password and $newpass mysql_query("UPDATE `Register` SET `Password` = '$password' WHERE `emailinput` = '$email'"); mysql_close($connect); However my LOGIN script is failing. I have it setup to TEST and echo if its login or not. It always returns FAILED. I entered the DB and changed the crypted salty pw to "TEST" and I got a SUCCESS. So my problem is somewhere in this LOGIN script I assume. Now I am not sure how to implement my $Salty in this. But also be advised that even without SALTY (just using crypt to store my pass) - I was still unable to perform a login successfully. And if you're gonna suggest i use blowfish - note that my webhost doesn't have it supported and i don't know how to install it. here's my login script: if (isset($_POST['formsubmitted'])) { include ('../../scripts/dbconnect.php'); $username = mysql_real_escape_string($_POST['username']); $password = crypt(mysql_real_escape_string($_POST['password'])); $qry = "SELECT ID FROM Register WHERE emailinput='$username' AND Password='$password'"; $result = mysql_query($qry); if(mysql_num_rows($result) > 0) { echo 'SUCCESS'; //START SESSION } else { echo 'FAILED'; //YOU ARE NOT LOGGED IN } } So what's wrong with this login? Why isn't it working just using the crypt/storing only crypt? How can i make it work storing both the crypt and randomly generated SALTY :) ? Ty advance

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  • SQL Server query problem

    - by user335160
    I want to achieved the results shown in the attached image. The Table Structure and Data are the ffg below: Table Relationship Overall IB Limit->one to many-> Facility Limit Facility Limit->one to many-> Facility Sub Limit Tables Structure and Data Overall IB Limit Id SCAF Reference Approval Date 1 NEW-001 January 1, 2011 2 NEW-002 January 2, 2011 3 NEW-003 January 3, 2011 ---------------------------- Facility Limit Id OverallIBLimitId Product Type 1 1 RPA 2 1 CG 3 2 RPA 4 3 CG ---------------------------- Facility Sub Limit Id FacilityLimitId Sub-Limit Type Amount Tenor Status Status Date 1 1 RPA at max 2,000,0000.00 2 months Approved January 5, 2011 2 1 Oil 3,000,0000.00 3 yrs Approved January 5, 2011 3 2 CG at minor 4,000,0000.00 1 yr Approved January 5, 2011 4 2 CG at max 5,000,0000.00 6 months Approved January 5, 2011 5 2 Flood Component 1 5,000,0000.00 6 months Approved January 5, 2011 6 2 Flood Component 2 6,000,0000.00 3 yrs Approved January 5, 2011 7 3 RPA at minor 1,000,0000.00 6 months Approved January 5, 2011 8 4 One-Off 1,000,0000.00 6 months Approved January 5, 2011

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  • Tree data in MySql database table

    - by Robert Koritnik
    I have a table that uses Adjacency list model for hierarchy storage. My most relevant columns in this table are therefore: ItemId // is auto_increment ParentId Level ParentTrail // in the form of "parentId/../parentId/itemId" then I created a before insert tigger, that populates columns Level and ParentTrail. Since the last column also includes current item's ID I had to use a trick in my trigger because auto_increment columns are not available in the before insert trigger. So I get that value from the information_schema.tables table. All works fine, until I try to write an update trigger, that would update my item and its descendants when the item changes its parent (ParentId has changed). But I can't make an update on my table inside the update trigger. All I can do is to change current record's values but not other's. I could use a separate table for hierarchy data, but that would mean that I would also have to create a view that would combine these two tables (1:1 relation) and I would like to avoid this is at all possible. Is there a way to have all these in the same table so that these fields (Level and ParetTrail) set/update themselves automagically using triggers?

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  • mysql display each day in a month

    - by Jason
    during a month, display the infor each date, order by date, but this infor is empty in some day. how can i still display each day as a row? Product date ----------------- 20 2008-01-01 10 2008-01-02 20 2008-01-03 10 2008-01-05 09 2008-01-08 30 2008-01-09 result: Product date ----------------- 20 2008-01-01 10 2008-01-02 20 2008-01-03 0 2008-01-04 10 2008-01-05 0 2008-01-06 0 2008-01-07 09 2008-01-08 30 2008-01-09

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  • Mysql partition error?

    - by drake
    I have a kinda table like this: CREATE TABLE test ( id MEDIUMINT NOT NULL AUTO_INCREMENT, user VARCHAR(30), time VARCHAR(30), status VARCHAR(30), origin VARCHAR(30), PRIMARY KEY (id) ) ENGINE=MyISAM; PARTITION BY RANGE(id) ( PARTITION p0 VALUES LESS THAN (500000), PARTITION p1 VALUES LESS THAN (1000000), PARTITION p2 VALUES LESS THAN (1500000), PARTITION p3 VALUES LESS THAN (2000000), PARTITION p4 VALUES LESS THAN (2500000) ) I have Three questions: 1) I have here an #1064 error; 2) How can i set split test.user as alphabetic range in partition; 3) How can i check that the partition was successful;

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  • Creating a [materialised]view from generic data in Oracle/Mysql

    - by Andrew White
    I have a generic datamodel with 3 tables CREATE TABLE Properties ( propertyId int(11) NOT NULL AUTO_INCREMENT, name varchar(80) NOT NULL ) CREATE TABLE Customers ( customerId int(11) NOT NULL AUTO_INCREMENT, customerName varchar(80) NOT NULL ) CREATE TABLE PropertyValues ( propertyId int(11) NOT NULL, customerId int(11) NOT NULL, value varchar(80) NOT NULL ) INSERT INTO Properties VALUES (1, 'Age'); INSERT INTO Properties VALUES (2, 'Weight'); INSERT INTO Customers VALUES (1, 'Bob'); INSERT INTO Customers VALUES (2, 'Tom'); INSERT INTO PropertyValues VALUES (1, 1, '34'); INSERT INTO PropertyValues VALUES (2, 1, '80KG'); INSERT INTO PropertyValues VALUES (1, 2, '24'); INSERT INTO PropertyValues VALUES (2, 2, '53KG'); What I would like to do is create a view that has as columns all the ROWS in Properties and has as rows the entries in Customers. The column values are populated from PropertyValues. e.g. customerId Age Weight 1 34 80KG 2 24 53KG I'm thinking I need a stored procedure to do this and perhaps a materialised view (the entries in the table "Properties" change rarely). Any tips?

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  • Need help with a SQL CTE Query

    - by Chuck
    I have a table that I need to get some specific data from for a view. Here's the base table structure with some sample data: | UserID | ReportsToUserID | Org ID | ------------------------------------- | 1 | NULL | 1 | ------------------------------------- | 2 | 1 | 1 | ------------------------------------- | 3 | 2 | 1 | ------------------------------------- | 4 | 3 | 1 | ------------------------------------- The users will be entering reports and users can see the reports of users who report to them and any users who report to those users. Users who report to no one can see everything in their organization Given my sample data above, user 1 can see the reports of 2, 3, & 4; user 2 can see the reports of 3 & 4; and user 3 can see the reports of 4. For the view, I'd like to have the data returned as follows: | UserID | CanSeeUserID | OrgID | -------------------------------------------- | 1 | 2 | 1 | -------------------------------------------- | 1 | 3 | 1 | -------------------------------------------- | 1 | 4 | 1 | -------------------------------------------- | 2 | 3 | 1 | -------------------------------------------- etc... Below is my current code, any help is greatly appreciated. WITH CTEUsers (UserID, CanSeeUserID, OrgID) AS ( SELECT e.ID, e.ReportsToUserID, e.OrgID FROM Users e WITH(NOLOCK) WHERE COALESCE(ReportsToUserID,0) = 0 --ReportsToUserID can be NULL or 0 UNION ALL SELECT e.ReportsToUserID, e.ID,e.OrgID FROM Users e WITH(NOLOCK) JOIN CTEUsers c ON e.ID = c.UserID ) SELECT * FROM CTEUsers

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  • how i group do this mysql query

    - by moustafa
    i want to make charts system and i think it must be like that 1 jan 2009 = 10 post 2 jan 2009 = 2 post 4 jan 2009 = 10 post 6 jan 2009 = 60 post and i have posts table that has id,user_id,date how i can select from posts to show it like that

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  • How can I perform this query between related tables without using UNION?

    - by jeremy
    Suppose I have two separate tables that I watch to query. Both of these tables has a relation with a third table. How can I query both tables with a single, non UNION based query? I want the result of the search to rank the results by comparing a field on each table. Here's a theoretical example. I have a User table. That User can have both CDs and books. I want to find all of that user's books and CDs with a single query matching a string ("awesome" in this example). A UNION based query might look like this: SELECT "book" AS model, name, ranking FROM book WHERE name LIKE 'Awesome%' UNION SELECT "cd" AS model, name, ranking FROM cd WHERE name LIKE 'Awesome%' ORDER BY ranking DESC How can I perform a query like this without the UNION? If I do a simple left join from User to Books and CDs, we end up with a total number of results equal to the number of matching cds timse the number of matching books. Is there a GROUP BY or some other way of writing the query to fix this?

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  • MySQL Database Design with Internationalization

    - by Some name
    Hello, I'm going to start work on a medium sized application, and i'm planning it's db design. One thing that I'm not sure about is this. I will have many tables which will need internationalization, such as: "membership_options, gender_options, language_options etc" Each of these tables will share common i18n fields, like: "title, alternative_title, short_description, description" In your opinion which is the best way to do it? Have an i18n table with the same fields for each of the tables that will need them? or do something like: Membership table Gender table ---------------- -------------- id | created_at id | created_at 1 - 22.03.2001 1 - 14.08.2002 2 - 22.03.2001 2 - 14.08.2002 General translation table ------------------------- record_id | table_name | string_name | alternative_title| .... |id_language 1 - membership regular null 1 (english) 1 - membership normale null 2 (italian) 1 - gender man null 1(english) 1 -gender uomo null 2(italian) This would avoid me repeating something like: membership_translation table ----------------------------- membership_id | name | alternative_title | id_lang 1 regular null 1 1 normale null 2 gender_translation table ----------------------------- gender_id | name | alternative_title | id_lang 1 man null 1 1 uomo null 2 and so on, so i would probably reduce the number of db tables, but i'm not sure about performance.I'm not much of a DB designer, so please let me know.

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  • Help with a MySQL SELECT WHERE Clause

    - by Dr. DOT
    A column in my table contains email addresses. I have a text string that contains the a few usernames of email addresses separated by commas. I can make text sting into an array if necessary to get my SELECT WHERE clause to work correctly. Text string search argument is 'bob,sally,steve' I want to produce a WHERE clause that only returns rows where the username portion of the email address in the table matches one of the usernames in my text string search argument. Thus a row with [email protected] would not be returned but [email protected] would be. Does anyone have a WHERE clause sample that produces this result? Thanks.

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  • MySQL foreign key constraint disappearing

    - by Bramjam
    This is my table: /* oefenreeks leerplan */ CREATE TABLE leerplan_oefenreeks ( leerplan_oefenreeks_id INT PRIMARY KEY AUTO_INCREMENT NOT NULL, leerplan_id INT NOT NULL, oefenreeks_id INT NOT NULL, plaats INT NOT NULL ); /* fk */ ALTER TABLE leerplan_oefenreeks ADD CONSTRAINT fk_leerp_oefenr_leerplan FOREIGN KEY(leerplan_id) REFERENCES leerplan (leerplan_id) ON DELETE CASCADE; ALTER TABLE leerplan_oefenreeks ADD CONSTRAINT fk_leerp_oefenr_oefenreeks FOREIGN KEY(oefenreeks_id) REFERENCES oefenreeks (oefenreeks_id) ON DELETE CASCADE; /* when I execute the nexline, my fk_leerp_oefenr_leerplan constraint vanishes/disappears*/ ALTER TABLE leerplan_oefenreeks ADD CONSTRAINT un_leerp_oefenr UNIQUE(leerplan_id, oefenreeks_id); ALTER TABLE leerplan_oefenreeks ADD CONSTRAINT un_leerp_oefenr_plaats UNIQUE(leerplan_id, plaats); When I go and check only 3 constraints exist. fk_leerp_oefenr_leerplan disappears. I don't understand why this happens.

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