Search Results

Search found 17980 results on 720 pages for 'mod auth mysql'.

Page 184/720 | < Previous Page | 180 181 182 183 184 185 186 187 188 189 190 191  | Next Page >

  • What is wrong with this trigger in mysql?

    - by Jimit
    Hi all, Below is trigger that I need to create but It is not getting created.Please any buddy can explain me what is wrong with this trigger ? Help me please. DELIMITER $$ CREATE TRIGGER property_history_update AFTER UPDATE ON `properties` FOR EACH ROW BEGIN IF OLD.ListPrice != NEW.ListPrice THEN INSERT INTO `property_history` SET ListingKey=OLD.ListingKey,ListPrice = NEW.ListPrice, ListingStatus = OLD.ListingStatus,LastUpdatedTime = NEW.LocalLastModifiedOn; END IF; END$$ DELIMITER ;

    Read the article

  • 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.

    Read the article

  • PHP/Mysql issues

    - by queryne
    Php/my sql newbie question. I have a database I've imported into my local phpmyadmin. However it seems I can't access it from my a php application. The connection string seems right and when i try to authenticate user credentials to access database information, no problems. However authenticate everyone and knows when i put in fake credentials. Still it won't pull any other information from the database. For instance, once a users login they should see something like, "Hello username"... that kind of thing. At this point I see "Hello" without the username. Any ideas what i might be missing?

    Read the article

  • MySQL: Count occurrences of known (or enumerated) distinct values

    - by Eilidh
    After looking at how to count the occurrences of distinct values in a field, I am wondering how to count the occurrences of each distinct value if the distinct values are known (or enumerated). For example, if I have a simple table - TrafficLight Colour ------------ ------ 1 Red 2 Amber 3 Red 4 Red 5 Green 6 Green where one column (in this case Colour) has known (or enumerated) distinct values, how could I return the count for each colour as a separate value, rather than as an array, as in the linked example. To return an array with a count of each colour (using the same method as in the linked example), the query would be something like SELECT Colour COUNT(*) AS ColourCount FROM TrafficLights GROUP BY Colour, and return an array - Colour ColourCount ------ ----------- Red 3 Amber 1 Green 2 What I would like to do is to return the count for each Colour AS a separate total (e.g. RedCount). How can I do this?

    Read the article

  • Testing + production server and syncing MySQL data

    - by Matthew
    I have a web application running on LAMP with a testing server and a production server. Is there a standard practice for keeping the data on the testing server in sync with the production server? The data on the testing server gets out of date pretty quick and I feel like there must be an easier way than just dumping the production server and copying it onto the testing server every so often. It's not important that the data is in total sync, just that the testing server represents the production enviornment as accurately as possible.

    Read the article

  • Generating Unordered List with PHP + CodeIgniter from a MySQL Database

    - by Tim
    Hello Everyone, I am trying to build a dynamically generated unordered list in the following format using PHP. I am using CodeIgniter but it can just be normal php. This is the end output I need to achieve. <ul id="categories" class="menu"> <li rel="1"> Arts &amp; Humanities <ul> <li rel="2"> Photography <ul> <li rel="3"> 3D </li> <li rel="4"> Digital </li> </ul> </li> <li rel="5"> History </li> <li rel="6"> Literature </li> </ul> </li> <li rel="7"> Business &amp; Economy </li> <li rel="8"> Computers &amp; Internet </li> <li rel="9"> Education </li> <li rel="11"> Entertainment <ul> <li rel="12"> Movies </li> <li rel="13"> TV Shows </li> <li rel="14"> Music </li> <li rel="15"> Humor </li> </ul> </li> <li rel="10"> Health </li> And here is my SQL that I have to work with. -- -- Table structure for table `categories` -- CREATE TABLE IF NOT EXISTS `categories` ( `id` mediumint(8) NOT NULL auto_increment, `dd_id` mediumint(8) NOT NULL, `parent_id` mediumint(8) NOT NULL, `cat_name` varchar(256) NOT NULL, `cat_order` smallint(4) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ; So I know that I am going to need at least 1 foreach loop to generate the first level of categories. What I don't know is how to iterate inside each loop and check for parents and do that in a dynamic way so that there could be an endless tree of children. Thanks for any help you can offer. Tim

    Read the article

  • Strange behavior of MySQL UPDATE query in PHP?

    - by Prashant
    When I am executing following query then its not updating views column by 1 instead sometimes its updating it by 2 or 3. Say currently views count is 24 then after executing this query it becomes 26 or sometimes its 27. $views = $views + 1; $_SQL = ''; $_SQL = 'UPDATE videos SET views = '.$views.' WHERE VideoId= "'.$videoid.'";'; @mysql_query($_SQL); I am not getting why this is happening, am I missing something or the query is executing 2 times automatically? Please help me to figure out the issue. Thanks

    Read the article

  • How to build unlimited level of menu through PHP and mysql

    - by Starx
    Well, to build my menu my menu I use a db similar structure like this To assign another submenu for existing submenu I simply assign its parent's id as its value of parent field. parent 0 means top menu now there is not problem while creating submenu inside another submenu now this is way I fetch the submenu for the top menu <ul class="topmenu"> <? $list = $obj -> childmenu($parentid); //this list contains the array of submenu under $parendid foreach($list as $menu) { extract($menu); echo '<li><a href="#">'.$name.'</a></li>'; } ?> </ul> What I want to do is. I want to check if a new menu has other child menu and I want to keep on checking until it searches every child menu that is available and I want to display its child menu inside its particular list item like this Home ........

    Read the article

  • generic Mysql stored procedure

    - by psu
    Hi, I have the fallowing stored procedure: CREATE PROCEDURE `get`(IN tb VARCHAR(50), IN id INTEGER) BEGIN SELECT * FROM tb WHERE Indx = id; END// When I call get(user,1) I get the following: ERROR 1054 (42S22): Unknown column 'user' in 'field list'

    Read the article

  • MySQL parameter resource error

    - by Derek
    Here is my error: Warning: mysql_query() expects parameter 2 to be resource, null given... This refers to line 23 of my code which is: $result = mysql_query($sql, $connection) My entire query code looks like this: $query = "SELECT * from users WHERE userid='".intval( $_SESSION['SESS_USERID'] )."'"; $result = mysql_query($query, $connection) or die ("Couldn't perform query $query <br />".mysql_error()); $row = mysql_fetch_array($result); I don't have a clue what has happpened here. All I wanted to do was to have the value of the users 'fullname' displayed in the header section of my web page. So I am outputting this code immediately after to try and achieve this: echo 'Hello '; echo $row['fullname']; Before this change, I had it working perfectly, where the session variable of fullname was echoed $_SESSION['SESS_NAME']. However, because my user can update their information (including their name), I wanted the name displayed in the header to be updated accordingly, and not displaying the session value.

    Read the article

  • (mySQL) Unable to query 2 tables properly for data

    - by Devner
    I have 2 tables. One is 'page_links' and the other is 'rpp'. Table page_links is the superset of table rpp. The following is the schema of my tables: -- Table structure for table `page_links` -- CREATE TABLE IF NOT EXISTS `page_links` ( `page` varchar(255) NOT NULL, `page_link` varchar(100) NOT NULL, `heading_id` tinyint(3) unsigned NOT NULL, PRIMARY KEY (`page`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1; -- -- Dumping data for table `page_links` -- INSERT INTO `page_links` (`page`, `page_link`, `heading_id`) VALUES ('a1.php', 'A1', 8), ('b1.php', 'B1', 8), ('c1.php', 'C1', 5), ('d1.php', 'D1', 5), ('e1.php', 'E1', 8), ('f1.php', 'F1', 8), ('g1.php', 'G1', 8), ('h1.php', 'H1', 1), ('i1.php', 'I1', 1), ('j1.php', 'J1', 8), ('k1.php', 'K1', 8), ('l1.php', 'L1', 8), ('m1.php', 'M1', 8), ('n1.php', 'N1', 8), ('o1.php', 'O1', 8), ('p1.php', 'P1', 4), ('q1.php', 'Q1', 5), ('r1.php', 'R1', 4); -- Table structure for table `rpp` -- CREATE TABLE IF NOT EXISTS `rpp` ( `role_id` tinyint(3) unsigned NOT NULL, `page` varchar(255) NOT NULL, `is_allowed` tinyint(1) NOT NULL, PRIMARY KEY (`role_id`,`page`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1; -- -- Dumping data for table `rpp` -- INSERT INTO `rpp` (`role_id`, `page`, `is_allowed`) VALUES (3, 'a1.php', 1), (3, 'b1.php', 1), (3, 'c1.php', 1), (3, 'd1.php', 1), (3, 'e1.php', 1), (3, 'f1.php', 1), (3, 'h1.php', 1), (3, 'i1.php', 1), (3, 'l1.php', 1), (3, 'm1.php', 1), (3, 'n1.php', 1), (4, 'a1.php', 1), (4, 'b1.php', 1), (4, 'q1.php', 1), (5, 'r1.php', 1); WHAT I AM TRYING TO DO: I am trying to query both the above tables (in a single query) in such a way that all the pages from page_links are displayed along with the is_allowed value from rpp for a particular role. For example, I want to get the is_allowed value of all the pages from rpp for role_id = 3 and at the same time, list all the available pages from page_links. A clear example of my expected result would be: page is_allowed role_id ---------------------------------------- a1.php 1 3 b1.php 1 3 c1.php 1 3 d1.php 1 3 e1.php 1 3 f1.php 1 3 g1.php NULL NULL h1.php 1 3 i1.php 1 3 j1.php NULL NULL k1.php NULL NULL l1.php 1 3 m1.php 1 3 n1.php 1 3 o1.php NULL NULL p1.php NULL NULL q1.php NULL NULL r1.php NULL NULL One more example of my desired result could be achieved by doing a LEFT JOIN rpp ON page_links.page = rpp.page but we need to omit using role_id = 3 (or any value) to be able to get that. But I do want to specify the role_id as well and get the results. I need the query to be able to get this result. I would appreciate any replies that could help me with this. If you can suggest me any changes as well to the table(s) design to be able to achieve the desired result, that's good as well. Thanks in advance.

    Read the article

  • Mysql SQL join question

    - by David
    I am trying to find all deals information along with how many comments they have received. My query select deals.*, count(comments.comments_id) as counts from deals left join comments on comments.deal_id=deals.deal_id where cancelled='N' But now it only shows the deals that have at least one comment. What is the problem?

    Read the article

  • Why does this MySQL Query hang?

    - by zzapper
    SELECT * FROM tbl_order_head AS o INNER JOIN tbl_orders_log AS c ON o.PAYMENT_TRANSACTION_LOG_ID=c.TRANSACTION_ID WHERE o.VISUAL_ID = '77783'; tbl_order_head 67,000 (30 fields) records, tbl_orders_log 17000 (5 fields) records. I don't know if it would eventually return as I am running it on a live server and fear overloading. I am doing similar queries and much more complex queries successfully.

    Read the article

  • 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.

    Read the article

  • Mysql - multiple values and between

    - by realshadow
    Hey, I need to select 10 products and display them. Each product has 3 different prices. The select to get the prices looks like this: SELECT * FROM products_loans WHERE CODE IN('10X15/12', '10X15/Q10-10', '10X15/Q20-10') AND 550 BETWEEN PRICE_FROM AND PRICE_TO; Where 550 is the base price. Now this select returns 3 rows, but I want to modify it so it will return 30 results. I dont like the idea to execute 10 queries at once. I know I can easily achieve that with "OR", but I would like to ask if there is some other more elegant way to this. The select "should" look like this: SELECT * FROM products_loans WHERE KOD IN('10X15/12', '10X15/Q10-10', '10X15/Q20-10') AND (550, 325, 780) BETWEEN CENA_OD AND CENA_DO; Note that there is no "price" column or anything in the table which I could use to do a JOIN and I cant modify the table.

    Read the article

  • Java/MySQL: Working with data in classes

    - by skiwi
    What is the best way to deal with accessing/modifying tables in a database? I have read about the Data Access Object approach, but none of the resources I have found so far indicate a clear implementation of it. So assume you have a database with a table called accounts that has columns id, name, password and email. How would you properly access it within Java? I mean most people know how to do SQL statements, but that is not really the point. I hope people here can be of help. Regards.

    Read the article

  • Mysql - Grouping the result based on a mathematical operation and SUM() function

    - by SpikETidE
    Hi all... I'm having the following two tables... Table : room_type type_id type_name no_of_rooms max_guests rate 1 Type 1 15 2 1254 2 Type 2 10 1 3025 Table : reservation reservation_id start_date end_date room_type booked_rooms 1 2010-04-12 2010-04-15 1 8 2 2010-04-12 2010-04-15 1 2 Now... I have this query SELECT type_id, type_name FROM room_type WHERE id NOT IN (SELECT room_type FROM reservation WHERE start_date >= '$start_date' AND end_date <= '$end_date') What the query does is it selects the rooms that are not booked between the start date and end date. Also, as you can see from the reservation table, we also have 'number of rooms booked between the two dates' factor also... I need to add this 'no.of booked rooms between the two dates' factor also in to the query... The query should return the type of rooms for which at least one room is free between the two dates. I worked out the logic but just can't represent it as a query....! How will you do this...? Thanks for your suggestions..!

    Read the article

  • define mysql indexing

    - by Bharanikumar
    Hi Am not sure, This is the right place to post this question , But in our stack overflow only am getting clear vision solutions , What is indexing and what is fulltext , for the above both questions i know the ans, but i cant expose that ans in the exact way to the interviewer , (indexing means somthing like index in book) (fulltext means for search string), Can please give me very simple defination for this questions , Advance thanks

    Read the article

  • PHP/mySQL - using result from 'CONCAT' and 'AS' in 'LIKE' clause

    - by Phil Jackson
    Hi I have the following code; if( ! empty( $post['search-bar'] ) ) { $search_data = preg_replace("#\s\s#is", '', preg_replace("#[^\w\d\s+]#is", '', $post['search-bar'] ) ); $data_array = explode( " ", $search_data ); $data_array = "'%" . implode( "%' OR '%", $data_array ) . "%'"; $query = "SELECT CONCAT( PROFILE_PROFFESION, FIRST_NAME, LAST_NAME, DISPLAY_NAME) AS 'STRING' FROM `" . ACCOUNT_TABLE . "` WHERE STRING LIKE ( " . $data_array . " ) AND BUSINESS_POST_CODE LIKE '" . substr(P_BUSINESS_POST_CODE, 0, 4) . "%'"; $q = mysql_query( $query, $CON ) or die( "_error_" . mysql_error() ); if( mysql_num_rows( $q ) != 0 ) { die(); } } Problem is I want to use the temp col 'STRING' in the where clause but is returning 'unknown coloumn STRING Can any one point me in the right direction, regards Phil

    Read the article

  • MySQL sub query

    - by Juddling
    UPDATE members SET money=money+100 WHERE username IN (SELECT username FROM forum); Lets say I wanted to give each of my members 100 money for each post in my forum. This query works but if one member has posted more than once, they only get 100. Could someone correct this query please?

    Read the article

< Previous Page | 180 181 182 183 184 185 186 187 188 189 190 191  | Next Page >