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  • MySQL pivot tables - covert rows to columns

    - by user2723490
    This is the structure of my table: Then I run a query SELECT `date`,`index_name`,`results` FROM `mst_ind` WHERE `index_name` IN ('MSCI EAFE Mid NR USD', 'Alerian MLP PR USD') AND `time_period`='M1' and get a table like How can I convert "index_name" rows to columns like: date | MSCI EAFE Mid NR USD | Alerian MLP PR USD etc In other words I need each column to represent an index and rows to represent date-result. I understand that MySQL doesn't have pivot table functions. What is the easiest way of doing this? I've tried this code, but it generates an error: SELECT `date`, MAX(IF(index_name = 'Alerian MLP PR USD' AND `time_period`='M1', results, NULL)) AS res1, MAX(IF(index_name = 'MSCI EAFE Mid NR USD' AND `time_period`='M1', results, NULL)) AS res2 FROM `mst_ind` GROUP BY `date I need to make the conversion on the query level - not PHP. Please suggest a nice and elegant solution. Thanks!

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  • How to get table cells evenly spaced?

    - by DaveDev
    I'm trying to create a page with a number of static html tables on them. What do I need to do to get them to display each column the same size as each other column in the table? The HTML is as follows: <span class="Emphasis">Interest rates</span><br /> <table cellpadding="0px" cellspacing="0px" class="PerformanceTable"> <tr><th class="TableHeader"></th><th class="TableHeader">Current rate as at 31 March 2010</th><th class="TableHeader">31 December 2009</th><th class="TableHeader">31 March 2009</th></tr> <tr class="TableRow"><td>Index1</td><td class="PerformanceCell">1.00%</td><td>1.00%</td><td>1.50%</td></tr> <tr class="TableRow"><td>index2</td><td class="PerformanceCell">0.50%</td><td>0.50%</td><td>0.50%</td></tr> <tr class="TableRow"><td>index3</td><td class="PerformanceCell">0.25%</td><td>0.25%</td><td>0.25%</td></tr> </table> <span>Source: Bt</span><br /><br /> <span class="Emphasis">Stock markets</span><br /> <table cellpadding="0px" cellspacing="0px" class="PerformanceTable"> <tr><th class="TableHeader"></th><th class="TableHeader">As at 31 March 2010</th><th class="TableHeader">1 month change</th><th class="TableHeader">QTD change</th><th class="TableHeader">12 months change</th></tr> <tr class="TableRow"><td>index1</td><td class="PerformanceCell">1169.43</td><td class="PerformanceCell">5.88%</td><td class="PerformanceCell">4.87%</td><td class="PerformanceCell">46.57%</td></tr> <tr class="TableRow"><td>index2</td><td class="PerformanceCell">1958.34</td><td class="PerformanceCell">7.68%</td><td class="PerformanceCell">5.27%</td><td class="PerformanceCell">58.31%</td></tr> <tr class="TableRow"><td>index3</td><td class="PerformanceCell">5679.64</td><td class="PerformanceCell">6.07%</td><td class="PerformanceCell">4.93%</td><td class="PerformanceCell">44.66%</td></tr> <tr class="TableRow"><td>index4</td><td class="PerformanceCell">2943.92</td><td class="PerformanceCell">8.30%</td><td class="PerformanceCell">-0.98%</td><td class="PerformanceCell">44.52%</td></tr> <tr class="TableRow"><td>index5</td><td class="PerformanceCell">978.81</td><td class="PerformanceCell">9.47%</td><td class="PerformanceCell">7.85%</td><td class="PerformanceCell">26.52%</td></tr> <tr class="TableRow"><td>index6</td><td class="PerformanceCell">3177.77</td><td class="PerformanceCell">10.58%</td><td class="PerformanceCell">6.82%</td><td class="PerformanceCell">44.84%</td></tr> </table> <span>Source: B</span><br /><br /> I'm also open to suggestion on how to tidy this up, if there are any? :-)

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  • MySQL "ERROR 1005 (HY000): Can't create table 'foo.#sql-12c_4' (errno: 150)"

    - by Ankur Banerjee
    Hi, I was working on creating some tables in database foo, but every time I end up with errno 150 regarding the foreign key. Firstly, here's my code for creating tables: CREATE TABLE Clients ( client_id CHAR(10) NOT NULL , client_name CHAR(50) NOT NULL , provisional_license_num CHAR(50) NOT NULL , client_address CHAR(50) NULL , client_city CHAR(50) NULL , client_county CHAR(50) NULL , client_zip CHAR(10) NULL , client_phone INT NULL , client_email CHAR(255) NULL , client_dob DATETIME NULL , test_attempts INT NULL ); CREATE TABLE Applications ( application_id CHAR(10) NOT NULL , office_id INT NOT NULL , client_id CHAR(10) NOT NULL , instructor_id CHAR(10) NOT NULL , car_id CHAR(10) NOT NULL , application_date DATETIME NULL ); CREATE TABLE Instructors ( instructor_id CHAR(10) NOT NULL , office_id INT NOT NULL , instructor_name CHAR(50) NOT NULL , instructor_address CHAR(50) NULL , instructor_city CHAR(50) NULL , instructor_county CHAR(50) NULL , instructor_zip CHAR(10) NULL , instructor_phone INT NULL , instructor_email CHAR(255) NULL , instructor_dob DATETIME NULL , lessons_given INT NULL ); CREATE TABLE Cars ( car_id CHAR(10) NOT NULL , office_id INT NOT NULL , engine_serial_num CHAR(10) NULL , registration_num CHAR(10) NULL , car_make CHAR(50) NULL , car_model CHAR(50) NULL ); CREATE TABLE Offices ( office_id INT NOT NULL , office_address CHAR(50) NULL , office_city CHAR(50) NULL , office_County CHAR(50) NULL , office_zip CHAR(10) NULL , office_phone INT NULL , office_email CHAR(255) NULL ); CREATE TABLE Lessons ( lesson_num INT NOT NULL , client_id CHAR(10) NOT NULL , date DATETIME NOT NULL , time DATETIME NOT NULL , milegage_used DECIMAL(5, 2) NULL , progress CHAR(50) NULL ); CREATE TABLE DrivingTests ( test_num INT NOT NULL , client_id CHAR(10) NOT NULL , test_date DATETIME NOT NULL , seat_num INT NOT NULL , score INT NULL , test_notes CHAR(255) NULL ); ALTER TABLE Clients ADD PRIMARY KEY (client_id); ALTER TABLE Applications ADD PRIMARY KEY (application_id); ALTER TABLE Instructors ADD PRIMARY KEY (instructor_id); ALTER TABLE Offices ADD PRIMARY KEY (office_id); ALTER TABLE Lessons ADD PRIMARY KEY (lesson_num); ALTER TABLE DrivingTests ADD PRIMARY KEY (test_num); ALTER TABLE Applications ADD CONSTRAINT FK_Applications_Offices FOREIGN KEY (office_id) REFERENCES Offices (office_id); ALTER TABLE Applications ADD CONSTRAINT FK_Applications_Clients FOREIGN KEY (client_id) REFERENCES Clients (client_id); ALTER TABLE Applications ADD CONSTRAINT FK_Applications_Instructors FOREIGN KEY (instructor_id) REFERENCES Instructors (instructor_id); ALTER TABLE Applications ADD CONSTRAINT FK_Applications_Cars FOREIGN KEY (car_id) REFERENCES Cars (car_id); ALTER TABLE Lessons ADD CONSTRAINT FK_Lessons_Clients FOREIGN KEY (client_id) REFERENCES Clients (client_id); ALTER TABLE Cars ADD CONSTRAINT FK_Cars_Offices FOREIGN KEY (office_id) REFERENCES Offices (office_id); ALTER TABLE Clients ADD CONSTRAINT FK_DrivingTests_Clients FOREIGN KEY (client_id) REFERENCES Clients (client_id); These are the errors that I get: mysql> ALTER TABLE Applications ADD CONSTRAINT FK_Applications_Cars FOREIGN KEY (car_id) REFERENCES Cars (car_id); ERROR 1005 (HY000): Can't create table 'foo.#sql-12c_4' (errno: 150) I ran SHOW ENGINE INNODB STATUS which gives a more detailed error description: ------------------------ LATEST FOREIGN KEY ERROR ------------------------ 100509 20:59:49 Error in foreign key constraint of table practice9/#sql-12c_4: FOREIGN KEY (car_id) REFERENCES Cars (car_id): Cannot find an index in the referenced table where the referenced columns appear as the first columns, or column types in the table and the referenced table do not match for constraint. Note that the internal storage type of ENUM and SET changed in tables created with >= InnoDB-4.1.12, and such columns in old tables cannot be referenced by such columns in new tables. See http://dev.mysql.com/doc/refman/5.1/en/innodb-foreign-key-constraints.html for correct foreign key definition. ------------ I searched around on StackOverflow and elsewhere online - came across a helpful blog post here with pointers on how to resolve this error - but I can't figure out what's going wrong. Any help would be appreciated!

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  • random data using php & mysql

    - by Prakash
    I have mysql database structure like below: CREATE TABLE test ( id int(11) NOT NULL auto_increment, title text NULL, tags text NULL, PRIMARY KEY (id) ); data on field tags is stored as a comma separated text like html,php,mysql,website,html etc... now I need create an array that contains around 50 randomly selected tags from random records. currently I am using rand() to select 15 random mysql data from database and then holding all the tags from 15 records in an array. Then I am using array_rand() for randomizing the array and selecting only 50 random records. $query=mysql_query("select * from test order by id asc, RAND() limit 15"); $tags=""; while ($eachData=mysql_fetch_array($query)) { $additionalTags=$eachData['tags']; if ($tags=="") { $tags.=$additionalTags; } else { $tags.=$tags.",".$additionalTags; } } $tags=explode(",", $tags); $newTags=array(); foreach ($tags as $tag) { $tag=trim($tag); if ($tag!="") { if (!in_array($tag, $newTags)) { $newTags[]=$tag; } } } $random_newTags=array_rand($newTags, 50); Now I have huge records on the database, and because of that; rand() is performing very slow and sometimes it doesn't work. So can anyone let me know how to handle this situation correctly so that my page will work normally.

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  • PHP: table structure

    - by A3efan
    I'm developing a website that has some audio courses, each course can have multiple lessons. I want to display each course in its own table with its different lessons. This is my sql statement: Table: courses id, title Table: lessons id, cid (course id), title, date, file $sql = "SELECT lessons.*, courses.title AS course FROM lessons INNER JOIN courses ON courses.id = lessons.cid GROUP BY lessons.id ORDER BY lessons.id" ; Can someone help me with the PHP code? This is the I code I have written: mysql_select_db($database_config, $config); mysql_query("set names utf8"); $sql = "SELECT lessons.*, courses.title AS course FROM lessons INNER JOIN courses ON courses.id = lessons.cid GROUP BY lessons.id ORDER BY lessons.id" ; $result = mysql_query($sql) or die(mysql_error()); while ($row = mysql_fetch_assoc($result)) { echo "<p><span class='heading1'>" . $row['course'] . "</span> </p> "; echo "<p class='datum'>Posted onder <a href='*'>*</a>, latest update on " . strftime("%A %d %B %Y %H:%M", strtotime($row['date'])); } echo "</p>"; echo "<class id='text'>"; echo "<p>...</p>"; echo "<table border: none cellpadding='1' cellspacing='1'>"; echo "<tr>"; echo "<th>Nr.</th>"; echo "<th width='450'>Lesso</th>"; echo "<th>Date</th>"; echo "<th>Download</th>"; echo "</tr>"; echo "<tr>"; echo "<td>" . $row['nr'] . "</td>"; echo "<td>" . $row['title'] . "</td>"; echo "<td>" . strftime("%d/%m/%Y", strtotime($row['date'])) . "</td>"; echo "<td><a href='audio/" . rawurlencode($row['file']) . "'>MP3</a></td>"; echo "</tr>"; echo "</table>"; echo "<br>"; } ?>

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  • Print table data mysql php

    - by Marcelo
    Hi people, i'm having a problem trying to print some data of a table. I'm new at this php mysql stuff but i think my code is right. Here it is: <html> <body> <h1>Lista de usuários</h1> <?php $host="localhost"; // Host name $username="root"; // Mysql username $password=""; // Mysql password $db_name="sabs"; // Database name $tbl_name="doador"; // Table name // Connect to server and select database. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $sql="SELECT * FROM $tbl_name"; $result=mysql_query($sql); while($rows = mysql_fetch_array($result)){ echo $row['id'] . " " .$row['nome'] . " " . $row['sobrenome'] . " " . $row['email'] . " " . $row['login'] . " " . $row['senha'] . " " . $row['idade'] . " ". $row['peso'] . " " . $row['fuma'] . " " . $row['sexo'] . " " . $row['doencas']; echo "<BR/>"; } mysql_close(); ?> </body> </html> All columns of the echo command exist in my table in the database. Don't get why it's not printing those values. Thanks for the attention.

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  • MySQL, An Ideal Choice for The Cloud

    - by Bertrand Matthelié
    As the world's most popular web database, MySQL has quickly become the leading database for the cloud, with most providers offering MySQL-based services. 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:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; 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;} Access our Resource Kit to discover: Why MySQL has become the leading database in the cloud, and how it addresses the critical attributes of cloud-based deployments How ISVs rely on MySQL to power their SaaS offerings Best practices to deploy the world’s most popular open source database in public and private clouds Normal 0 false false false EN-US X-NONE X-NONE You will also find out how you can leverage MySQL together with Hadoop and other technologies to unlock the value of Big Data, either on-premise or in the cloud. Access white papers, webinars, case studies and other resources in /* 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:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; 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;} our Resource Kit now!

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  • How to proceed setting up a secondary mysql linux slave?

    - by Algorist
    I have a mysql database master and slave in production. I want to setup additional mysql slave. There is around 15 Terabyte of data in the database and there are MYISAM and InnoDB tables in the database. I am thinking of below options: Shutdown master database and copy the mysql data folder to secondary slave. Can Innodb tables be copied like this? Run flush table with read lock, scp the file to new slave and unlock the table and this is possible for myisam tables, can I do the same for innodb tables too? Thanks for looking at the question.

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  • How to generate the right password format for Apache2 authentication in use with DBD and MySQL 5.1?

    - by Walkman
    I want to authenticate users for a folder from a MySQL 5.1 database with AuthType Basic. The passwords are stored in plain text (they are not really passwords, so doesn't matter). The password format for apache however only allows for SHA1, MD5 on Linux systems as described here. How could I generate the right format with an SQL query ? Seems like apache format is a binary format with a lenght of 20, but the mysql SHA1 function return 40 long. My SQL query is something like this: SELECT CONCAT('{SHA}', BASE64_ENCODE(SHA1(access_key))) FROM user_access_keys INNER JOIN users ON user_access_keys.user_id = users.id WHERE name = %s where base64_encode is a stored function (Mysql 5.1 doesn't have TO_BASE64 yet). This query returns a 61 byte BLOB which is not the same format that apache uses. How could I generate the same format ? You can suggest other method for this too. The point is that I want to authenticate users from a MySQL5.1 database using plain text as password.

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  • What could cause sudden crash of a MySQL 5.0.67 installation?

    - by Alex R
    I have an old Ubuntu 8.10 32-bit with MySQL 5.0.67. There's 5.7GB of data in it and it grows by about 100MB every day. About 3 days ago, the MySQL instance begin dying suddenly and quitely (no log entry) during the nightly mysqldump. What could be causing it? Upgrading MySQL is a long-term project for me, unless there happens to be a specific bug in 5.0.67 then I guess I'll just need to reprioritize. I'm hoping somebody might be familiar with this problem since this is a fairly popular version bundled with Ubuntu 8.10. Thanks

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  • Update last child id in parent table using mysql

    - by Sam Saffron
    Given the following tables: Topic id, last_updated_child_id Response id, topic_id, updated_at How do I update the Topic table so the last_updated_child_id is equal to the latest response id (based on date). So for example given: Topic id last_updated_child_id -- ----------------------- 1 null 2 null 3 null Response id topic_id updated_at -- ---- ---- 1 1 2010 2 1 2012 3 1 2011 4 2 2000 I would like to execute an UPDATE statement that would result in the Topic table being: id last_updated_child_id -- ----------------------- 1 2 2 4 3 null Note: I would like to avoid temp tables if possible and am happy for a MySQL specific solution.

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  • mysql create table help with unique

    - by Matt
    I'm trying to create a table, and can't figure out how to assign two columns to be unique.. I know how to alter a table thats already created, but how do you do it in the create.. im after a create if not exist col1 TEXT, col2 TEXT, col3 TEXT unique(col1, col2) ^very rough basic but you get the idea

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  • Select random row from MySQL (with probability)

    - by James Simpson
    I have a MySQL table that has a row called cur_odds which is a percent number with the percent probability that that row will get selected. How do I make a query that will actually select the rows in approximately that frequency when you run through 100 queries for example? I tried the following, but a row that has a probability of 0.35 ends up getting selected around 60-70% of the time. SELECT * FROM table ORDER BY RAND()*cur_odds DESC

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  • Create a Cumulative Sum Column in MySQL

    - by Kirk
    I have a table that looks like this: id count 1 100 2 50 3 10 I want to add a new column called cumulative_sum, so the table would look like this: id count cumulative_sum 1 100 100 2 50 150 3 10 160 Is there a MySQL update statement that can do this easily? What's the best way to accomplish this?

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  • Advice on how to complete specific MySQL JOIN

    - by Tim
    Hello, I have a mysql table jobs. This is the basic structure of jobs. id booked_user_id assigned_user_id I then also have another table, meta. Meta has the structure: id user_id first_name last_name How can I join these tables so that both booked_user_id and assigned_user_id can access meta.first_name? Thanks for your advice Tim

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  • MySQL Search Query on two different fields

    - by user181677
    I need to search on two fields using LIKE function and should match also in reverse order. My table uses InnoDB which dont have Full text search. For example, I have users table with first_name and last_name column. I tried this SQL statement but no luck. SELECT CONCAT(first_name, ' ', last_name) as fullname FROM users WHERE fullname LIKE '%doe%';

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  • A SELECT statement for Mysql

    - by Hossein
    I have this table: id,bookmarkID,tagID I want to fetch the top N bookmarkIDs for a given list of tags. Does anyone know a very fast solution for this? the table is quite large(12 million records) I am using MySql

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  • mysql query to get unique value from one column

    - by vesselyp
    i have a table named locations of which i want to select and get values in such a way that it should select only distinct values from a column but select all other values . table name: locations column names 1: country values : America, India, India, India column names 2: state/Province : Newyork, Punjab, Karnataka, kerala when i select i should get India only once and all the three states listed under India . is ther any way..??? sombody please help

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  • Complex MySQL table select/join with pre-condition

    - by Howard
    Hello, I have the schema below CREATE TABLE `vocabulary` ( `vid` int(10) unsigned NOT NULL auto_increment, `name` varchar(255), PRIMARY KEY vid (`vid`) ); CREATE TABLE `term` ( `tid` int(10) unsigned NOT NULL auto_increment, `vid` int(10) unsigned NOT NULL default '0', `name` varchar(255), PRIMARY KEY tid (`tid`) ); CREATE TABLE `article` ( `aid` int(10) unsigned NOT NULL auto_increment, `body` text, PRIMARY KEY aid (`aid`) ); CREATE TABLE `article_index` ( `nid` int(10) unsigned NOT NULL default '0', `tid` int(10) unsigned NOT NULL default '0' ) INSERT INTO `vocabulary` values (1, 'vocabulary 1'); INSERT INTO `vocabulary` values (2, 'vocabulary 2'); INSERT INTO `term` values (1, 1, 'term v1 t1'); INSERT INTO `term` values (2, 1, 'term v1 t2 '); INSERT INTO `term` values (3, 2, 'term v2 t3'); INSERT INTO `term` values (4, 2, 'term v2 t4'); INSERT INTO `term` values (5, 2, 'term v2 t5'); INSERT INTO `article` values (1, ""); INSERT INTO `article` values (2, ""); INSERT INTO `article` values (3, ""); INSERT INTO `article` values (4, ""); INSERT INTO `article` values (5, ""); INSERT INTO `article_index` values (1, 1); INSERT INTO `article_index` values (1, 3); INSERT INTO `article_index` values (2, 2); INSERT INTO `article_index` values (3, 1); INSERT INTO `article_index` values (3, 3); INSERT INTO `article_index` values (4, 3); INSERT INTO `article_index` values (5, 3); INSERT INTO `article_index` values (5, 4); Example. Select term of a defiend vocabulary (with non-zero article index), e.g. vid=2 select a.tid, count(*) as article_count from term t JOIN article_index a ON t.tid = a.tid where t.vid = 2 group by t.tid; +-----+---------------+ | tid | article_count | +-----+---------------+ | 3 | 4 | | 4 | 1 | +-----+------------ Question: Select terms a. of a defiend vocabulary (with non-zero article index, e.g. vid=1 = term {1,2}) b. given that those terms are linked with articles which are linked with terms under vid=2, e.g. = {1}, term with tid=2 is excluded since no linkage to terms under vid=2 SQL: Any idea? Expected result: +-----+---------------+ | tid | article_count | +-----+---------------+ | 1 | 2 | +-----+---------------+

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  • MySql mutliple tables

    - by Chris Harrison
    I've been looking into JOIN, subqueries and other ways of doing this, but I can't work out the best way to do this is... I have a table (ps_category_product): id_product, id_category I want to perform a query on it like: SELECT id_product FROM ps_category_product WHERE id_category='$this_cat' BUT, I only want to perform this query where the ID's are returned by a query on another table (ps_product): id_product, active SELECT id_product FROM ps_product WHERE active='1' Can anyone help me with getting these two queries working together?

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  • Open Source Survey: Oracle Products on Top

    - by trond-arne.undheim
    Oracle continues to work with the open source community to bring the most innovative and productive software to market (more). Oracle products received the most votes in several key categories of the 2010 Linux Journal Reader's Choice Awards. With over 12,000 technologists reporting, these product earned top spots: Best Office Suite: OpenOffice.org Best Single Office Program: OpenOffice.org Writer Best Database: MySQL Best Virtualization Solution: VirtualBox "As the leading open source technology and service provider, Oracle continues to work with the community stakeholders to rapidly innovate many open source products for use in fully tested production environments," says Edward Screven, Oracle's chief corporate architect. "Supporting open source is important to Oracle and our customers, and we continue to invest in it." According to a recent report by the Linux Foundation, Oracle is one of the top ten contributors to the Linux Kernel. Oracle also contributes millions of lines of code to these important projects: OpenJDK: 7,002,579 Eclipse: 1,800,000 (#3 in active committers) MySQL: 5,073,113 NetBeans: 7,870,446 JSF: 701,980 Apache MyFaces Trinidad: 1,316,840 Hudson: 1,209,779 OpenOffice.org: 7,500,000

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  • Pitfalls of using MySQL as your database choice?

    - by Sergio
    I've read online on multiple occassions that MySQL is a bad database. The places I've read this include some threads on Reddit, but they never seem to delve in on why it's a poor product. Is there any truth to this claim? I've never used it beyond a very simple CRUD scenario, and that was for a university project during my second year. What pitfalls, if any, are there when choosing MySQL as your database?

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  • Tap into MySQL's Amazing Performance Results with the Performance Tuning Course

    - by Antoinette O'Sullivan
    Want to leverage the high-speed load utilities, distinctive memory caches, full text indexes, and other performance-enhancing mechanisms that MySQL offers to fuel today's critical business systems. The authentic MySQL Performance Tuning course, in 4 days, teaches you to evaluate the MySQL architecture, learn to use the tools, configure the database for performance, tune application and SQL code, tune the server, examine the storage engines, assess the application architecture, and learn general tuning concepts. You can take this course in one the following three ways: Training-on-Demand: Access the streaming video, instructor delivery of this course from your own desk, at your own pace. Book time for hands-on practice when it suits you. Live-Virtual Class: Take this instructor-led class live from your own desk. With 700 events on the schedule you are sure to find a time and date to suit you! In-Class: Travel to a classroom to take this class. A sample of events on the schedule are as follows.  Location  Date  Delivery Language  Hamburg, Germany  22 October 2012  German  Prague, Czech Republic  1 October 2012  Czech  Warsaw, Poland  3 December 2012  Polish  London, England  19 November 2012  English  Rome, Italy  23 October 2012  Italian Lisbon, Portugal  6 November 2012  European Portugese  Aix en Provence, France  4 September 2012   French  Strasbourg, France 16 October 2012   French  Nieuwegein, Netherlands 26 November 2012   Dutch  Madrid, Spain 17 December 2012   Spanish  Mechelen, Belgium  1 October 2012  English  Riga, Latvia  10 December 2012  Latvian  Petaling Jaya, Malaysia  10 September 2012 English   Edmonton, Canada 10 December 2012   English  Vancouver, Canada 10 December 2012   English  Ottawa, Canada 26 November 2012   English  Toronto, Canada 26 November 2012   English  Montreal, Canada 26 November 2012   English  Mexico City, Mexico 10 September 2012   Spanish  Sao Paolo, Brazil 26 November 2012  Brazilian Portugese   Tokyo, Japan 19 November 2012   Japanese  Tokyo, Japan  19 November 2012  Japanese For further information on this class, or to register your interest in additional events, go to the Oracle University Portal: http://oracle.com/education/mysql

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