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  • insert array to mysql db function

    - by ganjan
    Hi. I have an array where the keys represent each column in my database. Now I want a function that makes a mysql update query. Something like $db['money'] = $money_input + $money_db; $db['location'] = $location $query = 'UPDATE tbl_user SET '; for($x = 0; $x < count($db); $x++ ){ $query .= $db something ".=." $db something } $query .= "WHERE username=".$username." ";

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  • PHP MySQL Syntax Error 'You have an error in your SQL syntax'

    - by Alec
    I cannot figure out the issue with my code here. I am trying to take info from the table, then subtract 1 second from Current_Time which looks like '2:00'. The problem is, I get: "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Current_Time) VALUES('22')' at line 1" I don't even understand where it gets 22 from. Thanks, I really appreciate it. if (isset($_GET['id']) && isset($_GET['time'])) { mysql_select_db("aleckaza_pennyauction", $connection); $query = "SELECT Current_Time FROM Live_Auctions WHERE ID='1'"; $results = mysql_query($query) or die(mysql_error()); while ($row = mysql_fetch_array($results)) { $newTime = $row['Current_Time'] - 1; $query = "INSERT INTO Live_Auctions(Current_Time) VALUES('".$newTime."')"; $results = mysql_query($query) or die(mysql_error()); } }

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  • Mysql Query - Order By Not Working

    - by jwzk
    I'm running Mysql 5.0.77 and I'm pretty sure this query should work? SELECT * FROM purchases WHERE time_purchased BETWEEN '2010-04-15 00:00:00' AND '2010-04-18 23:59:59' ORDER BY time_purchased ASC, order_total DESC time_purchased is DATETIME, and an index. order_total is DECIMAL(10,2), and not an index. I want to order all purchases by the date (least to greatest), and then by the order total (greatest to least). So I would output similar to: 2010-04-15 $100 2010-04-15 $80 2010-04-15 $20 2010-04-16 $170 2010-04-16 $45 2010-04-16 $15 2010-04-17 $274 .. and so on. The output I am getting from that query has the dates in order correctly, but it doesn't appear to sort the order total column at all. Thoughts? Thanks.

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  • PHP MySQL Insert Data

    - by happyCoding25
    Hello, Im trying to insert data into a table in MySQL. I found/modified some code from w3Schools and still couldn't get it working. Heres what I have so far: <?php $rusername=$_POST['username']; $rname=$_POST['name']; $remail=$_POST['emailadr']; $rpassword=$_POST['pass']; $rconfirmpassword=$_POST['cpass']; if ($rpassword==$rconfirmpassword) { $con = mysql_connect("host","user","password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("mydbname ", $con); } mysql_query("INSERT INTO members (id, username, password) VALUES ('4', $rusername, $rpassword)"); ?> Did I mistype something? To my understanding "members" is the name of the table. If anyone knows whats wrong I appreciate the help. Thanks

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  • Experience of some php projects.PHP PROJECT FLOW

    - by user106726
    I'm new to the terms PHP, JS, HTML..bt now I'm willing to apply for a php developer position in a company and before I face an interview I just want to make sure that I sound confident enough to take up the job.Could any one please help me with this.What are all the things I've to keep in mind before attending the interview? can anyone just share with me their project experiences so that I can tlk abt that? pls help me people

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  • Link mysql with php on apache

    - by Cristy
    THE STORY: I've installed Apache 2.2.17 , PHP 5.2.16 , MySQL 5.5.8 . The PHP woks great, the phpinfo() works on the localhost. THE PROBLEM: The thing is that in phpinfo() there is no mysql mentionened. I've done the following: moved php.ini to C:\windows removed the ";" in front of the mysql extension line ( extension=php_mysql.dll ) copied the libmysql.dll to php folder & windows\system32 checked the extension path in php.ini to be " C:\php\ext" searched the internet for a solution for about 2 hours... ADDITIONAL INFO: In the Apache Enviorment path I have the following: C:\Program Files (x86)\PHP\;C:\Windows\system32;C:\Windows;C:\Windows\System32\Wbem;C:\Windows\System32\WindowsPowerShell\v1.0\;C:\Program Files (x86)\QuickTime\QTSystem\;C:\OJI\MinGWStudio\work\mingw\bin;C:\Program Files (x86)\Microsoft SQL Server\100\Tools\Binn\;C:\Program Files\Microsoft SQL Server\100\Tools\Binn\;C:\Program Files\Microsoft SQL Server\100\DTS\Binn\; I think the Microsoft SQL path should be replaced with the MySQL one, but I don't know where to change that...

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  • mysql service do not launch

    - by ted
    Sorry for my English; I was trying to create db with rake in RoR application that has been configured for MySQL(gem installed, settings changed). After that attempt mysql-server broke: d@calister:~$ mysql ERROR 2002 (HY000): Can't connect to local MySQL server through socket '/var/run/mysqld/mysqld.sock' (2) mysqld is not running at all: d@calister:~$ ps aux | grep mysql d 3769 0.0 0.0 4368 832 pts/0 S+ 18:03 0:00 grep --color=auto mysql And also it doesn't seem it would like to run: d@calister:~$ sudo service mysql start start: Job failed to start Any suggestions? Thanks EDIT: d@calister:~$ sudo -u mysql mysqld 120520 18:45:11 [Note] Plugin 'FEDERATED' is disabled. mysqld: Table 'mysql.plugin' doesn't exist 120520 18:45:11 [ERROR] Can't open the mysql.plugin table. Please run mysql_upgrade to create it. 120520 18:45:11 InnoDB: The InnoDB memory heap is disabled 120520 18:45:11 InnoDB: Mutexes and rw_locks use GCC atomic builtins 120520 18:45:11 InnoDB: Compressed tables use zlib 1.2.3.4 120520 18:45:11 InnoDB: Initializing buffer pool, size = 128.0M 120520 18:45:11 InnoDB: Completed initialization of buffer pool 120520 18:45:11 InnoDB: highest supported file format is Barracuda. 120520 18:45:12 InnoDB: Waiting for the background threads to start 120520 18:45:13 InnoDB: 1.1.8 started; log sequence number 1589459 120520 18:45:13 [ERROR] Fatal error: Can't open and lock privilege tables: Table 'mysql.host' doesn't exist

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  • Attend MySQL Webinars This Week

    - by Bertrand Matthelié
    Interested in learning more about MySQL as embedded database? In building highly available MySQL applications with MySQL and DRBD? Join our webinars this week! All information below. Tuesday next week (November 20) we will provide an update about what's new in MySQL Enterprise Edition. We have live Q&A during the webinars so you'll get the chance to ask all your questions. Top 10 Reasons to Use MySQL as an Embedded Database Tuesday, November 13 9:00 a.m. PT Review the top 10 reasons why MySQL is technically well-suited for embedded use, as well as the related business reasons vendors choose MySQL initially, over time, and across product-lines. Register for the Webcast. MySQL High Availability with Distributed Replicated Block Device Thursday, November 15 9:00 a.m. PT Learn how to build highly available services with MySQL and distributed replicated block device (DRBD). The DRBD high-availability solution comprises a complete stack of open source software that delivers high-availability database clusters on commodity hardware, with the option of 24/7 support from Oracle. Register for the Webcast. Technology Update: What's New in MySQL Enterprise Edition Tuesday, November 20 9:00 a.m. PT Find out what's new in MySQL Enterprise Edition. Register for the Webcast.

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  • PHP 5.4 turn display_errors off php.ini

    - by Ethan H
    I need PHP errors not to be displayed but logged. I am using PHP 5.4 My current code to log errors in my php.ini is: log_errors = 1 error_log = "/path-to-file/error_log.txt" Which works however I am getting a 500 internal server error trying to turn error displaying off using display_errors. I have tried using the following, all returning 500 errors. display_errors = 0 display_errors = "0" display_errors = false display_errors = "false" display_errors = Off display_errors = "Off" According to the PHP documentation, as of PHP 5.4, it is a string. What am I suppose to set display_errors to to turn error displaying off?

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  • Getting the final value to this MySQL query...

    - by Jack W-H
    I've got my database set up with three tables - code, tags, and code_tags for tagging posts. This will be the SQL query processed when a post is submitted. Each tag is sliced up by PHP and individually inserted using these queries. INSERT IGNORE INTO tags (tag) VALUES ('$tags[1]'); SELECT tags.id FROM tags WHERE tag = '$tags[1]' ORDER BY id DESC LIMIT 1; INSERT INTO code_tags (code_id, tag_id) VALUES ($codeid, WHAT_GOES_HERE?) The WHAT_GOES_HERE? value at the end is what I need to know. It needs to be the ID of the tag that the second query fetched. How can I put that ID into the third query? I hope I explained that correctly. I'll rephrase if necessary.

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  • Huge mysql table with Zend Framework

    - by Uffo
    I have a mysql table with over 4 million of data; well the problem is that SOME queries WORK and SOME DON'T it depends on the search term, if the search term has a big volume of data in the table than I get the following error: Fatal error: Allowed memory size of 1048576000 bytes exhausted (tried to allocate 75 bytes) in /home/****/public_html/Zend/Db/Statement/Pdo.php on line 290 I currently have Zend Framework cache for metadata enabled, I have index on all the fields from that table.The site is running on a dedicated server with 2gb of ram. I've also set memory limit to: ini_set("memory_limit","1000M"); Any other things that I can optimize? Those are the types of query that I'm currently using: $do = $this->select() ->where('branche LIKE ?','%'.mysql_escape_string($branche).'%') ->order('premium DESC'); } //For name if(empty($branche) && empty($plz)) { $do = $this->select("MATCH(`name`) AGAINST ('{$theString}') AS score") ->where('MATCH(`name`) AGAINST( ? IN BOOLEAN MODE)', $theString) ->order('premium DESC, score'); } And a few other, but they are pretty much the same. Best Regards

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  • create new db in mysql with php syntax

    - by Jacksta
    I am trying to create a new db called testDB2, below is my code. Once running the script all I am getting an error on line 7 Fatal error: Call to undefined function mysqlquery() in /home/admin/domains/domain.com.au/public_html/db_createdb.php on line 7 This is my code <? $sql = "CREATE database testDB2"; $connection = mysql_connect("localhost", "admin_user", "pass") or die(mysql_error()); $result = mysqlquery($sql, $connection) or die(mysql_error()); if ($result) { $msg = "<p>Databse has been created!</p>"; } ?> <HTML> <head> <title>Create MySQL database</title> </head> <body> <? echo "$msg"; ?> </body> </HTML>

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  • Combinationally unique MySQL tables

    - by Jack Webb-Heller
    So, here's the problem (it's probably an easy one :P) This is my table structure: CREATE TABLE `users_awards` ( `user_id` int(11) NOT NULL, `award_id` int(11) NOT NULL, `duplicate` int(11) NOT NULL DEFAULT '0', UNIQUE KEY `award_id` (`award_id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 So it's for a user awards system. I don't want my users to be granted the same award multiple times, which is why I have a 'duplicate' field. The query I'm trying is this (with sample data of 3 and 2) : INSERT INTO users_awards (user_id, award_id) VALUES ('3','2') ON DUPLICATE KEY UPDATE duplicate=duplicate+1 So my MySQL is a little rusty, but I set user_id to be a primary key, and award_id to be a UNIQUE key. This (kind of) created the desired effect. When user 1 was given award 2, it entered. If he/she got this twice, only one row would be in the table, and duplicate would be set to 1. And again, 2, etc. When user 2 was given award 1, it entered. If he/she got this twice, duplicate updated, etc. etc. But when user 1 is given award 1 (after user 2 has already been awarded it), user 2 (with award 1)'s duplicate field increases and nothing is added to user 1. Sorry if that's a little n00bish. Really appreciate the help! Jack

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  • php warning mysql_fetch_assoc

    - by death the kid
    I am trying to access some information from mysql, but am getting the warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource for the second line of code below, any help would be much appreciated. $musicfiles=getmusicfiles($records['m_id']); $mus=mysql_fetch_assoc($musicfiles); for($j=0;$j<2;$j++) { if(file_exists($mus['musicpath'])) { echo '<a href="'.$mus['musicpath'].'">'.$mus['musicname'].'</a>'; } else { echo 'Hello world'; } } function getmusicfiles($m_id) { $music="select * from music WHERE itemid=".$s_id; $result=getQuery($music,$l); return $result; }

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  • Can't use PHP extension Mcrypt in Ubuntu 13.10 (Nginx, PHP-FPM)

    - by Marc-François
    I installed a fresh Ubuntu 13.10 on my laptop. Like I usually do, I install the packages I need for Web development, which are nginx, php5-fpm, mysql, php5-mysql, php5-mcrypt and a few others. After editing some configuration files, this usually works. But today, since 13.10, an error appears instead of the Web page I expected. Laravel requires the Mcrypt PHP extension. The package php5-mcrypt has been installed and reinstalled. The command php -m doesn't seem to show mcrypt. Any idea where the problem could come from? I've done this setup many times and it always worked.

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  • displaying a group once in php mysql

    - by JPro
    I have some data like this : 1 TC1 PASS 2 TC2 FAIL 3 TC3 INCONC 4 TC1 FAIL 5 TC21 FAIL 6 TC4 PASS 7 TC3 PASS 8 TC2 FAIL 9 TC1 TIMEOUT 10 TC21 FAIL If I try the below code : <?php mysql_connect("localhost", "root", "pop") or die(mysql_error()); mysql_select_db("jpd") or die(mysql_error()); $oustanding_fails = mysql_query("select * from SELECT_PASS ") or die(mysql_error()); $resultSetArray = array(); $platform; while($row1 = mysql_fetch_array( $oustanding_fails )) { if(trim($row1['TESTCASE']) <> trim($platform)) { echo $row1['TESTCASE']."-"; $platform = $row1['TESTCASE']; } echo $row1['RESULT'] ."<br>"; } ?> to get a result like this : TC1 PASS FAIL TIMEOUT TC2 FAIL FAIL TC3 INCONC PASS TC4 PASS AND SO ON. I am unable to get the result I want. Any ideas where exactly I am making mistake? Thanks.

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  • getting mysql_insert_id() while using ON DUPLICATE KEY UPDATE with PHP

    - by julio
    Hi-- I've found a few answers for this using mySQL alone, but I was hoping someone could show me a way to get the ID of the last inserted or updated row of a mysql DB when using PHP to handle the inserts/updates. Currently I have something like this, where column3 is a unique key, and there's also an id column that's an autoincremented primary key: $query ="INSERT INTO TABLE (column1, column2, column3) VALUES (value1, value2, value3) ON DUPLICATE KEY UPDATE SET column1=value1, column2=value2, column3=value3"; mysql_query($query); $my_id = mysql_insert_id(); $my_id is correct on INSERT, but incorrect when it's updating a row (ON DUPLICATE KEY UPDATE). I have seen several posts with people advising that you use something like INSERT INTO table (a) VALUES (0) ON DUPLICATE KEY UPDATE id=LAST_INSERT_ID(id) to get a valid ID value when the ON DUPLICATE KEY is invoked-- but will this return that valid ID to the PHP "mysql_insert_id()" function? Thanks for any advice.

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  • Jokes search engine / PHP based search engine on database

    - by matt74tm
    I'm looking for a script, functioning like the Google homepage that fetches data from a database rather than the internet. This is not intended to be a search engine, but a repository of jokes that can be pulled depending on the keywords typed. No sophisticated search techniques are required - keyword based is perfectly fine. If some mechanism of up/down-voting jokes can be incorporated, that would be fantastic, but I'm presuming that will be an entirely different game.

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  • SELECT INTO or Stored Procedure?

    - by Kerry
    Would this be better as a stored procedure or leave it as is? INSERT INTO `user_permissions` ( `user_id`, `object_id`, `type`, `view`, `add`, `edit`, `delete`, `admin`, `updated_by_user_id` ) SELECT `user_id`, $object_id, '$type', 1, 1, 1, 1, 1, $user_id FROM `user_permissions` WHERE `object_id` = $object_id_2 AND `type` = '$type_2' AND `admin` = 1 You can think of this with different objects, lets say you have groups and subgroups. If someone creates a subgroup, it is making everyone who had access to the parent group now also have access to the subgroup. I've never made a stored procedure before, but this looks like it might be time. This call be probably be called very often. Should I be creating a procedure or will the performance be insignificant?

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  • Jokes search engine / PHP based search engine on database

    - by matt_tm
    I'm looking for a script, functioning like the Google homepage that fetches data from a database rather than the internet. This is not intended to be a search engine, but a repository of jokes that can be pulled depending on the keywords typed. No sophisticated search techniques are required - keyword based is perfectly fine. If some mechanism of up/down-voting jokes can be incorporated, that would be fantastic, but I'm presuming that will be an entirely different game.

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  • I am not able to update form data to MySQL using PHP and jQuery

    - by Jimson Jose
    My problem is that I am unable to update the values entered in the form. I have attached all the files. I'm using MYSQL database to fetch data. What happens is that I'm able to add and delete records from form using jQuery and PHP scripts to MYSQL database, but I am not able to update data which was retrieved from the database. The file structure is as follows: index.php is a file with jQuery functions where it displays form for adding new data to MYSQL using save.php file and list of all records are view without refreshing page (calling load-list.php to view all records from index.php works fine, and save.php to save data from form) - Delete is an function called from index.php to delete record from MySQL database (function calling delete.php works fine) - Update is an function called from index.php to update data using update-form.php by retriving specific record from MySQL table, (works fine) Problem lies in updating data from update-form.php to update.php (in which update query is written for MySQL) I have tried in many ways - at last I had figured out that data is not being transferred from update-form.php to update.php; there is a small problem in jQuery AJAX function where it is not transferring data to update.php page. Something is missing in calling update.php page it is not entering into that page. I am new bee in programming. I had collected this script from many forums and made this one. So I was limited in solving this problem. I came to know that this is good platform for me and many where we get a help to create new things. Please find the link below to download all files which is of 35kb (virus free assurance): download mysmallform files in ZIPped format, including mysql query

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  • PHP - SQL query to get update time from table status

    - by Tribalcomm
    This is my php code (I already have a connection to the db): $array = mysql_query("SHOW TABLE STATUS FROM mytable;"); while ($array = mysql_fetch_array($result)) { $updatetime = $array['Update_time']; } echo $updatetime; I get: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource. I am running MySQL 5.0.89 and PHP5. I do not want to add a new field to the table... I want to use the table status... Any help? Thanks!

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  • Why is this PHP loop rendering every row twice?

    - by Christopher
    I'm working on a real frankensite here not of my own design. There's a rudimentary CMS and one of the pages shows customer records from a MySQL DB. For some reason, it has no probs picking up the data from the DB - there's no duplicate records - but it renders each row twice. <?php $limit = 500; $area = 'customers_list'; $prc = 'customer_list.php'; if($_GET['page']) { include('inc/functions.php'); $page = $_GET['page']; } else { $page = 1; } $limitvalue = $page * $limit - ($limit); $customers_check = get_customers(); $customers = get_customers($limitvalue, $limit); $totalrows = count($customers_check); ?> <!-- pid: customer_list --> <table border="0" width="100%" cellpadding="0" cellspacing="0" style="float: left; margin-bottom: 20px;"> <tr> <td class="col_title" width="200">Name</td> <td></td> <td class="col_title" width="200">Town/City</td> <td></td> <td class="col_title">Telephone</td> <td></td> </tr> <?php for ($i = 0; $i < count($customers); $i++) { ?> <tr> <td colspan="2" class="cus_col_1"><a href="customer_details.php?id=<?php echo $customers[$i]['customer_id']; ?>"><?php echo $customers[$i]['surname'].', '.$customers[$i]['first_name']; ?></a></td> <td colspan="2" class="cus_col_2"><?php echo $customers[$i]['town']; ?></td> <td class="cus_col_1"><?php echo $customers[$i]['telephone']; ?></td> <td class="cus_col_2"> <a href="javascript: single_execute('prc/customers.prc.php?delete=yes&id=<?php echo $customers[$i]['customer_id']; ?>')" onClick="return confirmdel();" class="btn_maroon_small" style="margin: 0px; float: right; margin-right: 10px;"><div class="btn_maroon_small_left"> <div class="btn_maroon_small_right">Delete Account</div> </div></a> <a href="customer_edit.php?id=<?php echo $customers[$i]['customer_id']; ?>" class="btn_black" style="margin: 0px; float: right; margin-right: 10px;"><div class="btn_black_left"> <div class="btn_black_right">Edit Account</div> </div></a> <a href="mailto: <?php echo $customers[$i]['email']; ?>" class="btn_black" style="margin: 0px; float: right; margin-right: 10px;"><div class="btn_black_left"> <div class="btn_black_right">Email Customer</div> </div></a> </td> </tr> <tr><td class="col_divider" colspan="6"></td></tr> <?php }; ?> </table> <!--///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////--> <!--// PAGINATION--> <!--///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////--> <div class="pagination_holder"> <?php if($page != 1) { $pageprev = $page-1; ?> <a href="javascript: change('<?php echo $area; ?>', '<?php echo $prc; ?>?page=<?php echo $pageprev; ?>');" class="pagination_left">Previous</a> <?php } else { ?> <div class="pagination_left, page_grey">Previous</div> <?php } ?> <div class="pagination_middle"> <?php $numofpages = $totalrows / $limit; for($i = 1; $i <= $numofpages; $i++) { if($i == $page) { ?> <div class="page_number_selected"><?php echo $i; ?></div> <?php } else { ?> <a href="javascript: change('<?php echo $area; ?>', '<?php echo $prc; ?>?page=<?php echo $i; ?>');" class="page_number"><?php echo $i; ?></a> <?php } } if(($totalrows % $limit) != 0) { if($i == $page) { ?> <div class="page_number_selected"><?php echo $i; ?></div> <?php } else { ?> <a href="javascript: change('<?php echo $area; ?>', '<?php echo $prc; ?>?page=<?php echo $i; ?>');" class="page_number"><?php echo $i; ?></a> <?php } } ?> </div> <?php if(($totalrows - ($limit * $page)) > 0) { $pagenext = $page+1; ?> <a href="javascript: change('<?php echo $area; ?>', '<?php echo $prc; ?>?page=<?php echo $pagenext; ?>');" class="pagination_right">Next</a> <?php } else { ?> <div class="pagination_right, page_grey">Next</div> <?php } ?> </div> <!--///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////--> <!--// END PAGINATION--> <!--///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////--> I'm not the world's best PHP expert but I think I can see an error in a for loop when there is one... But everything looks ok to me. You'll notice that the customer name is clickable; clicking takes you to another page where you can view their full info as held in the DB - and for both rows, the customer ID is identical, and manually checking the DB shows there's no duplicate entries. The code is definitely rendering each row twice, but for what reason I have no idea. All pointers / advice appreciated.

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  • Mysql Cluster not working on Ubuntu

    - by user53864
    I am unable to setup MySQL Cluster on ubuntu servers. As a starting point I started from the link but I am not successful and the tar ball version I download is 6.3.45. As I wanted to test the mysql cluster, the Data node and SQL node are same but sql never appeared as connected in management node console and it looks like below. [ndbd(NDB)] 2 node(s) id=2 @192.168.1.107 (Version: version number, Nodegroup: 0, Master) id=3 @192.168.1.108 (Version: version number, Nodegroup: 0) [ndb_mgmd(MGM)] 1 node(s) id=1 @192.168.1.105 (Version: version number) [mysqld(API)] 2 node(s) id=4 (not connected, accepting connect from 192.168.1.107) id=5 (not connected, accepting connect from 192.168.1.108) On all the 3 machines mysql-server & client(apt-get install mysql-server mysql-client) were already installed and I completely stopped and also removed them at the system start up. Now the mysqld is from extracted cluster tar ball(/usr/local/mysql/support-files/mysql.server). As for testing, I created a test database on both the data nodes but the tables are also not syncing on other node. I checked many links, configurations are remained similar in all the links but somewhere it's going wrong. Anymore extra package is required?, Could anyone help me here..?. I am trying this for past 3 days... Thank you!

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  • MySQL Cluster 7.3: On-Demand Webinar and Q&A Available

    - by Mat Keep
    The on-demand webinar for the MySQL Cluster 7.3 Development Release is now available. You can learn more about the design, implementation and getting started with all of the new MySQL Cluster 7.3 features from the comfort and convenience of your own device, including: - Foreign Key constraints in MySQL Cluster - Node.js NoSQL API  - Auto-installation of higher performance distributed, clusters We received some great questions over the course of the webinar, and I wanted to share those for the benefit of a broader audience. Q. What Foreign Key actions are supported: A. The core referential actions defined in the SQL:2003 standard are implemented: CASCADE RESTRICT NO ACTION SET NULL Q. Where are Foreign Keys implemented, ie data nodes or SQL nodes? A. They are implemented in the data nodes, therefore can be enforced for both the SQL and NoSQL APIs Q. Are they compatible with the InnoDB Foreign Key implementation? A. Yes, with the following exceptions: - InnoDB doesn’t support “No Action” constraints, MySQL Cluster does - You can choose to suspend FK constraint enforcement with InnoDB using the FOREIGN_KEY_CHECKS parameter; at the moment, MySQL Cluster ignores that parameter. - You cannot set up FKs between 2 tables where one is stored using MySQL Cluster and the other InnoDB. - You cannot change primary keys through the NDB API which means that the MySQL Server actually has to simulate such operations by deleting and re-adding the row. If the PK in the parent table has a FK constraint on it then this causes non-ideal behaviour. With Restrict or No Action constraints, the change will result in an error. With Cascaded constraints, you’d want the rows in the child table to be updated with the new FK value but, the implicit delete of the row from the parent table would remove the associated rows from the child table and the subsequent implicit insert into the parent wouldn’t reinstate the child rows. For this reason, an attempt to add an ON UPDATE CASCADE where the parent column is a primary key will be rejected. Q. Does adding or dropping Foreign Keys cause downtime due to a schema change? A. Nope, this is an online operation. MySQL Cluster supports a number of on-line schema changes, ie adding and dropping indexes, adding columns, etc. Q. Where can I see an example of node.js with MySQL Cluster? A. Check out the tutorial and download the code from GitHub Q. Can I use the auto-installer to support remote deployments? How about setting up MySQL Cluster 7.2? A. Yes to both! Q. Can I get a demo Check out the tutorial. You can download the code from http://labs.mysql.com/ Go to Select Build drop-down box Q. What is be minimum internet speen required for Geo distributed cluster with synchronous replication? A. if you're splitting you cluster between sites then we recommend a network latency of 20ms or less. Alternatively, use MySQL asynchronous replication where the latency of your WAN doesn't impact the latency of your reads/writes. Q. Where you can one learn more about the PayPal project with MySQL Cluster? A. Take a look at the following - you'll find press coverage, a video and slides from their keynote presentation  So, if you want to learn more, listen to the new MySQL Cluster 7.3 on-demand webinar  MySQL Cluster 7.3 is still in the development phase, so it would be great to get your feedback on these new features, and things you want to see!

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