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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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  • MySQL script to delete data in chunks until everything lower then id has been deleted

    - by Chriswede
    I need an MySQL Skript which does the following: delete chunks of the database until it has deleted all link_id's greater then 10000 exmaple: x = 10000 DELETE FROM pligg_links WHERE link_id > x and link_id < x+10000 x = x + 10000 ... So it would delete DELETE FROM pligg_links WHERE link_id > 10000 and link_id < 20000 then DELETE FROM pligg_links WHERE link_id > 20000 and link_id < 30000 until all id's less then 10000 have been removed I need this because the database is very very big (more then a gig) thank in advance

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  • MySQL: filling empty fields with zeroes when using GROUP BY

    - by SaltLake
    I've got MySQL table CREATE TABLE cms_webstat ( ID int NOT NULL auto_increment PRIMARY KEY, TIMESTAMP_X timestamp DEFAULT CURRENT_TIMESTAMP, # ... some other fields ... ) which contains statistics about site visitors. For getting visits per hour I use SELECT hour(TIMESTAMP_X) as HOUR , count(*) AS HOUR_STAT FROM cms_webstat GROUP BY HOUR ORDER BY HOUR DESC which gives me | HOUR | HOUR_STAT | | 24 | 15 | | 23 | 12 | | 22 | 9 | | 20 | 3 | | 18 | 2 | | 15 | 1 | | 12 | 3 | | 9 | 1 | | 3 | 5 | | 2 | 7 | | 1 | 9 | | 0 | 12 | And I'd like to get following: | HOUR | HOUR_STAT | | 24 | 15 | | 23 | 12 | | 22 | 9 | | 21 | 0 | | 20 | 3 | | 19 | 0 | | 18 | 2 | | 17 | 0 | | 16 | 0 | | 15 | 1 | | 14 | 0 | | 13 | 0 | | 12 | 3 | | 11 | 0 | | 10 | 0 | | 9 | 1 | | 8 | 0 | | 7 | 0 | | 6 | 0 | | 5 | 0 | | 4 | 0 | | 3 | 5 | | 2 | 7 | | 1 | 9 | | 0 | 12 | How should I modify the query to get such result? Thanks.

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  • Mysql replace() function, help with query (what chars do I escape?)

    - by jyoseph
    I am trying to update an old cms where images were stored in /images/editor/, they are now stored in a bucket on amazon s3. I'm trying to update the database using mysql replace. I've done this in the past with replacing simple words, but now Mysql is reporting an error, I suspect because this is more than a simple word: UPDATE contents SET desc = replace(desc, '/images/editor/', 'http://s3.amazonaws.com/my_bucket/editor/') Do I need to escape the : or slashes? I've tried escaping it with a '\' to no avail. Can someone get me pointed in the right direction? Thanks! Edit Here's the error I am getting, nothing too telling error : 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 'desc = replace(desc, '/images/editor', 'http://s3.amazonaws.com/app_navigator/ed' at line 1

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  • MYSQL Inserting rows that reference main rows.

    - by Andrew M
    I'm transferring my access logs into a database. I've got two tables: urlRequests id : int(10) host : varchar(100) path: varchar(300) unique index (host, path) urlAccesses id : int(10) request : int(10) <-- reference to urlRequests row ip : int(4) query : varchar(300) time : timestamp I need to insert a row into urlAccesses for every page load, but first a row in urlRequests has to exist with the requested host and path so that urlAccesses's row can reference it. I know I can do it this way: A. check if a row exists in urlRequests B. insert a row in urlRequests if it needs it C. insert a row into urlAccesses with the urlRequests's row id referenced That's three queries for every page load if the urlRequests row doesn't exist. I'm very new to MySQL, so I'm guessing that there's a way to go about this that would be faster and use less queries.

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  • Mysql on duplicate key update + sub query

    - by jwzk
    Using the answer from this question: http://stackoverflow.com/questions/662877/need-mysql-insert-select-query-for-tables-with-millions-of-records new_table * date * record_id (pk) * data_field INSERT INTO new_table (date,record_id,data_field) SELECT date, record_id, data_field FROM old_table ON DUPLICATE KEY UPDATE date=old_table.data, data_field=old_table.data_field; I need this to work with a group by and join.. so to edit: INSERT INTO new_table (date,record_id,data_field,value) SELECT date, record_id, data_field, SUM(other_table.value) as value FROM old_table JOIN other_table USING(record_id) ON DUPLICATE KEY UPDATE date=old_table.data, data_field=old_table.data_field, value = value; I can't seem to get the value updated. If I specify old_table.value I get a not defined in field list error.

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  • Realtime MySQL search results on an advanced search page

    - by Andrew Heath
    I'm a hobbyist, and started learning PHP last September solely to build a hobby website that I had always wished and dreamed another more competent person might make. I enjoy programming, but I have little free time and enjoy a wide range of other interests and activities. I feel learning PHP alone can probably allow me to create 98% of the desired features for my site, but that last 2% is awfully appealing: The most powerful tool of the site is an advanced search page that picks through a 1000+ record game scenario database. Users can data-mine to tremendous depths - this advanced page has upwards of 50 different potential variables. It's designed to allow the hardcore user to search on almost any possible combination of data in our database and it works well. Those who aren't interested in wading through the sea of options may use the Basic Search, which is comprised of the most popular parts of the Advanced search. Because the advanced search is so comprehensive, and because the database is rather small (less than 1,200 potential hits maximum), with each variable you choose to include the likelihood of getting any qualifying results at all drops dramatically. In my fantasy land where I can wield AJAX as if it were Excalibur, my users would have a realtime Total Results counter in the corner of their screen as they used this page, which would automatically update its query structure and report how many results will be displayed with the addition of each variable. In this way it would be effortless to know just how many variables are enough, and when you've gone and added one that zeroes out the results set. A somewhat similar implementation, at least visually, would be the Subtotal sidebar when building a new custom computer on IBuyPower.com For those of you actually still reading this, my question is really rather simple: Given the time & ability constraints outlined above, would I be able to learn just enough AJAX (or whatever) needed to pull this one feature off without too much trouble? would I be able to more or less drop-in a pre-written code snippet and tweak to fit? or should I consider opening my code up to a trusted & capable individual in the future for this implementation? (assuming I can find one...) Thank you.

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  • MySQL Query still executing after a day..?

    - by Matt Jarvis
    Hi - I'm trying to isolate duplicates in a 500MB database and have tried two ways to do it. One creating a new table and grouping: CREATE TABLE test_table as SELECT * FROM items WHERE 1 GROUP BY title; But it's been running for an hour and in MySQL Admin it says the status is Locked. The other way I tried was to delete duplicates with this: DELETE bad_rows.* from items as bad_rows inner join ( select post_title, MIN(id) as min_id from items group by title having count(*) 1 ) as good_rows on good_rows.post_title = bad_rows.post_title; ..and this has been running for 24hours now, Admin telling me it's Sending data... Do you think either or these queries are actually still running? How can I find out if it's hung? (with Apple OS X 10.5.7)

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  • MySQL temp table issue

    - by AmyD
    Hi folks! I'm trying to use temp tables to speed up my MySQL 4.1.22-standard database and what seems like a simple operation is causing me all kinds of issues. My code is below.... CREATE TEMPORARY TABLE nonDerivativeTransaction_temp (accession_number varchar(30), transactionDateValue date)) TYPE=HEAP; INSERT INTO nonDerivativeTransaction_temp VALUES( SELECT accession_number, transactionDateValue FROM nonDerivativeTransaction WHERE transactionDateValue = "2010-06-15"); SELECT * FROM nonDerivativeTransaction_temp; The original table (nonDerivativeTransaction) has two fields, accession_number (varchar(30)) and transactionDateValue (date). Apparently I am getting an issue with the first two statements but I can't seem to nail down what it is. Any help would be appreciated. Amy D.

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  • Match two mysql cols on alpha chars (ignoring numbers in same field)

    - by Steve
    I was wondering if you know of a way I could filter a mysql query to only show the ‘alpha’ characters from a specific field So something like SELECT col1, col2, **alpha_chars_only(col3)** FROM table I am not looking to update only select. I have been looking at some regex but without much luck most of what turned up was searching for fields that only contain ‘alpha’ chars. In a much watered down context... I have col1 which contains abc and col two contains abc123 and I want to match them on alpha chars only. There can be any number of letters or numbers. Any help very much wel come

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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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  • MySQL query problem

    - by SaltLake
    I've got MySQL table CREATE TABLE stat ( ID int NOT NULL auto_increment PRIMARY KEY, TIMESTAMP_X timestamp DEFAULT CURRENT_TIMESTAMP, # ... some other fields ... ) which contains statistics about site visitors. For getting visits per hour I use SELECT hour(TIMESTAMP_X) as HOUR , count(*) AS HOUR_STAT FROM cms_webstat GROUP BY HOUR ORDER BY HOUR DESC which gives me | HOUR | HOUR_STAT | | 24 | 15 | | 23 | 12 | | 22 | 9 | | 20 | 3 | | 18 | 2 | | 15 | 1 | | 12 | 3 | | 9 | 1 | | 3 | 5 | | 2 | 7 | | 1 | 9 | | 0 | 12 | And I'd like to get following: | HOUR | HOUR_STAT | | 24 | 15 | | 23 | 12 | | 22 | 9 | | 21 | 0 | | 20 | 3 | | 19 | 0 | | 18 | 2 | | 17 | 0 | | 16 | 0 | | 15 | 1 | | 14 | 0 | | 13 | 0 | | 12 | 3 | | 11 | 0 | | 10 | 0 | | 9 | 1 | | 8 | 0 | | 7 | 0 | | 6 | 0 | | 5 | 0 | | 4 | 0 | | 3 | 5 | | 2 | 7 | | 1 | 9 | | 0 | 12 | How should I modify the query to get such result? 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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  • Error comparing hash to hashed mysql password (output values are equal)

    - by Charlie
    Im trying to compare a hashed password value in a mysql database with the hashed value of an inputted password from a login form. However, when I compare the two values it says they aren't equal. I removed the salt to simply, and then tested what the outputs were and got the same values $password1 = $_POST['password']; $hash = hash('sha256', $password1); ...connect to database, etc... $query = "SELECT * FROM users WHERE username = '$username1'"; $result = mysql_query($query); $userData = mysql_fetch_array($result); if($hash != $userData['password']) //incorrect password { echo $hash."|".$userData['password']; die(); } ...other code... Sample output: 7816ee6a140526f02289471d87a7c4f9602d55c38303a0ba62dcd747a1f50361| 7816ee6a140526f02289471d87a7c4f9602d55c38303a0ba62dcd747a1f50361 Any thoughts?

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  • Optimize MySQL database query

    - by rajeeesh
    I had a commenting application in my web site. The comments will store in a MySQL table . table structure as follows id | Comment | user | created_date ------------------------------------------------------ 12 | comment he | 1245 | 2012-03-30 12:15:00 ------------------------------------------------------ I need to run a query for listing all the comments after a specific time. ie .. a query like this SELECT * FROM comments WHERE created_date > "2012-03-29 12:15:00" ORDER BY created_date DESC Its working fine.. My question is if I got a 1-2 lakh entry in this table is this query is sufficient for the purpose ? or this query will take time to execute ? In most cases I have to show last 2 days data + periodically ( interval of 10 mins ) checking for updates with ajax from this table ... Please help Thanks

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  • Can't Connect To Local Mysql Using IP Address, but CAN connect from remote server

    - by user1782041
    Here's an interesting one that does not seem to fall into any of the mysql connection issues I've read about or searched for: On an Ubuntu 12.04 box I had some system updates waiting to install, and I took care of that this evening. After the install, I started seeing some errors in my syslog complaining about a particular php script that could no longer connect to the mysql instance on the box. Here is the specific error: PHP Warning: mysql_connect(): Can't connect to MySQL server on '192.168.0.40' (4) Now, the server's IP address is 192.168.0.40, and I've checked to make sure that I have mysql listening on 0.0.0.0 so that I can connect using either "localhost" or "192.168.0.40". Here's where things get odd: From the local machine, if I try the following: mysql -uroot -p -h192.168.0.40 I get this error: ERROR 2003 (HY000): Can't connect to MySQL server on '192.168.0.40' (110) I've checked, and error 110 indicates an OS timeout, and error 2003 is the mysql generic "can't connect" error. This indicates that it is not permissions with the user. However, if I do the same thing from a remote machine (say, from 192.168.0.30), I log right in with no problems. Futher, other scripts on the local machine that connect to mysql using "localhost" for the host rather than "192.168.0.40" connect with no problems. Also, I can connect via the mysql socket with no problems both from the command line and php scripts. So, this feels like a networking issue of some kind on the local box, but there are no iptables rules on this box (it is firewalled externally) and I can't figure out what else may be causing this. This problematic script worked perfectly prior to the latest system update. For now, I'll simply change the script to connect via localhost, but I'd really like to know why it broke for 2 reasons: There may be other scripts that connect using 192.168.0.40 that don't run very often which are now broken. Auditing them all will take more time than I feel like devoting at the moment. I'm curious, and want to know why it broke so I can fix it correctly. Any help?

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  • mysql command line not working

    - by Sandeepan Nath
    I have mysql running in my fedora system. I have xampp setup on the system and php projects present in the webspace are working fine. PhpMyAdmin is working fine. echoing phpinfo() in a PHP script also shows mysql enabled. But running mysql connect command mysql -u[username] -p[password] Gives this - bash: mysql: command not found How do I fix that? Any pointers? I guess I need to do some pointing (define some path in some file) so that my system knows that mysql is installed. What exactly do I have to do? Additional Details This system was someone else's and he is not available here. May be PHP/Mysql was setup already in the system. I just freshly extracted xampp for linux into /opt/lampp/ and have put all the above mentioned things (PHP projects and PhpMyAdmin) there. After doing that I had a socket problem (PhpMyAdmin was not working and showing this)- #2002 - The server is not responding (or the local MySQL server's socket is not correctly configured) I restarted lampp using ./lampp restart but problem remained. Then after turning on system today, I started lampp and everything worked just fine. No project issues anymore only command line Mysql not working

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  • MySQL partition "full"?

    - by gdea73
    I have a server that runs Debian 6.2, with Apache, PHP5, and MySQL. Well, I hadn't done anything with MySQL at all so far, just Apache and PHP; I must have installed it (mysql-server) at some point along the line, and I decided to login to the database for the first time a couple days ago as I was considering using the database for a future website project. I noticed that the "root" user had a password, and I didn't recall having set one. My usual root password was incorrect. So I attempted to reset the password. sudo service mysql stop (stopped successfully) sudo /usr/bin/mysqld_safe --skip-grant-tables --skip-networking & started successfully, from what I can tell. However, mysql itself returns "Can't connect to local MySQL server through socket '/var/run/mysqld/mysqld,sock' (2)", and additionally sudo service mysql start returns "/etc/init.d/mysql: ERROR: The partition with /var/lib/mysql is too full! ... failed!" df -h tells me that / is 26% used, a 20GB partition, and /home, roughly 900GB, has only 5% usage. On a potentially related note, I've been experiencing random hangs since I noticed this problem, my tty2 randomly froze several times while idle, and the entire system is suddenly unstable. gnome-terminal also does not open. (Gnome-terminal apparently works now, disregard that part, but the server is still being somewhat unstable, I randomly lost connection when I was SSHed into it from my laptop, twice now.)

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  • Spring-mvc project can't select from a particular mysql table

    - by Dan Ray
    I'm building a Spring-mvc project (using JPA and Hibernate for DB access) that is running just great locally, on my dev box, with a local MySQL database. Now I'm trying to put a snapshot up on a staging server for my client to play with, and I'm having trouble. Tomcat (after some wrestling) deploys my war file without complaint, and I can get some response from the application over the browser. When I hit my main page, which is behind Spring Security authentication, it redirects me to the login page, which works perfectly. I have Security configured to query the database for user details, and that works fine. In fact, a change to a password in the database is reflected in the behavior of the login form, so I'm confident it IS reaching the database and querying the user table. Once authenticated, we go to the first "real" page of the app, and I get a "data access failure" error. The server's console log gets this line (redacted): ERROR org.hibernate.util.JDBCExceptionReporter - SELECT command denied to user 'myDbUser'@'localhost' for table 'asset' However, if I go to MySQL from the shell using exactly the same creds, I have no problem at all selecting from the asset table: [development@tomcat01stg]$ mysql -u myDbUser -pmyDbPwd dbName ... mysql> \s -------------- mysql Ver 14.12 Distrib 5.0.77, for redhat-linux-gnu (i686) using readline 5.1 Connection id: 199 Current database: dbName Current user: myDbUser@localhost ... UNIX socket: /var/lib/mysql/mysql.sock -------------- mysql> select count(*) from asset; +----------+ | count(*) | +----------+ | 19 | +----------+ 1 row in set (0.00 sec) I've broken down my MySQL access settings, cleaned out the user and re-run the grant commands, set up a version of the user from 'localhost' and another from '%', making sure to flush permissions.... Nothing is changing the behavior of this thing. What gives?

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  • Upgrading PHP, MySQL old-passwords issue

    - by Rushyo
    I've inherited a Windows 2k3 server running an XAMPP-installation from the stone age. I needed to upgrade PHP to facilitate an upgrade to MediaWiki to facilitate a new MediaWiki extension (to facilitate some documentation to facilitate doing my job to facilitate getting paid to facilit... you get the idea). However... installing a new version of PHP resulted in PHP's MySQL libraries refusing to communicate using MySQL's 'old style' 152-bit passwords. Not a problem in theory. The MySQL installation is post-4.1, so it should have the functionality to upgrade the user's passwords from 152-bit to 328-bit (what a weird hashing algorithm...). I ran the following: SET PASSWORD = PASSWORD('foo'); on MySQL but querying: SELECT user, password FROM mysql.user; returned just the same password I started out with - 152-bit. Now... I suspect you're thinking 'AHA! old-passwords is on!'. Unfortunately it's not - I've disabled it in the configuration (explicitly set it to 0), made doubly sure I have an absolute reference to that configuration file and ensured the service isn't using the --old-passwords flag. The service was reset after each and every operation. So I went onto another system and generated the 328-bit hash on there, copying the hash over to the first MySQL instance. Unfortunately, that didn't work either (I did remember to FLUSH PRIVILEGES). The application error is: "'mysqlnd cannot connect to MySQL 4.1+ using the old insecure authentication. Please use an administration tool [...snip...] Is there anything else I can try to get PHP to recognise MySQL as not using the 'old insecure authentication'? MySQL seems to be stuck in 'old-passwords' mode and I can't get it out of it.

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  • How to find string in a string

    - by owca
    I somehow need to find the longest string in other string, so if string1 will be "Alibaba" and string2 will be "ba" , the longest string will be "baba". I have the lengths of strings, but what next ? char* fun(char* a, char& b) { int length1=0; int length2=0; int longer; int shorter; char end='\0'; while(a[i] != tmp) { i++; length1++; } int i=0; while(b[i] != tmp) { i++; length++; } if(dlug1 > dlug2){ longer = length1; shorter = length2; } else{ longer = length2; shorter = length1; } //logics here } int main() { char name1[] = "Alibaba"; char name2[] = "ba"; char &oname = *name2; cout << fun(name1, oname) << endl; system("PAUSE"); return 0; }

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  • string maniupulations, oops, how do I replace parts of a string

    - by Joe Gibson
    I am very new to python. Could someone explain how I can manipulate a string like this? This function receives three inputs: complete_fmla: has a string with digits and symbols but has no hyphens ('-') nor spaces. partial_fmla: has a combination of hyphens and possibly some digits or symbols, where the digits and symbols that are in it (other than hyphens) are in the same position as in the complete_formula. symbol: one character The output that should be returned is: If the symbol is not in the complete formula, or if the symbol is already in the partial formula, the function should return the same formula as the input partial_formula. If the symbol is in the complete_formula and not in the partial formula, the function should return the partial_formula with the symbol substituting the hyphens in the positions where the symbol is, in all the occurrences of symbol in the complete_formula. For example: generate_next_fmla (‘abcdeeaa’, ‘- - - - - - - - ’, ‘d’) should return ‘- - - d - - - -’ generate_next_fmla (‘abcdeeaa’, ‘- - - d - - - - ’, ‘e’) should return ‘- - - d e e - -’ generate_next_fmla (‘abcdeeaa’, ‘- - - d e e - - ’, ‘a’) should return ‘a - - d e e a a’ Basically, I'm working with the definition: def generate_next_fmla (complete_fmla, partial_fmla, symbol): Do I turn them into lists? and then append? Also, should I find out the index number for the symbol in the complete_fmla so that I know where to append it in the string with hyphens??

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  • What Will Happen to Real Estate Leases when Operating Leases are Gone?

    - by Theresa Hickman
    Many people are concerned about what will happen to real estate leases when FASB and IASB abolish operating leases. They plan to unveil the proposed standards on treating leases this summer as part of the convergence project but no "finalized ruling" is expected for at least a year because it will need to get formal consensus from many players, such as the SEC, American Association of Investors, Congress, the Big Four, American Associate of Realtors, the international equivalents of these, etc. If your accounting is a bit rusty, an Operating Lease is where you lease equipment or some asset for a shorter period than the actual (expected) life of the asset and then give the asset back while it still has some useful life in it. (Think leasing a car). Because an Operating Lease does not contain any of the provisions that would qualify it as a Capital Lease, the lease is not treated as a sale or purchase and hits the lessee's rental expense and the lessor's revenue. So it all stays on the P&L (assuming no prepayments are made). Capital Leases, on the other hand, hit lessee's and lessor's balance sheets because the asset is treated as a sale. (I'm ignoring interest and depreciation here to emphasize my point). Question: What will happen to real estate leases when Operating Leases go away and how will Oracle Financials address these changes? Before I attempt to address these questions, here's a real-life example to expound on some of the issues: Let's say a U.S. retailer leases a store in a mall for 15 years. Under U.S. GAAP, the lease is considered an operating or expense lease. Will that same lease be considered a capital lease under IFRS? Real estate leases are supposedly going to be capitalized under IFRS. If so, will everyone need to change all leases from operating to capital? Or, could we make some adjustments so we report the lease as an expense for operations reporting but capitalize it for SEC reporting? Would all aspects of the lease be capitalized, or would some line items still be expensed? For example, many retail store leases are defined to include (1) the agreed-to rent amount; (2) a negotiated increase in base rent, e.g., maybe a 5% increase in Year 5; (3) a sales rent component whereby the retailer pays a variable additional amount based on the sales generated in the prior month; (4) parking lot maintenance fees. Would the entire lease be capitalized, or would some portions still be expensed? To help answer these questions, I met up with our resident accounting expert and walking encyclopedia, Seamus Moran. Here's what he had to say: Oracle is aware of the potential changes specific to reporting/capitalization of real estate leases; i.e., we are aware that FASB and IASB have identified real estate leases as one of the areas for standards convergence. Oracle stays apprised of the on-going convergence through our domain expertise staff, our relationship with customers, our market awareness, and, of course, our relationships with the Big 4. This is part of our normal process with respect to regulatory compliance worldwide. At this time, Oracle expects that the standards convergence committee will make a recommendation about reporting standards for real estate leases in about a year. Following typical procedures, we also expect that the recommendation will be up for review for a year, and customers will then need to start reporting to the new standard about a year after that. So that means we would expect the first customer to report under the new standard in maybe 3 years. Typically, after the new standard is finalized and distributed, we find that our customers then begin to evaluate how they plan to meet the new standard. And through groups like the Customer Advisory Boards (CABs), our customers tell us what kind of product changes are needed in order to satisfy their new reporting requirements. Of course, Oracle is also working with the Big 4 and Accenture and other implementers in order to ascertain that these recommended changes will indeed meet new reporting standards. So the best advice we can offer right now is, stay apprised of the standards convergence committee; know that Oracle is also staying abreast of developments; get involved with your CAB so your voice is heard; know that Oracle products continue to be GAAP compliant, and we will continue to maintain that as our standard. But exactly what is that "standard"--we need to wait on the standards convergence committee. In a nut shell, operating leases will become either capital leases or month to month rentals, but it is still too early, too political and too uncertain to call out at this point.

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  • Recursive String Function (Java)

    - by Jake Brooks
    Hi, I am trying to design a function that essentially does as follows: String s = "BLAH"; store the following to an array: blah lah bah blh bla bl ba bh ah al So basically what I did there was subtract each letter from it one at a time. Then subtract a combination of two letters at a time, until there's 2 characters remaining. Store each of these generations in an array. Hopefully this makes sense, Jake

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