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  • MYSQL et le programme 1-Click

    - by swalker
    Les partenaires OPN Silver et Revendeurs réalisant des transactions par le biais de distributeurs agréés Oracle VAD peuvent désormais revendre des abonnements à MySQL Standard Edition et Enterprise Edition via le programme 1-Click. Les membres OPN Silver peuvent également revendre des Licences perpétuelles MySQL SE et EE. Pour accéder aux dernières informations en date, accédez à la Zone de connaissances des technologies Oracle 1-Click pour les PME.

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  • In Technology, Ignorance is NOT Bliss

    - by Tanu Sood
    Author: Debra Lilley, ACE Director, UK Proof I’m not technical -  I’ve just finished a Latin America tour with OTN and a funny thing happened that I want to share with you; because it is quite a good analogy for how many of us use technology today and you know how I love analogies. In Costa Rica we had a really long journey up through the mountains to where our conference was to be. The road was windy and narrow and once it got dark there was no scenery to see, boredom set in. At one stage I looked at my watch to see the time, but in the dark I couldn’t make it out, so I thought I would be clever and use the torch in my smartphone! Even though as soon as I switched on the phone it showed the time, I ignored it and used the torch to read my watch. That’s us when we pay maintenance on software, ask for enhancements, and either chose not to upgrade or as I have seen so many times, upgrade but don’t use the new features. I know there are always other factors not least the upgrade costs themselves but in the later releases of all the Oracle family of applications Oracle have done a lot to make the interoperability of them with Oracle Fusion Middleware more successful and in many cases for the first time. My heritage is Oracle E Business Suite (EBS) and the availability of Oracle Weblogic for EBS is fantastic for an Oracle powered organisation that can move away from supporting multiple flavours of application server. The same release made available  - the no downtime patching that Oracle Database 11g introduced with Edition Based Redefinition. I am not saying you must use these features but you must be aware of what each release of your application brings and make a business based decision as to whether it is for you or not. I like to have a simple spreadsheet of features with no-value, nice-to-have, must-have ratings, but make the spreadsheet cumulative so that when you do upgrade you have all the features listed you previously didn’t take up. That way you can avoid the ‘using your phone to read your watch’ scenario. About the Author: Debra Lilley, Fusion Champion, UKOUG Board Member, Fusion User Experience Advocate and ACE Director. Lilley has 18 years experience with Oracle Applications, with E Business Suite since 9.4.1, moving to Business Intelligence Team Lead and Oracle Alliance Director. She has spoken at over 100 conferences worldwide and posts at debrasoraclethoughts  

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  • Making Room for Innovation - Oracle Interactive eBook

    - by Javier Puerta
    Innovation and complexity are two critical topics on the minds of business leaders. Innovation is what gives them a competitive edge; increased complexity is their greatest challenge. Learn how Oracle is helping customers change the game and make room for innovation by simplifying IT. Access the new Oracle interactive e-book, “Simplify IT and Unleash Innovation”. You can download it here.

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  • MySQL - Calculating fields on the fly vs storing calculated data

    - by Christian Varga
    Hi Everyone, I apologise if this has been asked before, but I can't seem to find an answer to a question that I have about calculating on the fly vs storing fields in a database. I read a few articles that suggested it was preferable to calculate when you can, but I would just like to know if that still applies to the following 2 examples. Example 1. Say you are storing data relating to a car. You store the fuel tank size in litres, and how many litres it uses per 100km. You also want to know how many KMs it can travel, which can be calculated from the tank size and economy. I see 2 ways of doing this: When a car is added or updated, calculate the amount of KMs and store this as a static field in the database. Every time a car is accessed, calculate the amount of KMs on the fly. Because the cars economy/tank size doesn't change (although it could be edited), the KMs is a pretty static value. I don't see why we would calculate it every single time the car is accessed. Wouldn't this waste cpu time as opposed to simply storing it in a separate field in the database and calculating only when a car is added or updated? My next example, which is almost an entirely different question (but on the same topic), relates to counting children. Let's say we have a app which has categories and items. We have a view where we display all the categories, and a count of all the items inside each category. Again, I'm wondering what's better. To perform a MySQL query to count all the items in each category every single time the page is accessed? Or store the count in a field in the categories table and update when an item is added / deleted? I know it is redundant to store anything that can be calculated, but I worry that calculating fields or counting records might be slow as opposed to storing the data in a field. If it's not then please let me know, I just want to learn about when to use either method. On a small scale I guess it wouldn't matter either way, but apps like Facebook, would they really count the amount of friends you have every time someone views your profile or would they just store it as a field? I'd appreciate any responses to both of these scenarios, and any resource that might explain the benefits of calculating vs storing. Thanks in advance, Christian

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  • Oracle CRM und Inquira Vertriebsskript für Partner

    - by swalker
    Informationen über Produktpositionierung und Funktionalität von Oracle CRM und InQuira Diese Skripts von Oracle CRM und InQuira (Sales Playbook) unterstützen Sie beim Vertrieb, bei der Identifizierung und Qualifizierung von Vertriebs-Chancen und bei der Entwicklung von Vertriebsszenarien. Setzen Sie Schwerpunkte bei der Verwendung Ihrer Ressourcen, und erweitern Sie Ihr Angebot mit den OPN Specialized-Optionen, die Ihrem Unternehmen zur Verfügung stehen.

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  • mysqli returns only one row instead of multiple rows

    - by Tristan
    Hello, i'm totally new to mysqli and i took a generated code and adapted it for my need. public function getServeurByName($string) { $stmt = mysqli_prepare($this->connection, "SELECT DISTINCT * FROM $this->tablename where GSP_nom=?"); $this->throwExceptionOnError(); mysqli_stmt_bind_param($stmt, 's', $string); $this->throwExceptionOnError(); mysqli_stmt_execute($stmt); $this->throwExceptionOnError(); mysqli_stmt_bind_result($stmt, $row->idServ, $row->timestamp, ........... ........... $row->email); if(mysqli_stmt_fetch($stmt)) { $row->timestamp = new DateTime($row->timestamp); return $row; } else { return null; } } the problem, this example i took the template on returns only one row instead of all the records. how to fix that please ?

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  • SQL for total count and count within that where condition is true

    - by twmulloy
    Hello, I have a single user table and I'm trying to come up with a query that returns the total count of all users grouped by date along with the total count of users grouped by date who are of a specific client. Here is what I have thus far, where there's the total count of users grouped by date, but can't seem to figure out how to get the count of those users where user.client_id = x SELECT user.created, COUNT(user.id) AS overall_count FROM user GROUP BY DATE(user.created) trying for a row result like this: [created] => 2010-05-15 19:59:30 [overall_count] => 10 [client_count] => (some fraction of overall count, the number of users where user.client_id = x grouped by date)

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  • inserting array into database table in single query

    - by Praveen Prasad
    iam having an array of items like [item1,itmem2,item3]; i have to insert these items at a particular userId: final results look like this UserId ItemId 2 || item1 2 || item2 2 || item3 currently iam looping through the array in php code and inserting each item one by one eg foreach($items as $item) { insert into items (UserId,ItemId) value (2,$item); } is it possible i can insert all entries in single query.

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  • MySQL LEFT OUTER JOIN virtual table

    - by user1707323
    I am working on a pretty complicated query let me try to explain it to you. Here is the tables that I have in my MySQL database: students Table --- `students` --- student_id first_name last_name current_status status_change_date ------------ ------------ ----------- ---------------- -------------------- 1 John Doe Active NULL 2 Jane Doe Retread 2012-02-01 students_have_courses Table --- `students_have_courses` --- students_student_id courses_course_id s_date e_date int_date --------------------- ------------------- ---------- ---------- ----------- 1 1 2012-01-01 2012-01-04 2012-01-05 1 2 2012-01-05 NULL NULL 2 1 2012-01-10 2012-01-11 NULL students_have_optional_courses Table --- `students_have_optional_courses` --- students_student_id optional_courses_opcourse_id s_date e_date --------------------- ------------------------------ ---------- ---------- 1 1 2012-01-02 2012-01-03 1 1 2012-01-06 NULL 1 5 2012-01-07 NULL Here is my query so far SELECT `students_and_courses`.student_id, `students_and_courses`.first_name, `students_and_courses`.last_name, `students_and_courses`.courses_course_id, `students_and_courses`.s_date, `students_and_courses`.e_date, `students_and_courses`.int_date, `students_have_optional_courses`.optional_courses_opcourse_id, `students_have_optional_courses`.s_date, `students_have_optional_courses`.e_date FROM ( SELECT `c_s_a_s`.student_id, `c_s_a_s`.first_name, `c_s_a_s`.last_name, `c_s_a_s`.courses_course_id, `c_s_a_s`.s_date, `c_s_a_s`.e_date, `c_s_a_s`.int_date FROM ( SELECT `students`.student_id, `students`.first_name, `students`.last_name, `students_have_courses`.courses_course_id, `students_have_courses`.s_date, `students_have_courses`.e_date, `students_have_courses`.int_date FROM `students` LEFT OUTER JOIN `students_have_courses` ON ( `students_have_courses`.`students_student_id` = `students`.`student_id` AND (( `students_have_courses`.`s_date` >= `students`.`status_change_date` AND `students`.current_status = 'Retread' ) OR `students`.current_status = 'Active') ) WHERE `students`.current_status = 'Active' OR `students`.current_status = 'Retread' ) `c_s_a_s` ORDER BY `c_s_a_s`.`courses_course_id` DESC ) `students_and_courses` LEFT OUTER JOIN `students_have_optional_courses` ON ( `students_have_optional_courses`.students_student_id = `students_and_courses`.student_id AND `students_have_optional_courses`.s_date >= `students_and_courses`.s_date AND `students_have_optional_courses`.e_date IS NULL ) GROUP BY `students_and_courses`.student_id; What I want to be returned is the student_id, first_name, and last_name for all Active or Retread students and then LEFT JOIN the highest course_id, s_date, e_date, and int_date for the those students where the s_date is since the status_change_date if status is 'Retread'. Then LEFT JOIN the highest optional_courses_opcourse_id, s_date, and e_date from the students_have_optional_courses TABLE where the students_have_optional_courses.s_date is greater or equal to the students_have_courses.s_date and the students_have_optional_courses.e_date IS NULL Here is what is being returned: student_id first_name last_name courses_course_id s_date e_date int_date optional_courses_opcourse_id s_date_1 e_date_1 ------------ ------------ ----------- ------------------- ---------- ---------- ------------ ------------------------------ ---------- ---------- 1 John Doe 2 2012-01-05 NULL NULL 1 2012-01-06 NULL 2 Jane Doe NULL NULL NULL NULL NULL NULL NULL Here is what I want being returned: student_id first_name last_name courses_course_id s_date e_date int_date optional_courses_opcourse_id s_date_1 e_date_1 ------------ ------------ ----------- ------------------- ---------- ---------- ------------ ------------------------------ ---------- ---------- 1 John Doe 2 2012-01-05 NULL NULL 5 2012-01-07 NULL 2 Jane Doe NULL NULL NULL NULL NULL NULL NULL Everything is working except one thing, I cannot seem to get the highest students_have_optional_courses.optional_courses_opcourse_id no matter how I form the query Sorry, I just solved this myself after writing this all out I think it helped me think of the solution. Here is the solution query: SELECT `students_and_courses`.student_id, `students_and_courses`.first_name, `students_and_courses`.last_name, `students_and_courses`.courses_course_id, `students_and_courses`.s_date, `students_and_courses`.e_date, `students_and_courses`.int_date, `students_optional_courses`.optional_courses_opcourse_id, `students_optional_courses`.s_date, `students_optional_courses`.e_date FROM ( SELECT `c_s_a_s`.student_id, `c_s_a_s`.first_name, `c_s_a_s`.last_name, `c_s_a_s`.courses_course_id, `c_s_a_s`.s_date, `c_s_a_s`.e_date, `c_s_a_s`.int_date FROM ( SELECT `students`.student_id, `students`.first_name, `students`.last_name, `students_have_courses`.courses_course_id, `students_have_courses`.s_date, `students_have_courses`.e_date, `students_have_courses`.int_date FROM `students` LEFT OUTER JOIN `students_have_courses` ON ( `students_have_courses`.`students_student_id` = `students`.`student_id` AND (( `students_have_courses`.`s_date` >= `students`.`status_change_date` AND `students`.current_status = 'Retread' ) OR `students`.current_status = 'Active') ) WHERE `students`.current_status = 'Active' OR `students`.current_status = 'Retread' ) `c_s_a_s` ORDER BY `c_s_a_s`.`courses_course_id` DESC ) `students_and_courses` LEFT OUTER JOIN ( SELECT * FROM `students_have_optional_courses` ORDER BY `students_have_optional_courses`.optional_courses_opcourse_id DESC ) `students_optional_courses` ON ( `students_optional_courses`.students_student_id = `students_and_courses`.student_id AND `students_optional_courses`.s_date >= `students_and_courses`.s_date AND `students_optional_courses`.e_date IS NULL ) GROUP BY `students_and_courses`.student_id;

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  • Perl like regular expression in Oracle DB

    - by user13136722
    There's regular expression support in Oracle DB Using Regular Expressions in Database Applications Oracle SQL PERL-Influenced Extensions to POSIX Standard But '\b' is not supported which I believe is quite wideliy used in perl and/or other tools perlre - perldoc.perl.org \b Match a word boundary So, I experimented with '\W' which is non-"word" character When combined with beginning-of-line and end-of-line like below, I think it works exactly the same as '\b' SELECT * FROM TAB1 WHERE regexp_like(TEXTCOL1, '(^|\W)a_word($|\W)', 'i')

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  • Customers Go On Record About Oracle ERP and HCM Cloud Services

    - by Kathryn Perry
    Listen to these Oracle customers from Red Robin, Herbalife, LendingClub, and Cricket.talk about how they're using Oracle ERP and HCM Cloud Services. Collectively they're driving cost savings, managing global, fast paced growth, automating processes, implementing quickly in the cloud, and much more. Here's the video link: http://www.youtube.com/user/FusionAppsAtOracle

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  • Parsing CSV File to MySQL DB in PHP

    - by Austin
    I have a some 350-lined CSV File with all sorts of vendors that fall into Clothes, Tools, Entertainment, etc.. categories. Using the following code I have been able to print out my CSV File. <?php $fp = fopen('promo_catalog_expanded.csv', 'r'); echo '<tr><td>'; echo implode('</td><td>', fgetcsv($fp, 4096, ',')); echo '</td></tr>'; while(!feof($fp)) { list($cat, $var, $name, $var2, $web, $var3, $phone,$var4, $kw,$var5, $desc) = fgetcsv($fp, 4096); echo '<tr><td>'; echo $cat. '</td><td>' . $name . '</td><td><a href="http://www.' . $web .'" target="_blank">' .$web.'</a></td><td>'.$phone.'</td><td>'.$kw.'</td><td>'.$desc.'</td>' ; echo '</td></tr>'; } fclose($file_handle); show_source(__FILE__); ?> First thing you will probably notice is the extraneous vars within the list(). this is because of how the excel spreadsheet/csv file: Category,,Company Name,,Website,,Phone,,Keywords,,Description ,,,,,,,,,, Clothes,,4imprint,,4imprint.com,,877-466-7746,,"polos, jackets, coats, workwear, sweatshirts, hoodies, long sleeve, pullovers, t-shirts, tees, tshirts,",,An embroidery and apparel company based in Wisconsin. ,,Apollo Embroidery,,apolloemb.com,,1-800-982-2146,,"hats, caps, headwear, bags, totes, backpacks, blankets, embroidery",,An embroidery sales company based in California. One thing to note is that the last line starts with two commas as it is also listed within "Clothes" category. My concern is that I am going about the CSV output wrong. Should I be using a foreach loop instead of this list way? Should I first get rid of any unnecessary blank columns? Please advise any flaws you may find, improvements I can use so I can be ready to import this data to a MySQL DB.

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  • how to tackle a custom forms database

    - by Neil Hickman
    I'm currently researching a project for the place that I work in. We are trying to create a system that will allow forms to be set up dynamically from a database. My question is what database structure would best suit something like this? I currently have a structure of: forms_form forms_formfields forms_formdata I don't think this is the most appropriate layout for this. Basically to make is make sense I need to be able to make a form within the database that can have infinite fields all customized and have the data when submitted stored in the database.

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  • It's Alive!

    - by Oracle OpenWorld Blog Team
    See what leading-edge, provocative, and fascinating new content will be featured at Oracle OpenWorld in 2012. by Karen Shamban It’s what you’ve been waiting for. The Oracle OpenWorld Content Catalog—the central repository for information on sessions, demos, labs, user groups, exhibitors, and more—is live. Right now. In the Content Catalog you can search on tracks, session types, session categories, keywords, and tags. Or, you can search for your favorite speakers to see what they’re presenting this year. And, directly from the catalog, you can share sessions you’re interested in with friends and colleagues through a broad array of social media channels. Start checking out Oracle OpenWorld content now to plan your week at the conference. Then you’ll be ready to sign up for all of your sessions in mid-July when the scheduling tool goes live. Thinking of cross-registering for JavaOne? The JavaOne Content Catalog is also live at this very minute so you can see what great content is on offer there.

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  • Help with SQL query (Calculate a ratio between two entitiess)

    - by Mestika
    Hi, I’m going to calculate a ratio between two entities but are having some trouble with the query. The principal is the same to, say a forum, where you say: A user gets points for every new thread. Then, calculate the ratio of points for the number of threads. Example: User A has 300 points. User A has started 6 thread. The point ratio is: 50:6 My schemas look as following: student(studentid, name, class, major) course(courseid, coursename, department) courseoffering(courseid, semester, year, instructor) faculty(name, office, salary) gradereport(studentid, courseid, semester, year, grade) The relations is a following: Faculity(name) = courseoffering(instructor) Student(studentid) = gradereport (studentid) Courseoffering(courseid) = course(courseid) Gradereport(courseid) = courseoffering(courseid) I have this query to select the faculty names there is teaching one or more students: SELECT COUNT(faculty.name) FROM faculty, courseoffering, gradereport, student WHERE faculty.name = courseoffering.instructor AND courseoffering.courseid = gradereport.courseid AND gradereport.studentid = student.studentid My problem is to find the ratio between the faculty members salary in regarding to the number of students they are teaching. Say, a teacher get 10.000 in salary and teaches 5 students, then his ratio should be 1:5. I hope that someone has an answer to my problem and understand what I'm having trouble with. Thanks Mestika

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  • ASP.Net / MySQL : Translating content into several languages

    - by philwilks
    I have an ASP.Net website which uses a MySQL database for the back end. The website is an English e-commerce system, and we are looking at the possibility of translating it into about five other languages (French, Spanish etc). We will be getting human translators to perform the translation - we've looked at automated services but these aren't good enough. The static text on the site (e.g. headings, buttons etc) can easily be served up in multiple languages via .Net's built in localization features (resx files etc). The thing that I'm not so sure about it how best to store and retrieve the multi-language content in the database. For example, there is a products table that includes these fields... productId (int) categoryId (int) title (varchar) summary (varchar) description (text) features (text) The title, summary, description and features text would need to be available in all the different languages. Here are the two options that I've come up with... Create additional field for each language For example we could have titleEn, titleFr, titleEs etc for all the languages, and repeat this for all text columns. We would then adapt our code to use the appropriate field depending on the language selected. This feels a bit hacky, and also would lead to some very large tables. Also, if we wanted to add additional languages in the future it would be time consuming to add even more columns. Use a lookup table We could create a new table with the following format... textId | languageId | content ------------------------------- 10 | EN | Car 10 | FR | Voiture 10 | ES | Coche 11 | EN | Bike 11 | FR | Vélo We'd then adapt our products table to reference the appropriate textId for the title, summary, description and features instead of having the text stored in the product table. This seems much more elegant, but I can't think of a simple way of getting this data out of the database and onto the page without using complex SQL statements. Of course adding new languages in the future would be very simple compared to the previous option. I'd be very grateful for any suggestions about the best way to achieve this! Is there any "best practice" guidance out there? Has anyone done this before?

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  • Need help with a SELECT statement

    - by Travis
    I express the relationship between records and searchtags that can be attached to records like so: TABLE RECORDS id name TABLE SEARCHTAGS id recordid name I want to be able to SELECT records based on the searchtags that they have. For example, I want to be able to SELECT all records that have searchtags: (1 OR 2 OR 5) AND (6 OR 7) AND (10) Using the above data structure, I am uncertain how to structure the SQL to accomplish this. Any suggestions? Thanks!

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  • How can get unique values from data table using dql?

    - by piemesons
    I am having a table in which there is a column in which various values are stored.i want to retrieve unique values from that table using dql. Doctrine_Query::create() ->select('rec.school') ->from('Records rec') ->where("rec.city='$city' ") ->execute(); Now i want only unique values. Can anybody tell me how to do that... Edit Table Structure: CREATE TABLE IF NOT EXISTS `records` ( `id` int(11) NOT NULL AUTO_INCREMENT, `state` varchar(255) COLLATE utf8_unicode_ci DEFAULT NULL, `city` varchar(255) COLLATE utf8_unicode_ci DEFAULT NULL, `school` varchar(255) COLLATE utf8_unicode_ci DEFAULT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci AUTO_INCREMENT=16334 ; This is the Query I am using: Doctrine_Query::create() ->select('DISTINCT rec.city') ->from('Records rec') ->where("rec.state = '$state'") // ->getSql(); ->execute(); Generting Sql for this gives me: SELECT DISTINCT r.id AS r__id, r.city AS r__city FROM records r WHERE r.state = 'AR' Now check the sql generated:::: DISTINCT is on 'id' column where as i want Distinct on city column. Anybody know how to fix this. EDIT2 Id is unique cause its an auto incremental value.Ya i have some real duplicates in city column like: Delhi and Delhi. Right.. Now when i am trying to fetch data from it, I am getting Delhi two times. How can i make query like this: select DISTINCT rec.city where state="xyz"; Cause this will give me the proper output. EDIT3: Anybody who can tell me how to figure out this query..???

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  • Write a SQL to meet my requirement.

    - by rgksugan
    I have been trying to solve this problem for a lot of days. But wouldn't. Please help me. I need a SQL to list product_code, product_name, qty_sold, last_order_date for all the products that have been sold within a date range sorted by the number of quantity sold. My Table structure: tbl_product(product_id,product_code,product_name) tbl_order_detail(order_item_id,order_id,product_id,quantity) tbl_order(order_id,order_date)

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  • mysql changing delimiter

    - by jimsmith
    I'm trying to add this function using php myadmin, first off I get on error line 5, which is apparently because you need to change the delimiter from ; to something else so i tried this DELIMITER | CREATE FUNCTION LEVENSHTEIN (s1 VARCHAR(255), s2 VARCHAR(255)) RETURNS INT DETERMINISTIC BEGIN DECLARE s1_len, s2_len, i, j, c, c_temp, cost INT; DECLARE s1_char CHAR; DECLARE cv0, cv1 VARBINARY(256); SET s1_len = CHAR_LENGTH(s1), s2_len = CHAR_LENGTH(s2), cv1 = 0x00, j = 1, i = 1, c = 0; IF s1 = s2 THEN RETURN 0; ELSEIF s1_len = 0 THEN RETURN s2_len; ELSEIF s2_len = 0 THEN RETURN s1_len; ELSE WHILE j <= s2_len DO SET cv1 = CONCAT(cv1, UNHEX(HEX(j))), j = j + 1; END WHILE; WHILE i <= s1_len DO SET s1_char = SUBSTRING(s1, i, 1), c = i, cv0 = UNHEX(HEX(i)), j = 1; WHILE j <= s2_len DO SET c = c + 1; IF s1_char = SUBSTRING(s2, j, 1) THEN SET cost = 0; ELSE SET cost = 1; END IF; SET c_temp = CONV(HEX(SUBSTRING(cv1, j, 1)), 16, 10) + cost; IF c > c_temp THEN SET c = c_temp; END IF; SET c_temp = CONV(HEX(SUBSTRING(cv1, j+1, 1)), 16, 10) + 1; IF c > c_temp THEN SET c = c_temp; END IF; SET cv0 = CONCAT(cv0, UNHEX(HEX(c))), j = j + 1; END WHILE; SET cv1 = cv0, i = i + 1; END WHILE; END IF; RETURN c; END DELIMITER ; But I get this error: #1064 - 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 'delimiter | Please help !?

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  • What's New in the latest release of Oracle User Productivity Kit 11.0

    Enterprises are always looking to reduce overall project timelines, optimize business processes, and increase acceptance of their enterprise applications to ensure maximum ROI. The latest release of Oracle User Productivity Kit helps customers streamline the workflow process for the creation of content and offers conceptual-based assessment options to increase user adoption. Discover what is great and innovative about the latest release of Oracle UPK and UPK Professional. Learn about the integration of the UPK Developer and the Knowledge Center, which provides developers with a centralized, web-based platform for content deployment, tracking, and reporting.

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