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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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  • Shrinking a Linux OEL 6 virtual Box image (vdi) hosted on Windows 7

    - by AndyBaker
    v\:* {behavior:url(#default#VML);} o\:* {behavior:url(#default#VML);} w\:* {behavior:url(#default#VML);} .shape {behavior:url(#default#VML);} Recently for a customer demonstration there was a requirement to build a virtual box image with Oracle Enterprise Manager Cloud Control 12c. This meant installing OEL Linux 6 as well as creating an 11gr2 database and Oracle Enterprise Manager Cloud Control 12c on a single virtual box. Storage was sized at 300Gb using dynamically allocated storage for the virtual box and about 10Gb was used for Linux and the initial build. After copying over all the binaries and performing all the installations the virtual box became in the region of 80Gb used size on the host operating system, however internally it only really needed around 20Gb. This meant 60Gb had been used when copying over all the binaries and although now free was not returned to the host operating system due to the growth of the virtual box storage '.vdi' file.  Once the ‘vdi’ storage had grown it is not shrunk automatically afterwards. Space is always tight on the laptop so it was desirable to shrink the virtual box back to a minimal size and here is the process that was followed. Install 'zerofree' Linux package into the OEL6 virtual box The RPM was downloaded and installed from a site similar to below; http://rpm.pbone.net/index.php3/stat/4/idpl/12548724/com/zerofree-1.0.1-5.el5.i386.rpm.html A simple internet search for ’zerofree Linux rpm’ was easy to perform and find the required rpm. Execute 'zerofree' package on the desired Linux file system To execute this package the desired file system needs to be mounted read only. The following steps outline this process. As root: # umount /u01 As root:# mount –o ro –t ext4 /u01 NOTE: The –o is options and the –t is the file system type found in the /etc/fstab. Next run zerofree against the required storage, this is located by a simple ‘df –h’ command to see the device associated with the mount. As root:# zerofree –v /dev/sda11   NOTE: This takes a while to run but the ‘-v’ option gives feedback on the process. What does Zerofree do? Zerofree’s purpose is to go through the file system and zero out any unused sectors on the volume so that the later stages can shrink the virtual box storage obtaining the free space back. When zerofree has completed the virtual box can be shutdown as the last stage is performed on the physical host where the virtual box vdi files are located. Compact the virtual box ‘.vdi’ files The final stage is to get virtual box to shrink back the storage that has been correctly flagged as free space after executing zerofree. On the physical host in this case a windows 7 laptop a DOS window was opened. At the prompt the first step is to put the virtual box binaries onto the PATH. C:\ >echo %PATH%   The above shows the current value of the PATH environment variable. C:\ >set PATH=%PATH%;c:\program files\Oracle\Virtual Box;   The above adds onto the existing path the virtual box binary location. C:\>cd c:\Users\xxxx\OEL6.1   The above changes directory to where the VDI files are located for the required virtual box machine. C:\Users\xxxxx\OEL6.1>VBoxManage.exe modifyhd zzzzzz.vdi compact  NOTE: The zzzzzz.vdi is the name of the required vdi file to shrink. Finally the above command is executed to perform the compact operation on the ‘.vdi’ file(s). This also takes a long time to complete but shrinks the VDI file back to a minimum size. In the case of the demonstration virtual box OEM12c this reduced the virtual box to 20Gb from 80Gb which was a great outcome to achieve.

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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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  • PASS Summit 2012, Day 1

    - by KKline
    One of the most positive experiences I can have, as a former leader of the PASS organization, is when I see a neophyte become a passionate support and champion for the community. On my first day in Seattle, before the event had even begun, I was stopped several by people who'd attended their first PASS Summit last year. But this year, they were excited to tell me that they'd started user groups in their own community, spoken for the first time at a PASS event, or even helped launch a SQL Saturday...(read more)

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  • What to leave when you're leaving

    - by BuckWoody
    There's already a post on this topic - sort of. I read this entry, where the author did a good job on a few steps, but I found that a few other tips might be useful, so if you want to check that one out and then this post, you might be able to put together your own plan for when you leave your job.  I once took over the system administrator (of which the Oracle and SQL Server servers were a part) at a mid-sized firm. The outgoing administrator had about a two- week-long scheduled overlap with me, but was angry at the company and told me "hey, I know this is going to be hard on you, but I want them to know how important I was. I'm not telling you where anything is or what the passwords are. Good luck!" He then quit that day. It took me about three days to find all of the servers and crack the passwords. Yes, the company tried to take legal action against the guy and all that, but he moved back to his home country and so largely got away with it. Obviously, this isn't the way to leave a job. Many of us have changed jobs in the past, and most of us try to be very professional about the transition to a new team, regardless of the feelings about a particular company. I've been treated badly at a firm, but that is no reason to leave a mess for someone else. So here's what you should put into place at a minimum before you go. Most of this is common sense - which of course isn't very common these days - and another good rule is just to ask yourself "what would I want to know"? The article I referenced at the top of this post focuses on a lot of documentation of the systems. I think that's fine, but in actuality, I really don't need that. Even with this kind of documentation, I still perform a full audit on the systems, so in the end I create my own system documentation. There are actually only four big items I need to know to get started with the systems: 1. Where is everything/everybody?The first thing I need to know is where all of the systems are. I mean not only the street address, but the closet or room, the rack number, the IU number in the rack, the SAN luns, all that. A picture here is worth a thousand words, which is why I really like Visio. It combines nice graphics, full text and all that. But use whatever you have to tell someone the physical locations of the boxes. Also, tell them the physical location of the folks in charge of those boxes (in case you aren't) or who share that responsibility. And by "where" in this case, I mean names and phones.  2. What do they do?For both the servers and the people, tell them what they do. If it's a database server, detail what each database does and what application goes to that, and who "owns" that application. In my mind, this is one of hte most important things a Data Professional needs to know. In the case of the other administrtors or co-owners, document each person's responsibilities.   3. What are the credentials?Logging on/in and gaining access to the buildings are things that the new Data Professional will need to do to successfully complete their job. This means service accounts, certificates, all of that. The first thing they should do, of course, is change the passwords on all that, but the first thing they need is the ability to do that!  4. What is out of the ordinary?This is the most tricky, and perhaps the next most important thing to know. Did you have to use a "special" driver for that video card on server X? Is the person that co-owns an application with you mentally unstable (like me) or have special needs, like "don't talk to Buck before he's had coffee. Nothing will make any sense"? Do you have service pack requirements for a specific setup? Write all that down. Anything that took you a day or longer to make work is probably a candidate here. This is my short list - anything you care to add? Share this post: email it! | bookmark it! | digg it! | reddit! | kick it! | live it!

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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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  • Bringing Alerts and Operators together with Notifications

    - by fatherjack
    I have covered SQL Server Alerts (Alerts are good, arent they?) on this blog before and I more recently did a post regarding Notifications (Are your Jobs talking to you) and how they should be configured. Now we need to check that these things are linked up so that when an Alert condition is met that you get the appropriate Notifications sent to Operators. Straight into the code we need and then a review of what it does ... DECLARE @ChosenOperator SYSNAME DECLARE @FailSafeOp TABLE ...(read more)

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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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  • 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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  • 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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  • 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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  • php connecting to mysql server(localhost) very slow

    - by Ahmad
    actually its little complicated: summary: the connection to DB is very slow. the page rendering takes around 10 seconds but the last statement on the page is an echo and i can see its output while the page is loading in firefox (IE is same). in google chrome the output becomes visible only when the loading finishes. loading time is approximately the same across browsers. on debugging i found out that its the DB connectivity that is creating problem. the DB was on another machine. to debug further. i deployed the DB on my local machine .. so now the DB connection is at 127.0.0.1 but the connectivity still takes long time. this means that the issue is with APACHE/PHP and not with mysql. but then i deployed my code on another machine which connects to DB remotely.and everything seems fine. basically the application uses couple of mod_rewrite.. but i removed all the .htaccess files and the slow connectivity issue remains.. i installed another APACHE on my machine and used default settings. the connection was still very slow. i added following statements to measure the execution time $stime = microtime(); $stime = explode(" ",$stime); $stime = $stime[1] + $stime[0]; // my code -- it involves connection to DB $mtime = microtime(); $mtime = explode(" ",$mtime); $mtime = $mtime[1] + $mtime[0]; $totaltime = ($mtime - $stime); echo $totaltime; the output is 0.0631899833679 but firebug Net panel shows total loading time of 10-11 seconds. same is the case with google chrome i tried to turn off windows firewall.. connectivity is still slow and i just can't quite find the reason.. i've tried multiple DB servers.. multiple apaches.. nothing seems to be working.. any idea of what might be the problem?

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  • Copy Table to Another Database

    - by Derek Dieter
    There are few methods of copying a table to another database, depending on your situation. Same SQL Server Instance If trying to copy a table to a database that is on the same instance of SQL Server, The easiest solution is to use a SELECT INTO while using the fully qualifed database names.SELECT * INTO Database2.dbo.TargetTable FROM Database1.dbo.SourceTableThis will [...]

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  • Join unrelated tables through a second level connected table

    - by Andy M
    Hello! I have two tables of activities on a page: Views & Comments. Views id timestamp project_id user_id page_id Comments id timestamp project_id user_id page_id comment Pages id project_id title Now pages are related to projects: Projects id account_id title I am trying to create a summary page that combines views and comments ordered by time (so that the most recent views/comments are at the beginning, grouped by projects. Also, only projects for a specific account. So the result could potentially be: Project 1 View 5 (June 20th) View 4 (June 18th) Comment 5 (June 15th) Comment 4 (June 14th) Comment 3 (June 12th) Project 3 View 3 (June 10th) View 2 (June 8th) Comment 2 (June 7th) Project 2 View 1 (June 5th) Comment 1 (June 4th) If you could help with how to do this using SQL (or even doctrine) that would be awesome. Thank you.

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  • Table alias -- Unkown column in field list

    - by Jason
    Hi all, I have a sql query which is executing a LEFT JOIN on 2 tables in which some of the columns are ambiguous. I can prefix the joined tables but when I try to prefix one of the columns from the table in the FROM clause, it tells me Unknown column. I even tried giving that table an alias like so ...From points AS p and using "p" to prefix the tables but that didn't work either. Can someone tell me what I'm doing wrong. Here is my query: SELECT point_title, point_url, address, city, state, zip_code, phone, `points`.`lat`, `points`.`longi`, featured, kmlno, image_url, category.title, category_id, point_id, lat, longi, reviews.star_points, reviews.review_id, count(reviews.point_id) as totals FROM (SELECT *, ( 3959 * acos( cos( radians('37.7717185') ) * cos( radians( lat ) ) * cos( radians( longi ) - radians('-122.4438929') ) + sin( radians('37.7717185') ) * sin( radians( lat ) ) ) ) AS distance FROM points HAVING distance < '25') as distResults LEFT JOIN category USING ( category_id ) LEFT JOIN reviews USING ( point_id ) WHERE (point_title LIKE '%Playgrounds%' OR category.title LIKE '%Playgrounds%') GROUP BY point_id ORDER BY totals DESC, distance LIMIT 0 , 10

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  • $query returns results but not the ones i want: $query looks good to me :S

    - by Toni Michel Caubet
    I'll start again, Lets say My data is: Table element (id,name,....) 1, name element 1, .... 2, name element 2, .... 3, name element 3, .... Table tags (id,name,id_element, ....) 1, happy , 1 2, result, 1 3, very , 1 4, element, 2 5, another, 3 6, element, 1 7, happy, 2 So if search is 'very, happy,element,result': Results i would like 1) element with id = 2 because it has all tags 2) element with id = 1 because it has the tag 'element' and the tag 'happy' (only 2 less taggs) 3) .... (only 3 less taggs) So if search is 'happy,element': Results i would like 1) element with id = 1 because it has all tags (and no more) 2) element with id = 2 because it has the tag 'element' and the tag 'happy' (and two more tags) 3) .... and 3 more tags This is an echo to my query: (it doesn't fit al requirements i wrote, but its first test to find with matched tags) SELECT element.id as id_deseada,tagg.* FROM element,tagg WHERE tagg.id_element = element.id AND tagg.nombre IN ('happy','tagg','result') GROUP BY tagg.id_element ORDER BY element.votos This returns 10 duplicated elements... :S and doen't even have all taggs (and on database there are taggs with 'happy' results) if it helps, thats how i get the elements of a tag (by name and with only one tagg) $query = "SELECT element.id FROM element,tagg WHERE tagg.nombre = '$nombre_tagg' AND tagg.id_element = element.id AND lan = '$lan' GROUP BY tagg.id_element"; I hope it's a bit easier to understand now, excuse my english.. :) Thanks a lot for you possible aportation!

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  • How do I re-create a MySQL InnoDB table from an .ibd file?

    - by knorv
    Assume that the following MySQL files have been restored from a backup tape: tablename.frm tablename.ibd Furthermore, assume that the MySQL installation was running with innodb_file_per_table and that the database was cleanly shutdown with mysqladmin shutdown. Given a fresh install of the same MySQL version that the restored MySQL files were taken from, how do I import the data from tablename.ibd/tablename.frm into this new install?

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  • More Tables or More Databases?

    - by BuckWoody
    I got an e-mail from someone that has an interesting situation. He has 15,000 customers, and he asks if he should have a database for their data per customer. Without a LOT more data it’s impossible to say, of course, but there are some general concepts to keep in mind. Whenever you’re segmenting data, it’s all about boundary choices. You have not only boundaries around how big the data will get, but things like how many objects (tables, stored procedures and so on) that will be involved, if there are any cross-sections of data (do they share location or product information) and – very important – what are the security requirements? From the answer to these types of questions, you now have the choice of making multiple tables in a single database, or using multiple databases. A database carries some overhead – it needs a certain amount of memory for locking and so on. But it has a very clean boundary – everything from objects to security can be kept apart. Having multiple users in the same database is possible as well, using things like a Schema. But keeping 15,000 schemas can be challenging as well. My recommendation in complex situations like this is similar to a post on decisions that I did earlier – I lay out the choices on a spreadsheet in rows, and then my requirements at the top in the columns. I  give each choice a number based on how well it meets each requirement. At the end, the highest number wins. And many times it’s a mix – perhaps this person could segment customers into larger regions or districts or products, in a database. Within that database might be multiple schemas for the customers. Of course, he needs to query across all customers, that becomes another requirement. Share this post: email it! | bookmark it! | digg it! | reddit! | kick it! | live it!

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  • Java: Making concurrent MySQL queries from multiple clients synchronised

    - by Misha Gale
    I work at a gaming cybercafe, and we've got a system here (smartlaunch) which keeps track of game licenses. I've written a program which interfaces with this system (actually, with it's backend MySQL database). The program is meant to be run on a client PC and (1) query the database to select an unused license from the pool available, then (2) mark this license as in use by the client PC. The problem is, I've got a concurrency bug. The program is meant to be launched simultaneously on multiple machines, and when this happens, some machines often try and acquire the same license. I think that this is because steps (1) and (2) are not synchronised, i.e. one program determines that license #5 is available and selects it, but before it can mark #5 as in use another copy of the program on another PC tries to grab that same license. I've tried to solve this problem by using transactions and table locking, but it doesn't seem to make any difference - Am I doing this right? Here follows the code in question: public LicenseKey Acquire() throws SmartLaunchException, SQLException { Connection conn = SmartLaunchDB.getConnection(); int PCID = SmartLaunchDB.getCurrentPCID(); conn.createStatement().execute("LOCK TABLE `licensekeys` WRITE"); String sql = "SELECT * FROM `licensekeys` WHERE `InUseByPC` = 0 AND LicenseSetupID = ? ORDER BY `ID` DESC LIMIT 1"; PreparedStatement statement = conn.prepareStatement(sql); statement.setInt(1, this.id); ResultSet results = statement.executeQuery(); if (results.next()) { int licenseID = results.getInt("ID"); sql = "UPDATE `licensekeys` SET `InUseByPC` = ? WHERE `ID` = ?"; statement = conn.prepareStatement(sql); statement.setInt(1, PCID); statement.setInt(2, licenseID); statement.executeUpdate(); statement.close(); conn.commit(); conn.createStatement().execute("UNLOCK TABLES"); return new LicenseKey(results.getInt("ID"), this, results.getString("LicenseKey"), results.getInt("LicenseKeyType")); } else { throw new SmartLaunchException("All licenses of type " + this.name + "are in use"); } }

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