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• Turn A Flash Drive Into a Portable Web Server

  - by Matthew Guay

  Portable applications are very useful for getting work done on the go, but how about portable servers?  Here’s how you can turn your flash drive into a portable web server. Getting Started To put a full web server on our flash drive, we’re going to use XAMPP Lite.  This lightweight, preconfigured server includes recent versions of Apache, MySQL, and PHP so you can run most websites and webapps directly from it.  You could use the full XAMPP, which includes more features such as a FileZilla FTP server and OpenSSL, but for most purposes, the light version is plenty for a portable server. Download the latest version of XAMPP Lite (link below).  In this tutorial, we used the self-extracting EXE version; you could choose the ZIP file and extract the files yourself, but we found it easier to use the executable. Run the installer, and click Browse choose where to install your server. Select your flash drive, or a folder in it, and click Ok.  Make sure your flash drive has at least 250MB of available storage space.  XAMPP will create an xampplite folder and store all the files in it during the installation.   Click Install, and all of the files will be extracted to your flash drive.  This may take a few moments depending on your flash drive’s speed. When the extraction process is finished, a Command Prompt window will open to finish the installation.  The first prompt will ask if you want to add shortcuts to the start menu and desktop; enter “n” since we don’t want to create start menu links to our portable server. Now enter “y” to configure XAMPP’s directories automatically. Finally, enter “y” to make XAMPP fully portable.  It will set up the servers to run without specific drive letters so your server will run from any computer. XAMPP will finalize your changes; press Enter when everything is completed. Setup will automatically launch the command line version of XAMPP.  On first run, confirm that your time zone is correct. And that’s it!  You can now run XAMPP’s control panel by entering 1, or you can exit and run XAMPP from any other computer with your flash drive. To complete your portable webserver kit, you may want to install Portable Firefox or Iron Browser on your flash drive so you always have your favorite browser ready to use. Running your portable server Using your portable server is very simple.  Open the xampplite folder on your flash drive and launch xampp-control.exe. Click Start beside Apache and MySql to get your webserver running. Please note: Do not check the Svc box, as this will run the server as a Windows service.  To keep XAMPP portable, you do not want it running as a service! Windows Firewall may prompt you that it blocked the server; click Allow access to let your server run. Once they’re running, you can click Admin to open the default XAMPP admin page running from your local webserver.  Or, you can view it by browsing to http://localhost/ or http://127.0.0.1/ in your browser. If everything is working correctly, you should see this page in your browser.  Choose your default language… And then you’ll see the default XAMPP admin page.   Click the Status link on the left sidebar to make sure everything is running correctly. If you click the Admin button for MySql in the XAMPP Control Panel, it will open phpMyAdmin in your default browser.  Alternately, you can open the MySql admin page by entering http://localhost/phpmyadmin/ or http://127.0.0.1/phpmyadmin/ in your favorite browser. Now you can add your own webpages to your webserver.  Save all of your web files in the \xampplight\htdocs\ folder on your flash drive. Install WordPress in your portable server Since XAMPP Lite includes MySql and PHP, you can even run webapps such as WordPress, the popular CMS and blogging platform.  Download WordPress (link below), and extract the files to the \xampplite\htdocs folder on your flash drive. Now all of the WordPress files are stored in \xampplite\htdocs\wordpress on your flash drive. We still need to setup WordPress on our portable server.  Open your MySql admin page http://localhost/phpmyadmin/ to create a new database for WordPress.  Enter a name for your database in the “Create new database” box, and click Create. Click the Privileges tab on the top, and the select “Add a new User”.   Enter a username and password for the database, and then click the Go button on the bottom of the page. Using WordPress Now, in your browser, enter http://localhost/wordpress/wp-admin/install.php.  Click Create a Configuration File to continue. Make sure you have your Database name, username, and password we created previously, and click “Let’s Go!” Enter your WordPress database name, username, and password, leave the other two entries as default, and click Submit. You should now have the database all ready to go.  Click “Run the install” to finish installing WordPress. Enter a title, username, and password for your test blog, as well as your email address, and then click “Install WordPress”. You now have a portable install of WordPress.  Click “Log In” to  access your WordPress admin page. Enter your username and password, and click Log In. Here you can add pages, posts, themes, extensions, and anything else just like you would on a normal WordPress site.  This is a great way to experiment with WordPress without messing up your real website. You can view your portable WordPress site by entering http://localhost/wordpress/ in your address bar. Closing your server When you’re done running your test server, click the Stop button on each of the services and then click the Exit button in the XAMPP control panel.  If you press the exit button on the top of the window, it will just minimize the control panel to the tray.   Alternately, you can shutdown your server by running xampp_stop.exe from your xampplite folder. Conclusion XAMPP Lite gives you a great way to run a full webserver directly from your flash drive.  Now, anywhere you go, you can test and tweak your webpages and webapps from any Windows computer.  Links Download XAMPP Lite Download WordPress Similar Articles Productive Geek Tips BitLocker To Go Encrypts Portable Flash Drives in Windows 7How To Use BitLocker on Drives without TPMSpeed up Your Windows Vista Computer with ReadyBoostView and Manage Flash Cookies the Easy WayInstall and Run Applications from Your iPod, Flash Drive or Mp3 Player TouchFreeze Alternative in AutoHotkey The Icy Undertow Desktop Windows Home Server – Backup to LAN The Clear & Clean Desktop Use This Bookmarklet to Easily Get Albums Use AutoHotkey to Assign a Hotkey to a Specific Window Latest Software Reviews Tinyhacker Random Tips DVDFab 6 Revo Uninstaller Pro Registry Mechanic 9 for Windows PC Tools Internet Security Suite 2010 OutlookStatView Scans and Displays General Usage Statistics How to Add Exceptions to the Windows Firewall Office 2010 reviewed in depth by Ed Bott FoxClocks adds World Times in your Statusbar (Firefox) Have Fun Editing Photo Editing with Citrify Outlook Connector Upgrade Error

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• How to connect a WordPress contact form to another database which uses a form script on a static site?

  - by eirlymeyer

  Static Site B has two separate contact form scripts. One script processes leads via a script developed using Cold Fusion. Another script processes leads via a script using MySql Database. New Site A is being developed using WordPress. How do I use a WordPress Contact Form plug-in to integrate these two scripts (built on ColdFusion, and uses the existing MySQL database) to ensure the same functionality and processing of leads.

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• jQuery "Auto Post-back" Select/Drop-Down List

  - by Doug Lampe

  I have one common piece of jQuery code which I use to submit a form any time the selection changes on a drop-down list (HTML select tag).  This is similar to setting AutoPostBack = true in ASP.Net.  I use a single CSS class (autoSubmit) to annotate that I want the drop-down to force the form to submit on change so the HTML looks something like this: <select id="myAutoSubmitDropDown" name="myAutoSubmitDropDown" class="autoSubmit">     <option value="1">Option 1</option>     <option value="2">Option 2</option> </select> Then the following jQuery will look for any element with this CSS class and submit the parent form when the value is changed: function wireUpAutoSubmit() {   $(".autoSubmit").each(function (index) {     $(this).change(function () {       $(this).closest('form').submit();     })   }); } I put this in a separate function since I might need to wire this up explicitly after an ajax call.  Therefore I use the following code to set this method to fire when the DOM is loaded: $(document).ready(function () {   wireUpAutoSubmit(); });

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• How-to create a select one choice listing common time zones

  - by frank.nimphius

  Normal 0 false false false EN-US X-NONE X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} ADF Faces provides an option to query a list of common timezones for display in a Select One Choice component. The EL expression for this is #{af:getCommonTimeZoneSelectItems()}. To use this expression in a Single Select One Choice component, drag and drop the component from the Oracle JDeveloper Component Palette into a JSF page.  In the opened dialog, copy the expression into the Value property below the Bind to list (select items) header. <af:selectOneChoice label="TimeZones" id="soc1">  <f:selectItems value="#{af:getCommonTimeZoneSelectItems()}"                          id="si1"/></af:selectOneChoice>

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• Need Help getting perl module DBD::mysql installed for bugzilla on RedHat.

  - by Alos Diallo

  Hi everyone I am having some issues getting Bugzilla setup, I have the software on the server and am trying to get the pre-rec's setup. I am using RedHat 4.1.2-42. I have all of the required perl modules save one:DBD::mysql When I try: sudo perl install-module.pl DBD::mysql I get the following response(this is only an excerpt): rm -f blib/arch/auto/DBD/mysql/mysql.so LD_RUN_PATH="/usr/lib64/mysql:/usr/lib64:/lib64" /usr/bin/perl myld gcc -shared -O2 -g -pipe -Wall -Wp,-D_FORTIFY_SOURCE=2 -fexceptions -fstack-protector --param=ssp-buffer-size=4 -m64 -mtune=generic dbdimp.o mysql.o -o blib/arch/auto/DBD/mysql/mysql.so \ -L/usr/lib64/mysql -lmysqlclient -lz -lcrypt -lnsl -lm -L/usr/lib64 -lssl -lcrypto \ /usr/bin/ld: skipping incompatible /usr/lib/libssl.so when searching for -lssl /usr/bin/ld: skipping incompatible /usr/lib/libssl.a when searching for -lssl /usr/bin/ld: cannot find -lssl collect2: ld returned 1 exit status make: * [blib/arch/auto/DBD/mysql/mysql.so] Error 1 /usr/bin/make -- NOT OK Running make test Can't test without successful make Running make install make had returned bad status, install seems impossible I then tried the following: CFLAGS="-I/usr/lib64/mysql:/usr/lib64:/lib64" perl install-module.pl DBD::mysql I get the same result. I have also tried to install it using CPAN but also get the same result. Right now I have DBD-mysql v3.0007 but need (v4.00) Also when I try to install open ssl it says I have the latest version. Does anyone know what I have to do to get this to work? Any help would be greatly appreciated. Thank you

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• Building Queries Systematically

  - by Jeremy Smyth

  The SQL language is a bit like a toolkit for data. It consists of lots of little fiddly bits of syntax that, taken together, allow you to build complex edifices and return powerful results. For the uninitiated, the many tools can be quite confusing, and it's sometimes difficult to decide how to go about the process of building non-trivial queries, that is, queries that are more than a simple SELECT a, b FROM c; A System for Building Queries When you're building queries, you could use a system like the following:  Decide which fields contain the values you want to use in our output, and how you wish to alias those fields Values you want to see in your output Values you want to use in calculations . For example, to calculate margin on a product, you could calculate price - cost and give it the alias margin. Values you want to filter with. For example, you might only want to see products that weigh more than 2Kg or that are blue. The weight or colour columns could contain that information. Values you want to order by. For example you might want the most expensive products first, and the least last. You could use the price column in descending order to achieve that. Assuming the fields you've picked in point 1 are in multiple tables, find the connections between those tables Look for relationships between tables and identify the columns that implement those relationships. For example, The Orders table could have a CustomerID field referencing the same column in the Customers table. Sometimes the problem doesn't use relationships but rests on a different field; sometimes the query is looking for a coincidence of fact rather than a foreign key constraint. For example you might have sales representatives who live in the same state as a customer; this information is normally not used in relationships, but if your query is for organizing events where sales representatives meet customers, it's useful in that query. In such a case you would record the names of columns at either end of such a connection. Sometimes relationships require a bridge, a junction table that wasn't identified in point 1 above but is needed to connect tables you need; these are used in "many-to-many relationships". In these cases you need to record the columns in each table that connect to similar columns in other tables. Construct a join or series of joins using the fields and tables identified in point 2 above. This becomes your FROM clause. Filter using some of the fields in point 1 above. This becomes your WHERE clause. Construct an ORDER BY clause using values from point 1 above that are relevant to the desired order of the output rows. Project the result using the remainder of the fields in point 1 above. This becomes your SELECT clause. A Worked Example   Let's say you want to query the world database to find a list of countries (with their capitals) and the change in GNP, using the difference between the GNP and GNPOld columns, and that you only want to see results for countries with a population greater than 100,000,000. Using the system described above, we could do the following:  The Country.Name and City.Name columns contain the name of the country and city respectively.  The change in GNP comes from the calculation GNP - GNPOld. Both those columns are in the Country table. This calculation is also used to order the output, in descending order To see only countries with a population greater than 100,000,000, you need the Population field of the Country table. There is also a Population field in the City table, so you'll need to specify the table name to disambiguate. You can also represent a number like 100 million as 100e6 instead of 100000000 to make it easier to read. Because the fields come from the Country and City tables, you'll need to join them. There are two relationships between these tables: Each city is hosted within a country, and the city's CountryCode column identifies that country. Also, each country has a capital city, whose ID is contained within the country's Capital column. This latter relationship is the one to use, so the relevant columns and the condition that uses them is represented by the following FROM clause:  FROM Country JOIN City ON Country.Capital = City.ID The statement should only return countries with a population greater than 100,000,000. Country.Population is the relevant column, so the WHERE clause becomes:  WHERE Country.Population > 100e6  To sort the result set in reverse order of difference in GNP, you could use either the calculation, or the position in the output (it's the third column): ORDER BY GNP - GNPOld or ORDER BY 3 Finally, project the columns you wish to see by constructing the SELECT clause: SELECT Country.Name AS Country, City.Name AS Capital,        GNP - GNPOld AS `Difference in GNP`  The whole statement ends up looking like this:  mysql> SELECT Country.Name AS Country, City.Name AS Capital, -> GNP - GNPOld AS `Difference in GNP` -> FROM Country JOIN City ON Country.Capital = City.ID -> WHERE Country.Population > 100e6 -> ORDER BY 3 DESC; +--------------------+------------+-------------------+ | Country            | Capital    | Difference in GNP | +--------------------+------------+-------------------+ | United States | Washington | 399800.00 | | China | Peking | 64549.00 | | India | New Delhi | 16542.00 | | Nigeria | Abuja | 7084.00 | | Pakistan | Islamabad | 2740.00 | | Bangladesh | Dhaka | 886.00 | | Brazil | Brasília | -27369.00 | | Indonesia | Jakarta | -130020.00 | | Russian Federation | Moscow | -166381.00 | | Japan | Tokyo | -405596.00 | +--------------------+------------+-------------------+ 10 rows in set (0.00 sec) Queries with Aggregates and GROUP BY While this system might work well for many queries, it doesn't cater for situations where you have complex summaries and aggregation. For aggregation, you'd start with choosing which columns to view in the output, but this time you'd construct them as aggregate expressions. For example, you could look at the average population, or the count of distinct regions.You could also perform more complex aggregations, such as the average of GNP per head of population calculated as AVG(GNP/Population). Having chosen the values to appear in the output, you must choose how to aggregate those values. A useful way to think about this is that every aggregate query is of the form X, Y per Z. The SELECT clause contains the expressions for X and Y, as already described, and Z becomes your GROUP BY clause. Ordinarily you would also include Z in the query so you see how you are grouping, so the output becomes Z, X, Y per Z.  As an example, consider the following, which shows a count of  countries and the average population per continent:  mysql> SELECT Continent, COUNT(Name), AVG(Population)     -> FROM Country     -> GROUP BY Continent; +---------------+-------------+-----------------+ | Continent     | COUNT(Name) | AVG(Population) | +---------------+-------------+-----------------+ | Asia          |          51 |   72647562.7451 | | Europe        |          46 |   15871186.9565 | | North America |          37 |   13053864.8649 | | Africa        |          58 |   13525431.0345 | | Oceania       |          28 |    1085755.3571 | | Antarctica    |           5 |          0.0000 | | South America |          14 |   24698571.4286 | +---------------+-------------+-----------------+ 7 rows in set (0.00 sec) In this case, X is the number of countries, Y is the average population, and Z is the continent. Of course, you could have more fields in the SELECT clause, and  more fields in the GROUP BY clause as you require. You would also normally alias columns to make the output more suited to your requirements. More Complex Queries  Queries can get considerably more interesting than this. You could also add joins and other expressions to your aggregate query, as in the earlier part of this post. You could have more complex conditions in the WHERE clause. Similarly, you could use queries such as these in subqueries of yet more complex super-queries. Each technique becomes another tool in your toolbox, until before you know it you're writing queries across 15 tables that take two pages to write out. But that's for another day...

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• Difference between IN and FIND_IN_SET

  - by Madhan ayyasamy

  Hi Friends,You may be confused with IN() and FIND_IN_SET() MYSQL functions. There are specific case/situation for both functions where to use which Mysql function. Look at below explanation about IN() and FIND_IN_SET()IN() : This function is used when you have a list of possible values and a single value in your database.Example: WHERE memberid IN (1,2,3)FIND_IN_SET() : This function is used where you have comma separated list of values stored in database and want to see if a certain value exists in that comma seperated list.Example: WHERE FIND_IN_SET( ‘table column name like id’, 'dynamic idlist' )

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• What schema documentation tools exist for PostgreSQL

  - by Brad Koch

  MySQL has MySQL Workbench for designing and documenting your schema, and generates CREATE and ALTER scripts based on your design. We're looking at migrating to PostgreSQL in the near future, and we do need a practical way of documenting and modifying the schema structure. What similar tools exist for Postgres (that are OS X/Linux compatible)? Alternatively, what equivalent conventions would be followed for designing and documenting the structure of your Postgres database?

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• Select (loop) or command not working in shell-script

  - by user208098

  I've been tinkering with Linux and Unix for years but still a novice in my mind and recently find myself trying to be more pro with it as I work in IT. So with that notion I'm studying shell scripting. I've hit a snag in ubuntu using the latest version 13.10 Saucy. When I use the select command in a sh script it doesn't work, depending on how I format the command it will either return Unexpected "do" or Unexpected "done". See the following two examples: This section of code produces an unexpected "do" error: #/bin/bash PS3='Please enter your choice' select opt in option1 option2 option3 quit do case $opt in "option1") echo "you chose choice 1" ;; "option2") echo "you chose choice 2" ;; "option3") echo "you chose choice 3" ;; "quit") break ;; *) echo invalid option ;; esac done This section of code produces an unexpected "done" error. #/bin/bash PS3='Please enter your choice' select opt in option1 option2 option3 quit ; do case $opt in "Option1") echo "you chose choice 1" ;; "Option2") echo "you chose choice 2" ;; "Option3") echo "you chose choice 3" ;; "quit") break ;; *) echo invalid option ;; esac done When I enter these parameters into the command line interactively or manually I get the desired result which is a list of choices to choose from. However when executed from a script I get the before mentioned errors. Also a side note I have tried this in Fedora as a script and it worked perfectly so my question is why isn't it working in Ubuntu, is this a difference between RHL and Debian? Or is it a bug in the latest version of Ubuntu? Thanks in advance for any help! KG

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• ???????/???MySQL?? ???

  - by user788995

  ????? ??:2012/05/28 ??:??????/?? MySQL???????????????????????????????????????????????MySQL?? ????????????????MySQL????????????????????? MySQL??????????????????????? ????????? ????????????????? http://otndnld.oracle.co.jp/ondemand/otn-seminar/movie/120518_MySQL%20Nyumon%20Jissen.wmv http://otndnld.oracle.co.jp/ondemand/otn-seminar/movie/mp4/120518_MySQL%20Nyumon%20Jissen.mp4 http://www.oracle.com/technetwork/jp/ondemand/mysql/120518-mysql-nyumon-jissen-jp-1641036-ja.pdf

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• How to make MySQL utilize available system resources, or find "the real problem"?

  - by anonymous coward

  This is a MySQL 5.0.26 server, running on SuSE Enterprise 10. This may be a Serverfault question. The web user interface that uses these particular queries (below) is showing sometimes 30+, even up to 120+ seconds at the worst, to generate the pages involved. On development, when the queries are run alone, they take up to 20 seconds on the first run (with no query cache enabled) but anywhere from 2 to 7 seconds after that - I assume because the tables and indexes involved have been placed into ram. From what I can tell, the longest load times are caused by Read/Update Locking. These are MyISAM tables. So it looks like a long update comes in, followed by a couple 7 second queries, and they're just adding up. And I'm fine with that explanation. What I'm not fine with is that MySQL doesn't appear to be utilizing the hardware it's on, and while the bottleneck seems to be the database, I can't understand why. I would say "throw more hardware at it", but we did and it doesn't appear to have changed the situation. Viewing a 'top' during the slowest times never shows much cpu or memory utilization by mysqld, as if the server is having no trouble at all - but then, why are the queries taking so long? How can I make MySQL use the crap out of this hardware, or find out what I'm doing wrong? Extra Details: On the "Memory Health" tab in the MySQL Administrator (for Windows), the Key Buffer is less than 1/8th used - so all the indexes should be in RAM. I can provide a screen shot of any graphs that might help. So desperate to fix this issue. Suffice it to say, there is legacy code "generating" these queries, and they're pretty much stuck the way they are. I have tried every combination of Indexes on the tables involved, but any suggestions are welcome. Here's the current Create Table statement from development (the 'experimental' key I have added, seems to help a little, for the example query only): CREATE TABLE `registration_task` ( `id` varchar(36) NOT NULL default '', `date_entered` datetime NOT NULL default '0000-00-00 00:00:00', `date_modified` datetime NOT NULL default '0000-00-00 00:00:00', `assigned_user_id` varchar(36) default NULL, `modified_user_id` varchar(36) default NULL, `created_by` varchar(36) default NULL, `name` varchar(80) NOT NULL default '', `status` varchar(255) default NULL, `date_due` date default NULL, `time_due` time default NULL, `date_start` date default NULL, `time_start` time default NULL, `parent_id` varchar(36) NOT NULL default '', `priority` varchar(255) NOT NULL default '9', `description` text, `order_number` int(11) default '1', `task_number` int(11) default NULL, `depends_on_id` varchar(36) default NULL, `milestone_flag` varchar(255) default NULL, `estimated_effort` int(11) default NULL, `actual_effort` int(11) default NULL, `utilization` int(11) default '100', `percent_complete` int(11) default '0', `deleted` tinyint(1) NOT NULL default '0', `wf_task_id` varchar(36) default '0', `reg_field` varchar(8) default '', `date_offset` int(11) default '0', `date_source` varchar(10) default '', `date_completed` date default '0000-00-00', `completed_id` varchar(36) default NULL, `original_name` varchar(80) default NULL, PRIMARY KEY (`id`), KEY `idx_reg_task_p` (`deleted`,`parent_id`), KEY `By_Assignee` (`assigned_user_id`,`deleted`), KEY `status_assignee` (`status`,`deleted`), KEY `experimental` (`deleted`,`status`,`assigned_user_id`,`parent_id`,`date_due`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 And one of the ridiculous queries in question: SELECT users.user_name assigned_user_name, registration.FIELD001 parent_name, registration_task.status status, registration_task.date_modified date_modified, registration_task.date_due date_due, registration.FIELD240 assigned_wf, if(LENGTH(registration_task.description)>0,1,0) has_description, registration_task.* FROM registration_task LEFT JOIN users ON registration_task.assigned_user_id=users.id LEFT JOIN registration ON registration_task.parent_id=registration.id where (registration_task.status != 'Completed' AND registration.FIELD001 LIKE '%' AND registration_task.name LIKE '%' AND registration.FIELD060 LIKE 'GN001472%') AND registration_task.deleted=0 ORDER BY date_due asc LIMIT 0,20; my.cnf - '[mysqld]' section. [mysqld] port = 3306 socket = /var/lib/mysql/mysql.sock skip-locking key_buffer = 384M max_allowed_packet = 100M table_cache = 2048 sort_buffer_size = 2M net_buffer_length = 100M read_buffer_size = 2M read_rnd_buffer_size = 160M myisam_sort_buffer_size = 128M query_cache_size = 16M query_cache_limit = 1M EXPLAIN above query, without additional index: +----+-------------+-------------------+--------+--------------------------------+----------------+---------+------------------------------------------------+---------+-----------------------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+-------------------+--------+--------------------------------+----------------+---------+------------------------------------------------+---------+-----------------------------+ | 1 | SIMPLE | registration_task | ref | idx_reg_task_p,status_assignee | idx_reg_task_p | 1 | const | 1067354 | Using where; Using filesort | | 1 | SIMPLE | registration | eq_ref | PRIMARY,gbl | PRIMARY | 8 | sugarcrm401.registration_task.parent_id | 1 | Using where | | 1 | SIMPLE | users | ref | PRIMARY | PRIMARY | 38 | sugarcrm401.registration_task.assigned_user_id | 1 | | +----+-------------+-------------------+--------+--------------------------------+----------------+---------+------------------------------------------------+---------+-----------------------------+ EXPLAIN above query, with 'experimental' index: +----+-------------+-------------------+--------+-----------------------------------------------------------+------------------+---------+------------------------------------------------+--------+-----------------------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+-------------------+--------+-----------------------------------------------------------+------------------+---------+------------------------------------------------+--------+-----------------------------+ | 1 | SIMPLE | registration_task | range | idx_reg_task_p,status_assignee,NewIndex1,tcg_experimental | tcg_experimental | 259 | NULL | 103345 | Using where; Using filesort | | 1 | SIMPLE | registration | eq_ref | PRIMARY,gbl | PRIMARY | 8 | sugarcrm401.registration_task.parent_id | 1 | Using where | | 1 | SIMPLE | users | ref | PRIMARY | PRIMARY | 38 | sugarcrm401.registration_task.assigned_user_id | 1 | | +----+-------------+-------------------+--------+-----------------------------------------------------------+------------------+---------+------------------------------------------------+--------+-----------------------------+

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• Problem with rake db:migrate

  - by Shreyas Satish

  When I try rake db:migrate, I get the following error: !!! The bundled mysql.rb driver has been removed from Rails 2.2. Please install the mysql gem and try again: gem install mysql. rake aborted! no such file to load -- mysql And when I try to "gem install mysql" Building native extensions. This could take a while... ERROR: Error installing mysql: ERROR: Failed to build gem native extension. /usr/bin/ruby extconf.rb Can't find header files for ruby. Gem files will remain installed in /usr/lib/ruby/gems/1.8/gems/mysql-2.8.1 for inspection. I checked the rubygems folder and mysql gem has been installed. Any ideas? Cheers

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• how to iterate through multiple select options with jquery

  - by amir

  I was just wondering if it's possible to go through multiple select options and get their values and text(if one is selected get the value and text, if 2 is selected get both of their values and text and so on) I have 15 select boxes in one page? any help would be appreciated. <form> <select class="select" name="select3" id="select3"> <option value="0">0</option> <option value="1.99">1</option> <option value="1.99">2</option> <option value="1.99">3</option> <option value="1.99">4</option> <option value="1.99">5</option> <option value="1.99">6</option> <option value="1.99">7</option> <option value="1.99">8</option> </select> </form> <form> <select class="select" name="select" id="select"> <option value="0">0</option> <option value="1.99">1</option> <option value="1.99">2</option> <option value="1.99">3</option> <option value="1.99">4</option> <option value="1.99">5</option> <option value="1.99">6</option> <option value="1.99">7</option> <option value="1.99">8</option> </select> </form> all the select options have the same class. thanks

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• Query returns too few rows

  - by Tareq

  setup: mysql> create table product_stock( product_id integer, qty integer, branch_id integer); Query OK, 0 rows affected (0.17 sec) mysql> create table product( product_id integer, product_name varchar(255)); Query OK, 0 rows affected (0.11 sec) mysql> insert into product(product_id, product_name) values(1, 'Apsana White DX Pencil'); Query OK, 1 row affected (0.05 sec) mysql> insert into product(product_id, product_name) values(2, 'Diamond Glass Marking Pencil'); Query OK, 1 row affected (0.03 sec) mysql> insert into product(product_id, product_name) values(3, 'Apsana Black Pencil'); Query OK, 1 row affected (0.03 sec) mysql> insert into product_stock(product_id, qty, branch_id) values(1, 100, 1); Query OK, 1 row affected (0.03 sec) mysql> insert into product_stock(product_id, qty, branch_id) values(1, 50, 2); Query OK, 1 row affected (0.03 sec) mysql> insert into product_stock(product_id, qty, branch_id) values(2, 80, 1); Query OK, 1 row affected (0.03 sec) my query: mysql> SELECT IFNULL(SUM(s.qty),0) AS stock, product_name FROM product_stock s RIGHT JOIN product p ON s.product_id=p.product_id WHERE branch_id=1 GROUP BY product_name ORDER BY product_name; returns: +-------+-------------------------------+ | stock | product_name | +-------+-------------------------------+ | 100 | Apsana White DX Pencil | | 80 | Diamond Glass Marking Pencil | +-------+-------------------------------+ 1 row in set (0.00 sec) But I want to have the following result: +-------+------------------------------+ | stock | product_name | +-------+------------------------------+ | 0 | Apsana Black Pencil | | 100 | Apsana White DX Pencil | | 80 | Diamond Glass Marking Pencil | +-------+------------------------------+ To get this result what mysql query should I run?

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• Javascript: how to document.print option value in Select block?

  - by HH

  The below returns [object HTMLOptionElement] while I try to return the value. document.write(document.getElementById("allSelect").options[0]).value; by value, I mean the thing called value there (there is a form around it but not relevant now): <select name="allSelect" id="allSelect" size=3> <option value="1"selected">1</option> <option value="2">2</option> <option value="3">3</option> </select> How can I return the value of an option?

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• MySQL & PHP - select/option lists and showing data to users that still allows me to generate queries

  - by Andrew Heath

  Sorry for the unclear title, an example will clear things up: TABLE: Scenario_victories ID scenid timestamp userid side playdate 1 RtBr001 2010-03-15 17:13:36 7 1 2010-03-10 2 RtBr001 2010-03-15 17:13:36 7 1 2010-03-10 3 RtBr001 2010-03-15 17:13:51 7 2 2010-03-10 ID and timestamp are auto-insertions by the database when the other 4 fields are added. The first thing to note is that a user can record multiple playings of the same scenario (scenid) on the same date (playdate) possibly with the same outcome (side = winner). Hence the need for the unique ID and timestamps for good measure. Now, on their user page, I'm displaying their recorded play history in a <select><option>... list form with 2 buttons at the end - Delete Record and Go to Scenario My script takes the scenid and after hitting a few other tables returns with something more user-friendly like: (playdate) (from scenid) (from side) ######################################################### # 2010-03-10 Road to Berlin #1 -- Germany, Hungary won # # 2010-03-10 Road to Berlin #1 -- Germany, Hungary won # # 2010-03-10 Road to Berlin #1 -- Soviet Union won # ######################################################### [Delete Record] [Go To Scenario] in HTML: <select name="history" size=3> <option>2010-03-10 Road to Berlin #1 -- Germany, Hungary won</option> <option>2010-03-10 Road to Berlin #1 -- Germany, Hungary won</option> <option>2010-03-10 Road to Berlin #1 -- Soviet Union won</option> </select> Now, if you were to highlight the first record and click Go to Scenario there is enough information there for me to parse it and produce the exact scenario you want to see. However, if you were to select Delete Record there is not - I have the playdate and I can parse the scenid and side from what's listed, but in this example all three records would have the same result. I appear to have painted myself into a corner. Does anyone have a suggestion as to how I can get some unique identifying data (ID and/or timestamp) to ride along on this form without showing it to the user? PHP-only please, I must be NoScript compliant!

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• Starting MySQL . ERROR! Manager of pid-file quit without updating file.

  - by Joseph Silvashy

  This is totally odd, mysql refuses to start: Starting MySQL . ERROR! Manager of pid-file quit without updating file.

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• IDs necessary in update script not being stored (or even seen!?) (PHP MySQL)

  - by Derek

  Hi guys, I really need help with this one...have spent 3 hours trying to figure it out... Basically, I have 3 tables necessary for this function to work (the query and PHP)... Authors, Books and Users. An author can have many books, and a user can have many books - that's it. When the admin user selects to update a book, they are presented with a form, displaying the current data within the fields, very straight forward... However there is one tricky part, the admin user can change the author for a book (incase they make a mistake) and also change the user for which the book is associated with. When I select to update the single book information I am not getting any values what so ever for author_id or user_id. Meaning that when the user updates the book info, the associations with the user and author is being scrapped altogether (when before there was an association)... I cannot see why this is happening because I can clearly see the IDs for the users and authors for my option values (this is because they are in select dropdowns). Here is what my sql to retrieve the user ID is: SELECT user_id, name FROM users and then i have my select options which brings up all the users in the system: <label>This book belongs to:</label> <select name="name" id="name"> <option value="<?php echo $row['user_id']?>" SELECTED><?php echo $row['name']?> - Current</option> <?php while($row = mysql_fetch_array($result)) { ?> <option value="<?php echo $row['user_id']; if (isset($_POST['user_id']));?>"><?php echo $row['name']?></option> <?php } ?> In the presented HTML form, I can select the users (by name) and within the source code I can see the IDs (for the value) matching against the names of the users. Finally, in my script that performs the update, I have this: $book_id = $_POST['book_id']; $bookname = $_POST['bookname']; $booklevel = $_POST['booklevel']; $author_id = $_POST['author_id']; $user_id = $_POST['user_id']; $sql = "UPDATE books SET bookname= '".$bookname."', booklevel= '".$booklevel."', author_id='".$author_id."', user_id= '".$user_id."' WHERE book_id = ".$book_id; The result of this query returns no value for either author_id or user_id... Obviously in this question I have given the information for the user stuff (with the HTML being displayed) but im guessing that I have the same problem with authors aswell... How can I get these ID's passed to the script so that the change can be acknowledge!! :(

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• [SQL] How to SELECT TOP 5 and then the following 5?

  - by Crays

  Hi guys, as the comments always have it, you can select to show the top 5 comments (which i know how) and if there is more than 5, you can click a link and it will show the following 5 comments. The problem is that i don't know how the code should be for it to show the next 5. Better idea of what i'm saying Let's say i have 10 comments, i use SELECT * FROM news ORDER BY ID DESC LIMIT 5 that will show the TOP 5 comments say, the comments with ID 10,9,8,7,6 but what if i want the comments with ID 5,4,3,2,1 ?

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• How to update a table using a select group by in a second one as the data source in MySQL?

  - by Jader Dias

  I can't do this in MySQL UPDATE tableA, tableB SET tableA.column1 = SUM(tableB.column2) WHERE tableA.column3 = tableB.column4 GROUP BY tableB.column4 ; Neither can I UPDATE tableA, ( SELECT SUM(tableB.column2) sumB, tableB.column4 FROM tableB GROUP BY tableB.column4 ) t1 SET tableA.column1 = sumB WHERE tableA.column3 = column4 ; Besides it being illegal code, I think you can understand what I tried to do with the queries above. Both of them had the same intent. How can I do that in MySQL?

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• How should I build a privacy drop-down (select) menu?

  - by animuson

  I'm trying to build something similar to Facebook's privacy selection menu, except without the 'custom' option. It will only list a few options such as 'show to all', 'show to friends only', or 'completely hidden'. Right now I'm thinking of using simple JavaScript to change a hidden input field to the new value they click on, so if they clicked on the division for 'show to friends only' it would change the corresponding field, say 'email_privacy', to 1. Is there a better way to do this or am I pretty much on track? P.S. I am not planning on using a select element, I was planning on building a custom drop-down menu using CSS since select elements are so highly non-customizable. I'm doing it this way to save space, rather than having this massive selection menu at the right which takes up a bunch of space. Note: I'm not really interested in using jQuery, that's just extra libraries and crap that I don't want to load. I can do it in JavaScript just as easily so I might as well use that.

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• How can you tell MYSQL to TRIM the X number of characters, beginning from the Back?

  - by Tim

  How do I write the following in MYSQL? SELECT SUBSTRING(value - (1 TRAILING CHARACTER)) FROM table; Basically substring(value, 2) trims the first letters. But I need to trim the last letters. I can't use substring(value, -4, 3) because I don't know the length of the value. Here's another example: SELECT * FROM table WHERE SUBSTRING(value - (4 TRAILING CHARACTER)) in (SELECT SUBSTRING(value - (1 TRAILING CHARACTER)) FROM table);

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• Open Select using Javascript/jQuery?

  - by Newbie

  Hello! Is there a way to open a select box using Javascript (and jQuery)? <select style="width:150px;"> <option value="1">1</option> <option value="2">Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nunc arcu nunc, rhoncus ac dignissim at, rhoncus ac tellus.</option> <option value="3">3</option> </select> I have to open my select, cause of ie bug. All versions of IE (6,7,8) cut my options. As far as I know, there is no css bugfix for this. At the moment I try to do the following: var original_width = 0; var selected_val = false; if (jQuery.browser.msie) { $('select').click(function(){ if (selected_val == false){ if(original_width == 0) original_width = $(this).width(); $(this).css({ 'position' : 'absolute', 'width' : 'auto' }); }else{ $(this).css({ 'position' : 'relative', 'width' : original_width }); selected_val = false; } }); $('select').blur(function(){ $(this).css({ 'position' : 'relative', 'width' : original_width }); }); $('select').blur(function(){ $(this).css({ 'position' : 'relative', 'width' : original_width }); }); $('select').change(function(){ $(this).css({ 'position' : 'relative', 'width' : original_width }); }); $('select option').click(function(){ $(this).css({ 'position' : 'relative', 'width' : original_width }); selected_val = true; }); } But clicking on my select the first time will change the width of the select but I have to click again to open it.

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• rake db:create gives some problem when used from limited account

  - by Xinxua

  I am using mysql 5.1 and mysql gems version is 2.73 This is giving the following error message when I try to run it from a limited account in my XP. If try it using the admin account, it works fine. I think this is wierd because it cannot be the problem of mysql gem. (in F:/Temp/wassup) !!! The bundled mysql.rb driver has been removed from Rails 2.2. Please install the mysql gem and try again: gem install mysql. rake aborted! 5: Access is denied. - C:/Program Files/Ruby/lib/ruby/gems/1.8/gems/mysql-2.7. 3-x86-mswin32/ext/mysql.so (See full trace by running task with --trace) I need to work from the limited account. Can anyone let me know why is this happening?

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• PostgreSQL: return select count(*) from old_ids;

  - by Alexander Farber

  Hello, please help me with 1 more PL/pgSQL question. I have a PHP-script run as daily cronjob and deleting old records from 1 main table and few further tables referencing its "id" column: create or replace function quincytrack_clean() returns integer as $BODY$ begin create temp table old_ids (id varchar(20)) on commit drop; insert into old_ids select id from quincytrack where age(QDATETIME) > interval '30 days'; delete from hide_id where id in (select id from old_ids); delete from related_mks where id in (select id from old_ids); delete from related_cl where id in (select id from old_ids); delete from related_comment where id in (select id from old_ids); delete from quincytrack where id in (select id from old_ids); return select count(*) from old_ids; end; $BODY$ language plpgsql; And here is how I call it from the PHP script: $sth = $pg->prepare('select quincytrack_clean()'); $sth->execute(); if ($row = $sth->fetch(PDO::FETCH_ASSOC)) printf("removed %u old rows\n", $row['count']); Why do I get the following error? SQLSTATE[42601]: Syntax error: 7 ERROR: syntax error at or near "select" at character 9 QUERY: SELECT select count(*) from old_ids CONTEXT: SQL statement in PL/PgSQL function "quincytrack_clean" near line 23 Thank you! Alex

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