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  • 127.0.0.1:9051 doesnt work after apache, mysql, php installation?

    - by Rana Muhammad Waqas
    I have installed apache2, mysql, and php and now it doesnt let Vidalia run on localhost. i tried to change the TCP connection (controlport) to any other ip 192.168.0.40 and tried to change the default port 9051 to any other but that doesnt work. I thought apache is running so i used this command sudo service apache2 stop but that still doesnt work. So now when i type 127.0.0.1:9051 in browser it says and if i type only type 127.0.0.1 after stopping the apapche2 service with the command mentioned above it says unable to connect I am not sure what to do now Help!

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  • How does using a LGPL gem affect my MIT licensed application?

    - by corsen
    I am developing an open source ruby application under the MIT license. I am using this license because I don't want to place any restrictions on the users of the application. Also I can actually read and understand this license. I recently started using another ruby gem in my project (require "somegem"). This ruby gem is under the LGPL license. Do I have to change anything about my project because I am using this other ruby gem that is licensed with LGPL? My project does not contain the source code for the other gem and it is not shipped with my project. It is simply listed as a dependency so that ruby gems will install it and my project will call into it from my code. Additionally, it would be helpful to know if there are any licenses I need to "watch out for" because using them would affect the license of my project. There are some other post about this topic but phrased in different ways. Since I find this license stuff tricky I am hoping to get a answer directed at my situation. Thank you, Corsen

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  • How do you install less css command line compiler?

    - by chrisjlee
    From my understanding and correct me if I'm wrong, I have to get ruby or NPM installed to get the less css compiler working. I don't have any ruby installed and I'm not really sure how to get my computer to that point. I also want to minimize my footprint; installing the minimal amount of ruby libraries if possible (because i will never use ruby except for when i run less). What are the steps involved in getting less working and running? Before you down vote, I know there was this previous thread (Less CCS compiler install). This particular person already has some other packages installed. I'm trying to figure out all the packages needed to get to that point. Or if someone could point me to the right documentation I would be thrilled!

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  • Where is the best place to teach myself a language, and which one?

    - by Lorinda
    Hello, I do not know any programming languages at all. I will self teach myself and need to know the best place to do so where I can learn from a most basic level. Where is a great place to begin learning a language? What language is best to learn first? Is it silly to learn Ruby first? Here, I came across someone saying that learning some of the higher languages can make you 'lazy' if you learn them first. Like Ruby amongst others. For my first language, my husband is advising me to learn Ruby (for his own personal interests). However, I need some independent advice of how to get started and what language I should learn first. I will eventually learn Ruby and then Rails. Four months ago, my husband ordered a text of objective C because he thought he would take it on. I flipped through and it was clearly starting at a place more advanced than where I am coming from. I have dabbled with a Ruby tutorial and I don't get it. I get what I am putting in is what I get, but I don't understand what is leading up to that. I need to know ALL the rules first. I then looked up computer languages and stared researching binary code which helped a lot, but not where I want to start. I don't have a lot of time right now in my life (with four kids) to go back that far. If I were going to school, that would be different. Any advice you could give is most welcomed.

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  • What is the best way to INSERT a large dataset into a MySQL database (or any database in general)

    - by Joe
    As part of a PHP project, I have to insert a row into a MySQL database. I'm obviously used to doing this, but this required inserting into 90 columns in one query. The resulting query looks horrible and monolithic (especially inserting my PHP variables as the values): INSERT INTO mytable (column1, colum2, ..., column90) VALUES ('value1', 'value2', ..., 'value90') and I'm concerned that I'm not going about this in the right way. It also took me a long (boring) time just to type everything in and testing writing the test code will be equally tedious I fear. How do professionals go about quickly writing and testing these queries? Is there a way I can speed up the process?

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  • How can I make my Java desktop application to run on different computers in a network locally and updating same mysql database?

    - by JaMpa
    I have developed Java desktop application for registration of students in the computer laboratories in our college, but what I need is to make this application whenever it runs on any computer (Windows XP machines) on the laboratory each and every data filled by the college students goes to the same MySql database (through WAMP server) running on the server machine (Win Server 2003). Also, I need this application to run during booting of windows computers in the laboratory for the purpose when any student tries to switch on any PC in the comp lab before interfacing the desktop icons should sign in through user name and password when he has already registered or signing up in order for getting the access.

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  • showing null rows using join

    - by Pradyut Bhattacharya
    Hi, In mysql i m selecting from a table shouts having a foreign key to another table named "roleuser" with the matching column as user_id Now the user_id column in the shouts table for some rows is null (not actually null but with no inserts in mysql) How to show all the rows of the shouts table either with user_id null or not I m executing the sql statement SELECT s.*, r.firstname, r.lastname FROM shouts s left join roleuser r where r.user_id = s.user_id limit 50; which does not executes and shows 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 'where r.user_id = s.user_id limit 50' at line 2 but using inner join the sql executes which shows rows which only have user_id values in the shouts table. the nulls are not shown. SELECT s.*, r.firstname, r.lastname FROM shouts s inner join roleuser r where r.user_id = s.user_id limit 50; How can i show all the rows from the shouts table and null values in the firstname and lastname columns where the user_id is null in the shouts table. If not at all possible with sql may be using stored procedures... Thanks Pradyut

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  • Getting percentage of "Count(*)" to the number of all items in "GROUP BY"

    - by celalo
    Let's say I need to have the ratio of "number of items available from certain category" to "the the number of all items". Please consider a MySQL table like this: /* mysql> select * from Item; +----+------------+----------+ | ID | Department | Category | +----+------------+----------+ | 1 | Popular | Rock | | 2 | Classical | Opera | | 3 | Popular | Jazz | | 4 | Classical | Dance | | 5 | Classical | General | | 6 | Classical | Vocal | | 7 | Popular | Blues | | 8 | Popular | Jazz | | 9 | Popular | Country | | 10 | Popular | New Age | | 11 | Popular | New Age | | 12 | Classical | General | | 13 | Classical | Dance | | 14 | Classical | Opera | | 15 | Popular | Blues | | 16 | Popular | Blues | +----+------------+----------+ 16 rows in set (0.03 sec) mysql> SELECT Category, COUNT(*) AS Total -> FROM Item -> WHERE Department='Popular' -> GROUP BY Category; +----------+-------+ | Category | Total | +----------+-------+ | Blues | 3 | | Country | 1 | | Jazz | 2 | | New Age | 2 | | Rock | 1 | +----------+-------+ 5 rows in set (0.02 sec) */ What I need is basically a result set resembles this one: /* +----------+-------+-----------------------------+ | Category | Total | percentage to the all items | (Note that number of all available items is "9") +----------+-------+-----------------------------+ | Blues | 3 | 33 | (3/9)*100 | Country | 1 | 11 | (1/9)*100 | Jazz | 2 | 22 | (2/9)*100 | New Age | 2 | 22 | (2/9)*100 | Rock | 1 | 11 | (1/9)*100 +----------+-------+-----------------------------+ 5 rows in set (0.02 sec) */ How can I achieve such a result set in a single query? Thanks in advance.

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  • Creating PHP strings using other variables...works manually, can't figure out how automatically

    - by Matt
    Hello, I'm trying to get a variable to be formed automatically using data pulled from a mysql database. I know the data is being pulled from the database in some form resembling its original form, but that data does not act the same as data that is manually typed and assigned to a string. For example, if a cell in a mysql table says... I said "goodbye" before I left. She also said "goodbye." ...and I manually copy/paste it and add the necessary escapes... $string1 = " I said \"goodbye\" before I left. She also said \"goodbye.\" "; ...that does not equal... $string1 = $mysqlResultArray['specificCellWithQuoteShownAbove'] Interestingly, if I echo both versions of $string1 and view the output, they appear to be exactly the same. But they do not function the same when put through various functions I've created. The functions only work if I do the manual copy/paste method--which is not what I want, obviously. I'm not sure if it has to do with the line breaks or the escapes--or some combination of the two. But while both strings are superficially the same, they are apparently functionally different and I don't know why. So how can I create $string1 without manually copy/pasting the contents from the mysql entry and instead querying for the data and assigning it to $string1 in such a way that it's exactly functionally equivalent as the manual copy/pasted string?

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  • Ordering by number of rows?

    - by Rob
    Alright, so I have a table outputting data from a MySQL table in a while loop. Well one of the columns it outputs isn't stored statically in the table, instead it's the sum of how many times it appears in a different MySQL table. Sorry I'm not sure this is easy to understand. Here's my code: $query="SELECT * FROM list WHERE added='$addedby' ORDER BY time DESC"; $result=mysql_query($query); while($row=mysql_fetch_array($result, MYSQL_ASSOC)){ $loghwid = $row['hwid']; $sql="SELECT * FROM logs WHERE hwid='$loghwid' AND time < now() + interval 1 hour"; $query = mysql_query($sql) OR DIE(mysql_error()); $boots = mysql_num_rows($query); //Display the table } The above is the code displaying the table. As you can see it's grabbing data from two different MySQL tables. However I want to be able to ORDER BY $boots DESC. But as its a counting of a completely different table, I have no idea of how to go about doing that. I would appreciate any help, thank you.

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  • multiple mysql_real_query() in while loop

    - by Steve
    It seems that when I have one mysql_real_query() function in a continuous while loop, the query will get executed OK. However, if multiple mysql_real_query() are inside the while loop, one right after the other. Depending on the query, sometimes neither the first query nor second query will execute properly. This seems like a threading issue to me. I'm wondering if the mysql c api has a way of dealing with this? Does anyone know how to deal with this? mysql_free_result() doesn't work since I am not even storing the results. //keep polling as long as stop character '-' is not read while(szRxChar != '-') { // Check if a read is outstanding if (HasOverlappedIoCompleted(&ovRead)) { // Issue a serial port read if (!ReadFile(hSerial,&szRxChar,1, &dwBytesRead,&ovRead)) { DWORD dwErr = GetLastError(); if (dwErr!=ERROR_IO_PENDING) return dwErr; } } // Wait 5 seconds for serial input if (!(HasOverlappedIoCompleted(&ovRead))) { WaitForSingleObject(hReadEvent,RESET_TIME); } // Check if serial input has arrived if (GetOverlappedResult(hSerial,&ovRead, &dwBytesRead,FALSE)) { // Wait for the write GetOverlappedResult(hSerial,&ovWrite, &dwBytesWritten,TRUE); //load tagBuffer with byte stream tagBuffer[i] = szRxChar; i++; tagBuffer[i] = 0; //char arrays are \0 terminated //run query with tagBuffer if( strlen(tagBuffer)==PACKET_LENGTH ) { sprintf(query,"insert into scan (rfidnum) values ('"); strcat(query, tagBuffer); strcat(query, "')"); mysql_real_query(&mysql,query,(unsigned int)strlen(query)); i=0; } mysql_real_query(&mysql,"insert into scan (rfidnum) values ('2nd query')",(unsigned int)strlen("insert into scan (rfid) values ('2nd query')")); mysql_free_result(res); } }

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  • Split a long JSON string into lines in Ruby

    - by David J.
    First, the background: I'm writing a Ruby app that uses SendGrid to send mass emails. SendGrid uses a custom email header (in JSON format) to set recipients, values to substitute, etc. SendGrid's documentation recommends splitting up the header so that the lines are shorter than 1,000 bytes. My question, then, is this: given a long JSON string, how can I split it into lines < 1,000 so that lines are split at appropriate places (i.e., after a comma) rather than in the middle of a word? This is probably unnecessary, but here's an example of the sort of string I'd like to split: X-SMTPAPI: {"sub": {"pet": ["dog", "cat"]}, "to": ["[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]", "[email protected]"]} Thanks in advance for any help you can provide!

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  • MySQL Interview Questions

    - by Campbell
    Hi, I've been asked to screen some candidates for a MySQL DBA / Developer position for a role that requires an enterprise level skill set. I myself am a SQL Server person so I know what I would be looking for from that point of view with regards to scalability / design etc but is there anything specific I should be asking with regards to MySQL? I would ideally like to ask them about enterprise level features of MySQL that they would typically only use when working on a big database. Need to separate out the enterprise developers from the home / small website kind of guys. Thanks.

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  • Can't get heroku work with rails 3.x postgresql

    - by framomo86
    I followed Heroku official guides to push rails project to heroku. The application.rb file is ok, I added pg gem and database.yml in the right way. When I push to heroku I get: -----> Preparing app for Rails asset pipeline Detected manifest.yml, assuming assets were compiled locally But when I open heroku via heroku open I get an error. I put heroku logs and get this. Started GET "/" for 93.45.227.255 at 2012-10-11 13:28:04 +0000 2012-10-11T13:28:04+00:00 app[web.1]: Processing by ProductsController#index as HTML 2012-10-11T13:28:04+00:00 app[web.1]: : SELECT a.attname, format_type(a.atttypid, a.atttypmod), d.adsrc, a.attnotnull 2012-10-11T13:28:04+00:00 app[web.1]: ActiveRecord::StatementInvalid (PG::Error: ERROR: relation "products" does not exist 2012-10-11T13:28:04+00:00 app[web.1]: LINE 4: WHERE a.attrelid = '"products"'::regclass 2012-10-11T13:28:04+00:00 heroku[router]: GET gift4.herokuapp.com/ dyno=web.1 queue=0 wait=0ms service=203ms status=500 bytes=643 2012-10-11T13:28:04+00:00 app[web.1]: ): 2012-10-11T13:28:04+00:00 app[web.1]: ON a.attrelid = d.adrelid AND a.attnum = d.adnum 2012-10-11T13:28:04+00:00 app[web.1]: 2012-10-11T13:28:04+00:00 app[web.1]: 2012-10-11T13:28:04+00:00 app[web.1]: ^ 2012-10-11T13:28:04+00:00 app[web.1]: 2012-10-11T13:28:04+00:00 app[web.1]: Completed 500 Internal Server Error in 72ms 2012-10-11T13:28:04+00:00 app[web.1]: FROM pg_attribute a LEFT JOIN pg_attrdef d 2012-10-11T13:28:04+00:00 app[web.1]: WHERE a.attrelid = '"products"'::regclass 2012-10-11T13:28:04+00:00 app[web.1]: AND a.attnum > 0 AND NOT a.attisdropped 2012-10-11T13:28:04+00:00 app[web.1]: ORDER BY a.attnum 2012-10-11T13:28:04+00:00 app[web.1]: app/controllers/products_controller.rb:5:in `index' 2012-10-11T13:28:04+00:00 heroku[router]: GET gift4.herokuapp.com/favicon.ico d So I tried heroku run rake db:reset And get this Heroku client internal error. ! Search for help at: https://help.heroku.com ! Or report a bug at: https://github.com/heroku/heroku/issues/new Error: Operation timed out - connect(2) (Errno::ETIMEDOUT) Backtrace: /Users/francescochecco/.rvm/gems/ruby-1.9.3-p194/gems/heroku-2.32.6/lib/heroku/client/rendezvous.rb:39:in `initialize' /Users/francescochecco/.rvm/gems/ruby-1.9.3-p194/gems/heroku-2.32.6/lib/heroku/client/rendezvous.rb:39:in `open' /Users/francescochecco/.rvm/gems/ruby-1.9.3-p194/gems/heroku-2.32.6/lib/heroku/client/rendezvous.rb:39:in `block in start' /Users/francescochecco/.rvm/rubies/ruby-1.9.3-p194/lib/ruby/1.9.1/timeout.rb:68:in `timeout' /Users/francescochecco/.rvm/gems/ruby-1.9.3-p194/gems/heroku-2.32.6/lib/heroku/client/rendezvous.rb:31:in `start' /Users/francescochecco/.rvm/gems/ruby-1.9.3-p194/gems/heroku-2.32.6/lib/heroku/command/run.rb:125:in `rendezvous_session' /Users/francescochecco/.rvm/gems/ruby-1.9.3-p194/gems/heroku-2.32.6/lib/heroku/command/run.rb:112:in `run_attached' /Users/francescochecco/.rvm/gems/ruby-1.9.3-p194/gems/heroku-2.32.6/lib/heroku/command/run.rb:21:in `index' /Users/francescochecco/.rvm/gems/ruby-1.9.3-p194/gems/heroku-2.32.6/lib/heroku/command.rb:206:in `run' /Users/francescochecco/.rvm/gems/ruby-1.9.3-p194/gems/heroku-2.32.6/lib/heroku/cli.rb:28:in `start' /Users/francescochecco/.rvm/gems/ruby-1.9.3-p194/gems/heroku-2.32.6/bin/heroku:16:in `<top (required)>' /Users/francescochecco/.rvm/gems/ruby-1.9.3-p194/bin/heroku:19:in `load' /Users/francescochecco/.rvm/gems/ruby-1.9.3-p194/bin/heroku:19:in `<main>' /Users/francescochecco/.rvm/gems/ruby-1.9.3-p194/bin/ruby_noexec_wrapper:14:in `eval' /Users/francescochecco/.rvm/gems/ruby-1.9.3-p194/bin/ruby_noexec_wrapper:14:in `<main>' Command: heroku run rake db:reset Version: heroku-gem/2.32.6 (x86_64-darwin11.3.0) ruby/1.9.3 autoupdate I tried everything. Anyone could help?

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  • Can I force MySql table name case sensitivity on file systems that aren't case sensitive

    - by Brian Deacon
    So our target environment is linux, making mysql case-sensitive by default. I am aware that we can make our linux environment not case sensitive with the lower_case_table_names variable, but we would rather not. We have a few times been bitten with a case mismatch because our dev rigs are OSX, and mysql is not case sensitive there. Is there a way we can force table names to be case sensitive on my OSX install of MySql (5.0.83 if that matters) so that we catch a table name case mismatch prior to deploying to the integration servers running on linux?

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  • Cassandra or mysql 5

    - by saturngod
    Should I use cassandra in 100,000 users project ? In mysql 5 have full text search and partition table. I'm starting to make Question and answer system like stackoverflow with CodeIgniter. It's move from vbulletin to new system. In old vbulletin have around 100,000 users and total post is around 80,000. In next 3 or 4 year, users and posts will be more and more. So, Should I use cassandra instead of mysql 5 ? If I use cassandra, I need to change gridserver in mediatemple to DV server in mediatemple. Cassandra is not built in hosting system. So, I must use VPS or DV server. If I use mysql 5, hosting is not problem but how about speed and search. Btw, What database using in Stack Over ?

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  • Mysql gem troubles with Snow Leopard upgrade

    - by so1o
    I have done everything that is on the web (i think) i have the new 64 bit xcode that came with snow leopard installed completely removed mysql, removed gems compeltely, removed rails installed mysql 64 bit, installed gems, installed mysql gem with the env ARCHFLAGS set I still get this nasty NameError: uninitialized constant MysqlCompat::MysqlRes from /Users/Navara/Sites/tuosystems/vendor/rails/activesupport/lib/active_support/dependencies.rb:440:in `load_missing_constant' from /Users/Navara/Sites/tuosystems/vendor/rails/activesupport/lib/active_support/dependencies.rb:80:in `rake_original_const_missing' from /usr/local/lib/ruby/gems/1.8/gems/rake-0.8.7/lib/rake.rb:2503:in `const_missing' Im not sure how to debug this.. any pointers will be greatly appreciated!

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  • Passing object from PHP to Mysql Stored procedure

    - by user268982
    Hi All, Scenario :- I have to call MYSQL stored procedure from PHP and do some operations ( around 15 commands ) on the database Problem :- I have to call stored procedure with 36 parameters. Lot of parameters . I don't think it is a good idea to pass these many individual parameters and even heard passing individul parameters increases network traffic. Looking for :- I created a Data Object at PHP side and is there any way I can create similar kind of Object in MYSQL and pass this object as a parameter and extract the data from the object in MYSQL stored procedure Thanks for your help Regards Kiran

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  • Ruby Rails Mongrel Sever failing to serve OXS1.6

    - by Mark V
    Hi there I'm fairly new to Rails and the Mac, and doing my first deploy... I'm trying to set up my rails app on a brand new Apple mini-server running OXS1.6 (Snow Leopard). It is currently running fine on my new iMac i7 (same OS). I start mongrel with this command: mongrel_rails start -e production -p 3000 -d -a 127.0.0.1 --debug And it starts giving this output in the log/mongrel.log ** Daemonized, any open files are closed. Look at log/mongrel.pid and log/mongrel.log for info. ** Starting Mongrel listening at 127.0.0.1:3000 ** Installing debugging prefixed filters. Look in log/mongrel_debug for the files. ** Starting Rails with production environment... /Library/Ruby/Gems/1.8/gems/rails-2.3.5/lib/rails/gem_dependency.rb:119:Warning: Gem::Dependency#version_requirements is deprecated and will be removed on or after August 2010. Use #requirement /Users/danadmin/ServiceApp/ServiceApp/app/helpers/input_grid_manager.rb:9: warning: already initialized constant ID_PREFIX /Users/danadmin/ServiceApp/ServiceApp/app/helpers/input_grid_manager.rb:10: warning: already initialized constant ADD_ID ** Rails loaded. ** Loading any Rails specific GemPlugins ** Signals ready. TERM => stop. USR2 => restart. INT => stop (no restart). ** Rails signals registered. HUP => reload (without restart). It might not work well. ** Mongrel 1.1.5 available at 127.0.0.1:3000 ** Writing PID file to log/mongrel.pid The output is the same on my dev iMac (including the warnings). The difference is that accessing http://127.0.0.1:3000 on my iMac serves up the app's login page. Where as on the mac mini-server accessing the same results in this error 500 text from mongrel: "We're sorry, but something went wrong." It's as if rails is not working. I'm pretty good at figuring things out if I have some log file messages to direct me, but mongrel.log has no error message (the output remains the same as above), and the log/production.log is empty (which makes me think rails has not started?). My gems are all the same versions between machines and so is the app code; and there are no clues I can see in any of the mongrel_debug logs, except that rails.log on the mac mini-server and the iMac are different. After a start and single access, first is the rails.log from the mac mini-server: D, [2010-04-15T13:45:34.870406 #6914] DEBUG -- : TRACING ON Thu Apr 15 13:45:34 +1200 2010 Thu Apr 15 13:46:08 +1200 2010 REQUEST / --- !map:Mongrel::HttpParams SERVER_NAME: 127.0.0.1 HTTP_ACCEPT: application/xml,application/xhtml+xml,text/html;q=0.9,text/plain;q=0.8,image/png,*/*;q=0.5 HTTP_CACHE_CONTROL: max-age=0 HTTP_HOST: 127.0.0.1:3000 HTTP_USER_AGENT: Mozilla/5.0 (Macintosh; U; Intel Mac OS X 10_6_0; en-US) AppleWebKit/533.2 (KHTML, like Gecko) Chrome/5.0.342.9 Safari/533.2 REQUEST_PATH: / SERVER_PROTOCOL: HTTP/1.1 HTTP_ACCEPT_LANGUAGE: en-US,en;q=0.8 REMOTE_ADDR: 127.0.0.1 PATH_INFO: / SERVER_SOFTWARE: Mongrel 1.1.5 SCRIPT_NAME: / HTTP_VERSION: HTTP/1.1 REQUEST_URI: / SERVER_PORT: "3000" HTTP_ACCEPT_CHARSET: ISO-8859-1,utf-8;q=0.7,*;q=0.3 REQUEST_METHOD: GET GATEWAY_INTERFACE: CGI/1.2 HTTP_ACCEPT_ENCODING: gzip,deflate,sdch HTTP_CONNECTION: keep-alive While on my iMac it seems the same except for the addition of the HTTP_COOKIE and the HTTP_IF_NONE_MATCH, here is rails.log from my iMac # Logfile created on Thu Apr 15 13:41:42 +1200 2010 by logger.rb/22285 D, [2010-04-15T13:41:42.934088 #2070] DEBUG -- : TRACING ON Thu Apr 15 13:41:42 +1200 2010 Thu Apr 15 13:42:05 +1200 2010 REQUEST / --- !map:Mongrel::HttpParams SERVER_NAME: 127.0.0.1 HTTP_ACCEPT: application/xml,application/xhtml+xml,text/html;q=0.9,text/plain;q=0.8,image/png,*/*;q=0.5 HTTP_HOST: 127.0.0.1:3000 HTTP_USER_AGENT: Mozilla/5.0 (Macintosh; U; Intel Mac OS X 10_6_3; en-US) AppleWebKit/533.2 (KHTML, like Gecko) Chrome/5.0.342.9 Safari/533.2 REQUEST_PATH: / SERVER_PROTOCOL: HTTP/1.1 HTTP_IF_NONE_MATCH: "\"216cc63ce3c1f286ef8dd4f18f354f6e\"" HTTP_ACCEPT_LANGUAGE: en-US,en;q=0.8 REMOTE_ADDR: 127.0.0.1 PATH_INFO: / SERVER_SOFTWARE: Mongrel 1.1.5 SCRIPT_NAME: / HTTP_COOKIE: _ServiceApp_session=BAh7DDonY3VzdG9tZXJfbGlzdF9maWx0ZXJfam9iX3N0YXR1c19pZGn6Og9zZXNzaW9uX2lkIiU0ZTk1ZWZjMmViMGU3NjE2YzA0NDc2YTkxYzJlNDZiOToaY3VycmVudF9jdXN0b21lcl9uYW1lIilUSEUgQ1VTVE9NRVIgTkFNRSBORUVEUyBUTyBCRSBMT0FERUQ6EF9jc3JmX3Rva2VuIjFuT1JMUWk0NlZrWlM3c2lUN3BaWCs5NkhRajhxYnFwRnhzVHVTWXEvUWY0PToZam9iX2xpc3RfZmlsdGVyX3RleHQiADogam9iX2xpc3RfZmlsdGVyX2VtcGxveWVlX2lkafo6HmN1c3RvbWVyX2xpc3RfZmlsdGVyX3RleHQiAA%3D%3D--d01bc5d0b457ad524d16cb3402b5dfed9afce83d HTTP_VERSION: HTTP/1.1 REQUEST_URI: / SERVER_PORT: "3000" HTTP_ACCEPT_CHARSET: ISO-8859-1,utf-8;q=0.7,*;q=0.3 REQUEST_METHOD: GET GATEWAY_INTERFACE: CGI/1.2 HTTP_ACCEPT_ENCODING: gzip,deflate,sdch HTTP_CONNECTION: keep-alive Any direction or ideas would be greatly appreciated. Thanks.

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  • PHP, MySQL, and temporary tables

    - by dave
    Php novice. 1.Is there anything wrong with this PHP & MySQL code? include_once "db_login.php" ; $sql = "DROP TEMPORARY TABLE IF EXISTS temp_sap_id_select" ; mysql_query ( $sql ) or ( "Error " . mysql_error () ) ; $sql = " CREATE TEMPORARY TABLE temp_sap_id_select ( `current_page` INT NOT NULL, `total_pages` INT NOT NULL, `select_date` DATE NOT NULL, `select_schcode` CHAR(6) NOT NULL, `select_user` CHAR(30) NOT NULL, `select_id` CHAR(9) NOT NULL ) " ; mysql_query ( $sql ) or ( "Error " . mysql_error () ) ; 2.Admittedly, I'm a dull boy, but I'll ask anyway: If I'm using a MySQL GUI and have open the target database, will it be aware of the above temporary table created by PHP/MySQL (IF the temporary table is properly created)?

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  • Connecting Named SQL Server Express 2005 from MySQL Migration Toolkit 1.1.10

    - by KoolKabin
    Hi guys, I am trying to migrate SQL Server Express 2005 database to mysql. I came across the mysql migration toolkit. When i started with the tool it asked for my sql server express information. I provided all the information of the sql express but it still can't connect. My machine has got 1.) SQL Server 2000 [Default instance eg computername ] 2.) SQL Server Express 2005 [Default Named Instance eg computername$SQLExpress ] *I can make easy connection from Microsoft SQL Server Management Studio. I am getting problem only while connecting from MySQl Migration toolkit 1.1.10

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  • Get data to android app from mysql server

    - by Fabian
    Hi i have an mysql database with some sports results in it. I want to write an android application to display these data on mobile phones. I´ve searched on the internet for this issue and i think it is not possible to have a direct connection between the mysql database and the android application. (Is this right?) So my question is the following: How can i have access in the android application to the mysql database in order to display some of the data? Thank you for your answers! Fabian

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  • ClassNotFoundException in MySQL Connector/J

    - by Ephraim
    This has been asked before but I cannot find the answer I need. 1) Using Class.forName("com.mysql.java.Driver") in the eclipse IDE all works well. I load the correct jar (mysql-connector-java-5.1.20-bin.jar), no exception. When I create a jar for my app a1.jar and double click the jar, I get the ClassnotFoundException. I created a .bat file in Windows XP with java -classpath c:\temp\mysql-connector-java-5.1.20-bin.jar -jar c:\temp\a1.jar the app statrs with the same exception. Furthermore using System.getProperty ("java.class.path") shows c:\temp\a1.jar whilst in the IDE I can see several directories

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  • Installing Wordpress - constant PHP/MySQL extension appears missing

    - by Driss Zouak
    I've got Win2003 w/IIS6, PHP 5 and MySQL installed. I can confirm PHP is installed correctly because I have a testMe.php that runs properly. When I run the Wordpress setup, I get informed that Your PHP installation appears to be missing the MySQL extension which is required by WordPress. But in my PHP.ini in the DYNAMIC EXTENSIONS section I have extension=php_mysql.dll extension=php_mysqli.dll I verified that mysql.dll and libmysql.dll are both in my PHP directory. I copied my libmysql.dll to the C:\Windows\System32 directory. When I try to run the initial setup for WordPress, I get this answer. I've Googled setting this up, and everything comes down to the above. I'm missing something, but none of the instructions that I've found online seem to cover whatever that is.

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  • Entity Framework - MySQL - Datetime format issue

    - by effkay
    Hello; I have a simple table with few date fields. Whenever I run following query: var docs = ( from d in base.EntityDataContext.document_reviews select d ).ToList(); I get following exception: Unable to convert MySQL date/time value to System.DateTime. MySql.Data.Types.MySqlConversionException: Unable to convert MySQL date/time value to System.DateTime The document reviews table has two date/time fields. One of them is nullable. I have tried placing following in connection string: Allow Zero Datetime=true; But I am still getting exception. Anyone with a solution?

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