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  • MYSQL: Error: Cannot add or update a child row: a foreign key constraint fails

    - by DalivDali
    Hi all, Using MySQL on Windows OS, and am getting an error upon attempting to create a foreign key between two tables: CREATE TABLE tf_traffic_stats ( domain_name char(100) NOT NULL, session_count int(11) NULL, search_count int(11) NULL, click_count int(11) NULL, revenue float NULL, rpm float NULL, cpc float NULL, traffic_date date NOT NULL DEFAULT '0000-00-00', PRIMARY KEY(domain_name,traffic_date)) and CREATE TABLE td_domain_name ( domain_id int(10) UNSIGNED AUTO_INCREMENT NOT NULL, domain_name char(100) NOT NULL, update_date date NOT NULL, PRIMARY KEY(domain_id)) The following statement gives me the error present in the subject line (cannot add or update a child row: a foreign key constraint fails): ALTER TABLE td_domain_name ADD CONSTRAINT FK_domain_name FOREIGN KEY(domain_name) REFERENCES tf_traffic_stats(domain_name) ON DELETE RESTRICT ON UPDATE RESTRICT Can someone point me in the right direction of what may be causing the error. I also have a foreign key referencing td_domain_name.domain_id, but I don't think this should be interfering... Appreciate it!

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  • MySQL - Sort on a calculated value based on two dates

    - by Petter Magnusson
    I have the following problem that needs to be solved in a MySQL query: Fields info - textfield date1 - a date field date2 - a date field offset1 - a text field with a number in the first two positions, example "10-High" offset2 - a text field with a number in the first two positions, example "10-High" I need to sort the records by the calculated "sortvalue" based on the current date (today): If today=date2 then sortvalue=offset1*10+offset2*5+1000 else sortvalue=offset1*10+offset2*5 I have quite good understanding of basic SQL with joins etc, but this I am not even sure if its possible...if it helps I could perhaps live with a single formula giving the same sort of effect as the IFs do....ie. before date1 = low value, after date2 = high value... Rgds PM

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  • Searching a table MySQL & PHP.

    - by S1syphus
    I want to be able to search through a MySQL table using values from a search string, from the url and display the results as an XML output. I think I have got the formatting and declaring the variables from the search string down. The issue I have is searching the entire table, I've looked over SO for previous answers, and they all seem to have to declare each column in the table to search through. So for example my database layout is as follows: **filesindex** -filename -creation -length -wall -playlocation First of all would the following be appropriate: $query = "SELECT * FROM filesindex WHERE filename LIKE '".$searchterm."%' UNION SELECT * FROM filesindex WHERE creation LIKE '".$searchterm."%' UNION SELECT * FROM filesindex WHERE length LIKE '".$searchterm."%' UNION SELECT * FROM filesindex WHERE wall LIKE '".$searchterm."%' UNION SELECT * FROM filesindex WHERE location LIKE '".$searchterm."%'"; Or ideally, is there an easier way that involves less hardcoding to search a table. Any ideas? Thanks

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  • MySQL: Insert row on table2 if row in table1 exists

    - by Andrew M
    I'm trying to set up a MySQL query that will insert a row into table2 if a row in table1 exist already, otherwise it will just insert the row into table1. I need to find a way to adapt the following query into inserting a row into table2 with the existing row's id. INSERT INTO table1 (host, path) VALUES ('youtube.com', '/watch') IF NOT EXISTS ( SELECT * FROM table1 WHERE host='youtube.com' AND path='/watch' LIMIT 1); Something kind of like this: INSERT ... IF NOT EXISTS(..) ELSE INSERT INTO table2 (table1_id) VALUES(row.id); Except I don't know the syntax for this.

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  • Reversing column values in mysql command line

    - by user94154
    I have a table posts with the column published, which is either 0 (unpublished) or 1 (published). Say I want to make all the published posts into unpublished posts and all the unpublished posts into published posts. I know that running UPDATE posts SET published = 1 WHERE published = 0; UPDATE posts SET published = 0 WHERE published = 1; will end up turning all my posts into published posts. How can I run these queries in the mysql command line so that it truly "reverse" the values, as opposed to the mistake outlined above? Thanks

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  • insert array to mysql db function

    - by ganjan
    Hi. I have an array where the keys represent each column in my database. Now I want a function that makes a mysql update query. Something like $db['money'] = $money_input + $money_db; $db['location'] = $location $query = 'UPDATE tbl_user SET '; for($x = 0; $x < count($db); $x++ ){ $query .= $db something ".=." $db something } $query .= "WHERE username=".$username." ";

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  • MySQL Query like not returning correct results

    - by Herr Kaleun
    Hello friends, i've a MySQL query that should return some rows that have the letters Ö or Ü in it but it actually does not. The query code is this: $this->db->like('title', $text ); It's PHP CodeIgniter active query. Lets assume we have 2 rows. 1. Büm 2. Bom if i search for Bü, the 1. row has to be returned but it does not. When i search for Bo the second row gets returned successfully and when i search for B both rows are returned. How could i fix this? What may be the underlieng cause? Thanks for reading.

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

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

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

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

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  • MYSQL Inserting rows that reference main rows.

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

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  • Mysql on duplicate key update + sub query

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

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  • MySQL Query still executing after a day..?

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

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  • MySQL temp table issue

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

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  • PHP MySQL Syntax Error 'You have an error in your SQL syntax'

    - by Alec
    I cannot figure out the issue with my code here. I am trying to take info from the table, then subtract 1 second from Current_Time which looks like '2:00'. The problem is, I get: "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Current_Time) VALUES('22')' at line 1" I don't even understand where it gets 22 from. Thanks, I really appreciate it. if (isset($_GET['id']) && isset($_GET['time'])) { mysql_select_db("aleckaza_pennyauction", $connection); $query = "SELECT Current_Time FROM Live_Auctions WHERE ID='1'"; $results = mysql_query($query) or die(mysql_error()); while ($row = mysql_fetch_array($results)) { $newTime = $row['Current_Time'] - 1; $query = "INSERT INTO Live_Auctions(Current_Time) VALUES('".$newTime."')"; $results = mysql_query($query) or die(mysql_error()); } }

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  • Mysql Query - Order By Not Working

    - by jwzk
    I'm running Mysql 5.0.77 and I'm pretty sure this query should work? SELECT * FROM purchases WHERE time_purchased BETWEEN '2010-04-15 00:00:00' AND '2010-04-18 23:59:59' ORDER BY time_purchased ASC, order_total DESC time_purchased is DATETIME, and an index. order_total is DECIMAL(10,2), and not an index. I want to order all purchases by the date (least to greatest), and then by the order total (greatest to least). So I would output similar to: 2010-04-15 $100 2010-04-15 $80 2010-04-15 $20 2010-04-16 $170 2010-04-16 $45 2010-04-16 $15 2010-04-17 $274 .. and so on. The output I am getting from that query has the dates in order correctly, but it doesn't appear to sort the order total column at all. Thoughts? Thanks.

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

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

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  • MySQL query problem

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

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  • PHP MySQL Insert Data

    - by happyCoding25
    Hello, Im trying to insert data into a table in MySQL. I found/modified some code from w3Schools and still couldn't get it working. Heres what I have so far: <?php $rusername=$_POST['username']; $rname=$_POST['name']; $remail=$_POST['emailadr']; $rpassword=$_POST['pass']; $rconfirmpassword=$_POST['cpass']; if ($rpassword==$rconfirmpassword) { $con = mysql_connect("host","user","password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("mydbname ", $con); } mysql_query("INSERT INTO members (id, username, password) VALUES ('4', $rusername, $rpassword)"); ?> Did I mistype something? To my understanding "members" is the name of the table. If anyone knows whats wrong I appreciate the help. Thanks

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  • Optimize MySQL database query

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

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  • Five Ways Enterprise 2.0 Can Transform Your Business - Q&A from the Webcast

    - by [email protected]
    A few weeks ago, Vince Casarez and I presented with KMWorld on the Five Ways Enterprise 2.0 Can Transform Your Business. It was an enjoyable, interactive webcast in which Vince and I discussed the ways Enterprise 2.0 can transform your business and more importantly, highlighted key customer examples of how to do so. If you missed the webcast, you can catch a replay here. We had a lot of audience participation in some of the polls we conducted and in the Q&A session. We weren't able to address all of the questions during the broadcast, so we attempted to answer them here: Q: Which area within your firm focuses on Web 2.0? Meaning, do you find new departments developing just to manage the web 2.0 (Twitter, Facebook, etc.) user experience or are you structuring current departments? A: There are three distinct efforts within Oracle. The first is around delivery of these Web 2.0 services for enterprise deployments. This is the focus of the WebCenter team. The second effort is injecting these Web 2.0 services into use cases that drive the different enterprise applications. This effort is focused on how to manage these external services and bring them into a cohesive flow for marketing programs, customer care, and purchasing. The third effort is how we consume these services internally to enhance Oracle's business delivery. It leverages the technologies and use cases of the first two but also pushes the envelope with regards to future directions of these other two areas. Q: In a business, Web 2.0 is mostly like action logs. How can we leverage the official process practice versus the logs of a recent action? Example: a system configuration modified last night on a call out versus the official practice that everybody would use in the morning.A: The key thing to remember is that most Web 2.0 actions / activity streams today are based on collaboration and communication type actions. At least with public social sites like Facebook and Twitter. What we're delivering as part of the WebCenter Suite are not just these types of activities but also enterprise application activities. These enterprise application activities come from different application modules: purchasing, HR, order entry, sales opportunity, etc. The actions within these systems are normally tied to a business object or process: purchase order/customer, employee or department, customer and supplier, customer and product, respectively. Therefore, the activities or "logs" as you name them are able to be "typed" so that as a viewer, you can filter or decide to see only certain types of information. In your example, you could have a view that only showed you recent "configuration" changes and this could be right next to a view that showed off the items to be watched every morning. Q: It's great to hear about customers using the software but is there any plan for future webinars to show what the products/installs look like? That would be very helpful.A: We don't have a webinar planned to show off the install process. However, we have a viewlet that's posted on Oracle Technology Network. You can see it here:http://www.oracle.com/technetwork/testcontent/wcs-install-098014.htmlAnd we've got excellent documentation that walks you through the steps here:http://download.oracle.com/docs/cd/E14571_01/install.1111/e12001/install.htmAnd there's a whole set of demos and examples of what WebCenter can do at this URL:http://www.oracle.com/technetwork/middleware/webcenter/release11-demos-097468.html Q: How do you anticipate managing metadata across the enterprise to make content findable?A: We need to first make sure we are all talking about the same thing when we use a word like "metadata". Here's why...  For a developer, metadata means information that describes key elements of the portal or application and what the portal or application can do. For content systems, metadata means key terms that provide a taxonomy or folksonomy about the information that is being indexed, ordered, and managed. For business intelligence systems, metadata means key terms that provide labels to groups of data that most non-mathematicians need to understand. And for SOA, metadata means labels for parts of the processes that business owners should understand that connect development terminology. There are also additional requirements for metadata to be available to the team building these new solutions as well as requirements to make this metadata available to the running system. These requirements are often separated by "design time" and "run time" respectively. So clearly, a general goal of managing metadata across the enterprise is very challenging. We've invested a huge amount of resources around Oracle Metadata Services (MDS) to be able to provide a more generic system for all of these elements. No other vendor has anything like this technology foundation in their products. This provides a huge benefit to our customers as they will now be able to find content, processes, people, and information from a common set of search interfaces with consistent enterprise wide results. Q: Can you give your definition of terms as to document and content, please?A: Content applies to a broad category of information from Word documents, presentations and reports through attachments to invoices and/or purchase orders. Content is essentially any type of digital asset including images, video, and voice. A document is just one type of content. Q: Do you have special integration tools to realize an interaction between UCM and WebCenter Spaces/Services?A: Yes, we've dedicated a whole team of engineers to exploit the key features of Oracle UCM within WebCenter.  While ensuring that WebCenter can connect to other non-Oracle systems, we've made sure that with the combined set of Oracle technology, no other solution can match the combined power and integration.  This is part of the Oracle Fusion Middleware strategy which is to provide best in class capabilities for Content and Portals.  When combined together, the synergy between the two products enables users to quickly add capabilities when they are needed.  For example, simple document sharing is part of the combined product offering, but if legal discovery or archiving is required, Oracle UCM product includes these capabilities that can be quickly added.  There's no need to move content around or add another system to support this, it's just a feature that gets turned on within Oracle UCM. Q: All customers have some interaction with their applications and have many older versions, how do you see some of these new Enterprise 2.0 capabilities adding value to existing enterprise application deployments?A: Just as Service Oriented Architectures allowed for connecting the processes of different applications systems to work together, there's a need for a similar approach with regards to these enterprise 2.0 capabilities. Oracle WebCenter is built on a core architecture that allows for SOA of these Enterprise 2.0 services so that one set of scalable services can be used and integrated directly into any type of application. In this way, users can get immediate value out of the Enterprise 2.0 capabilities without having to wait for the next major release or upgrade. These centrally managed WebCenter services expose a set of standard interfaces that make it extremely easy to add them into existing applications no matter what technology the application has been implemented. Q: We've heard about Oracle Next Generation applications called "Fusion Applications", can you tell me how all this works together?A: Oracle WebCenter powers the core collaboration and social computing services found within Fusion Applications. It is the core user experience technology for how all the application screens have been implemented. And the core concept of task flows allows for all the Fusion Applications modules to be adaptable and composable by business users and IT without needing to be a professional developer. Oracle WebCenter is at the heart of the new Fusion Applications. In addition, the same patterns and technologies are now being added to the existing applications including JD Edwards, Siebel, Peoplesoft, and eBusiness Suite. The core technology enables all these customers to have a much smoother upgrade path to Fusion Applications. They get immediate benefits of injecting new user interactions into their existing applications without having to completely move to Fusion Applications. And then when the time comes, their users will already be well versed in how the new capabilities work. Q: Does any of this work with non Oracle software? Other databases? Other application servers? etc.A: We have made sure that Oracle WebCenter delivers the broadest set of development choices so that no matter what technology you developers are using, WebCenter capabilities can be quickly and easily added to the site or application. In addition, we have certified Oracle WebCenter to run against non-Oracle databases like DB2 and SQLServer. We have stated plans for certification against MySQL as well. Later in CY 2011, Oracle will provide certification on non-Oracle application servers such as WebSphere and JBoss. Q: How do we balance User and IT requirements in regards to Enterprise 2.0 technologies?A: Wrong decisions are often made because employee knowledge is not tapped efficiently and opportunities to innovate are often missed because the right people do not work together. Collaboration amongst workers in the right business context is critical for success. While standalone Enterprise 2.0 technologies can improve collaboration for collaboration's sake, using social collaboration tools in the context of business applications and processes will improve business responsiveness and lead companies to a more competitive position. As these systems become more mission critical it is essential that they maintain the highest level of performance and availability while scaling to support larger communities. Q: What are the ways in which Enterprise 2.0 can improve business responsiveness?A: With a wide range of Enterprise 2.0 tools in the marketplace, CIOs need to deploy solutions that will meet the requirements from users as well as address the requirements from IT. Workers want a next-generation user experience that is personalized and aggregates their daily tools and tasks, while IT needs to ensure the solution is secure, scalable, flexible, reliable and easily integrated with existing systems. An open and integrated approach to deploying portals, content management, and collaboration can enhance your business by addressing both the needs of knowledge workers for better information and the IT mandate to conserve resources by simplifying, consolidating and centralizing infrastructure and administration.  

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

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

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  • mysql command line not working

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

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  • MySQL partition "full"?

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

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

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

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  • Upgrading PHP, MySQL old-passwords issue

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

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