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  • SQL Join only returning 1 row.

    - by kevin
    Not quite sure what I'm missing, but my SQL statement is only returning one row. SELECT tl.*, (tl.topic_total_rating/tl.topic_rates) as topic_rating, COUNT(pl.post_id) - 1 as reply_count, MIN(pl.post_time) AS topic_time, MAX(pl.post_time) AS topic_bump FROM topic_list tl JOIN post_list pl ON tl.topic_id=pl.post_parent WHERE tl.topic_board_link = %i AND topic_hidden != 1 ORDER BY %s I have two tables (post_list and topic_list), and post_list's post_parent links to a topic_list's topic_id. Instead of returning all the topics (where their board's topic_board_link is n), it only returns one topic.

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  • reterview data from two tables using inner join in cakephp

    - by user3593884
    I two tables from database one as user(id,first_name,last_name) and the second table location(id,country). I need to perform inner join with this two tables and the list should display first_name,last_name,country with condition user.id=location.id I have written sql queries in cakephp $this->set('users',$this->User->find('list', array( 'fields' => array('User.id', 'User.first_name','location.country'), array('joins' => array(array('table' => 'location', 'alias' => 'location', 'type' => 'INNER', 'conditions' => array('User.id = location.id'))))))); i get error -Unknown column 'location.country' in 'field list' Please help!

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  • I'm making a resume...what would you consider as intermediate python programming skills?

    - by user285884
    I've made a couple of scripts. One is a stock screener that can search through every stock. Another creates a heatmap that tells you what's performed well and badly over the past day. They aren't really that useful, just did them to work on my programming skills. I was able to throw some SQL in my scripts too. Would you call that intermediate? Thanks? How do you guys list your programming skills on your resume? Maybe there's a better way of putting it on my resume than "intermediate" or "beginner."

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  • why does InnoDB keep on growing without for every update?

    - by Akash Kava
    I have a table which consists of heavy blobs, and I wanted to conduct some tests on it. I know deleted space is not reclaimed by innodb, so I decided to reuse existing records by updating its own values instead of createing new records. But I noticed, whether I delete and insert a new entry, or I do UPDATE on existing ROW, InnoDB keeps on growing. Assuming I have 100 Rows, each Storing 500KB of information, My InnoDB size is 10MB, now when I call UPDATE on all rows (no insert/ no delete), the innodb grows by ~8MB for every run I do. All I am doing is I am storing exactly 500KB of data in each row, with little modification, and size of blob is fixed. What can I do to prevent this? I know about optimize table, but I cant do it because on regular usage, the table is going to be 60-100GB big, and running optimize will just stall entire server.

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  • how to increase or decrease the quantity in a field in my sql when we enter the new quantity

    - by madhu
    I'm doing a project I'm encountering a problem in my project......... the problem is i would like to increase or decrease the quantity in my sql quantity field when i pass the issued or delivered quantity i.e.... initially the quantity in the quantity filed is 1 when i intake the product of a quantity 10 then automatically it should update the quantity as 10+1=11 so it must update as 11, if i remove a 1 quantity it should update as 0........ how to write a code in jsp.......... pls do help

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  • Reverse wildcard search in codeigniter

    - by Andy Platt
    I am implementing a snippet-based content management system into my current project. Snippets can be associated with a page either by exact match of the full url falling back to a wildcard match on a partial url or finally a default snippet. To implement this I have a created table of page associations with a wildcard flag, the system first checks the current url against the non-wildcard associations and if it doesn't find a match it checks against the partial url's where the wildcard flag is set. In order to achieve this I am getting all the partial url's from the database and putting them into an array then walking the array to check for a match against the current url: protected function _check_wildcard($url = NULL) { if($url) { $q = $this->db->where('wildcard' ,'Y') ->from('content') ->get(); $wildcards = $q->result_array(); foreach($wildcards AS $wildcard) { if(strpos($url,$wildcard['url']) > 0) { return $wildcard['snipppet_id']; } } } else { return NULL; } } Can anyone suggest a better way to do this - preferably one that doesn't involve having to constantly download the full list of all the wildcards each time I load a page as I am afraid that this will have a negative effect on the scalability of the system down the line?

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  • Is this a good database design?

    - by alokpatil
    I am planning to make a railway reservation project... I am maintaining following tables: trainTable (trainId,trainName,trainFrom,trainTo,trainDate,trainNoOfBoogies)...PK(trainId) Boogie (trainId,boogieId,boogieName,boogieNoOfseats)...CompositeKey(trainId,boogieId)... Seats (trainId,boogieId,seatId,seatStatus,seatType)...CompositeKey(trainId,boogieId,seatId)... user (userId,name...personal details) userBooking (userId,trainId,boogieId,seatId)... Is this good design?

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  • Autopopulate from Select box from database

    - by Chris Spalton
    hope you can help, please forgive any poor coding or anytihng, I'm new to this and just hacking my way through to get things to work. That said, on one of my projects I have this code, which successfully populates the dropdown from a database when the page is loaded: <select name="Region" id="Region"> <option value="">-- Select Region --</option> <?php $region=$POST['Region']; if ($region); { $regionquery = "SELECT DISTINCT REGION FROM Sales_Execs "; $regionresult = mysql_query($regionquery); while($row = mysql_fetch_array($regionresult)) { echo "<option value=\"".$row['REGION']."\">".$row['REGION']."</option>\n "; } } ?> <script type="text/javascript"> document.getElementById('Region').value = <?php echo json_encode(trim($_POST['Region']));?>; </script> </select> On my next project that I'm working on now, I need to do the same thing, so I copied the above code amended, and placed in my new project: <select name="Sales_Exec" id="Sales_Exec"> <option value="">-- Select SE --</option> <?php $salesexec=$POST['Sales_Exec']; if ($salesexec); { $salesexecquery = "SELECT DISTINCT Assigned FROM Data "; $salesexecresult = mysql_query($salesexecquery); while($row = mysql_fetch_array($salesexecresult)) { echo "<option value=\"".$row['ASSIGNED']."\">".$row['ASSIGNED']."</option>\n "; } } ?> <script type="text/javascript"> document.getElementById('Sales_Exec').value = <?php echo json_encode(trim($_POST['Sales_Exec']));?>; </script> </select> This second chunk of code doesn't work... and I can't work out why as it seems I've copied it all and amended all the neccersary parts, can anyone spot what is wrong? Thankyou!

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  • Database Formatting for Album Tracks

    - by Sev
    I would like to store album's track names in a single field in a database. The number of tracks are arbitrary for each album. Each album is one record in the table. Each track must be linked to a specific URL which also should be stored in the database somewhere. Is it possible to do this by storing them in a single field, or is a relational table for the track names/urls the only way to go?

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  • How to update mutual entries in DB?

    - by Sthita
    I am not able to perform mutual update to my Database. My Requirement is like this : Insert IDS UPDate DB Entries ID ConnectedTo ID connectedTo connectedVia 2 1 --- No Entry ----- 2 3 -- No Entry ----- When ID 1 Comes in to picture (Enties happened to Table) 1 4 1 3 2 1 5 2 4 1 1 2 2 5 1 When ID 3 Comes into picture (Enties happened to Table) 3 4 3 5 1 3 1 ---- 3 2 ---- Similarly when ID 4 Comes in to picture 4 9 4 2 3 4 10 4 5 1 4 3 4 2 1 4 1 1 9 4 1 10 4 Its basically Updating the mutual connections.No Duplicate entries should happen. Like for example 2 3 and 3 2 are mutually connected and 1 is associated with them so there is no need to insert (3 2 via 1). I think this is my requirement. Anything possible combinations i have missed please let me know. i will Update. Please help me writing the logic or any example using jdbc and java.

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  • reading into table: comma values and quotes SQL

    - by every_answer_gets_a_point
    i have a string like this something = "something, something1, "something2, something else", something3" i need it to be read into a table like this: field1 = "something" field2= "something2" field3 = "something2, something else" field4 = "something3" please notice that the double quotes in the something string signified that the string inside the quotes is to be placed in one field anyone know how to do this with an insert into statement or some other way? the answer can be purely sql or can be vba with sql. thanks!

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  • INSERT ... ON DUPLICATE KEY UPDATE with WHERE?

    - by Raven
    Hi! I'm doing a INSERT ... ON DUPLICATE KEY UPDATE but I need the update part to be conditional, only doing the update if some extra condition has changed. However, WHERE is not allowed on this UPDATE. Is there any workaround for this? I can't do combinations of INSERT/UPDATE/SELECT since this needs to work over a replication.

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  • How do I do Textbox Submit

    - by Newb
    Hello everyone, I have a search box and a buttion. currently a user enter some text and press the search button. But I want to add another feature that instead of clicking the search button people can hit enter to search. How can I do that? Here is my code sample: <form method="post" action=""> <input id="search" name="search" type="text" /> <input id="search_btn" name="search_btn" type="submit" /> </form> Thanks in advance

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  • Empty files generated from running `mysqldump` using PHP

    - by alex
    I keep getting empty files generated from running $command = 'mysqldump --opt -h localhost -u username -p \'password\' dbname > \'backup 2009-04-15 09-57-13.sql\''; command($command); Anyone know what might be causing this? My password has strange characters in it, but works fine with connecting to the db. I've ran exec($command, $return) and outputted the $return array and it is finding the command. I've also ran it with mysqldump > file.sql and the file contains Usage: mysqldump [OPTIONS] database [tables] OR mysqldump [OPTIONS] --databases [OPTIONS] DB1 [DB2 DB3...] OR mysqldump [OPTIONS] --all-databases [OPTIONS] For more options, use mysqldump --help So it would seem like the command is working.

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  • how can I check data has been inserted successfully

    - by Piyush
    I have two insert statements.Second one will be executed only after successful execution of First one.What I wd like to do is- $sqlone="Insert into ....."; $sqltwo="Insert into....."; If(mysql_query($sqlone)) { If(mysql_query($sqltwo)) { show message Data inserted in both tables. } }

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  • Set AUTO_INCREMENT value programmatically

    - by Tim
    So this works... ALTER TABLE variation AUTO_INCREMENT = 10; But I want to do this; ALTER TABLE variation AUTO_INCREMENT = (SELECT MAX(id)+1 FROM old_db.varaition); but that doesnt work, and neither does; SELECT MAX(id)+1 INTO @old_auto_inc FROM old_db.variation ALTER TABLE variation AUTO_INCREMENT = @old_auto_inc; So does anyone know how to do this? ( I'm trying to ensure that AUTO_INCREMENT keys dont collide between an old and a new site and need to do this automatically. So I can just run a script when the new db goes live )

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  • Locking a detail view if a user is editing the item...

    - by BenTheDesigner
    Hi All I am developing a user manager which must control access to the detail view of editable items. At present, when a user clicks 'edit', the application queries the link table to check if a user is currently editing that page, if not, it allows access to the page and then inserts a record into the link table preventing another user from editing the same page at the same time. My question is what would the best way to handle the removal of records if say a user exists the browser without saving etc, therefore no action to remove the record. I have a couple of ideas but would like other input before I decide. BenTheDesigner

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  • Changing CCK content-types details results in numerous DB calls for the menu system

    - by Paul Strugger
    Every time I make a change in the details of a content-type it takes too long. I though it had to do with the fact that I had too many content-types and fields (~500), but when I load the devel module to see the queries that take that long I see: Executed 32212 queries in 12267.57 milliseconds. Queries taking longer than 5 ms and queries executed more than once, are highlighted. Page execution time was 55763.32 ms When I see the details I notice that the vast majority of db calls come from the menu system, e.g.: _menu_route menu_local_tasks admin_menu_link_save Why is that? Can I avoid some of these? It doesn't seem logical!

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  • Are Triggers Based On Queries Atomic?

    - by David
    I have a table that has a Sequence number. This sequence number will change and referencing the auto number will not work. I fear that the values of the trigger will collide. If two transactions read at the same time. I have ran simulated tests on 3 connections @ ~1 million records each and no collisions. CREATE TABLE `aut` ( `au_id` int(10) NOT NULL AUTO_INCREMENT, `au_control` int(10) DEFAULT NULL, `au_name` varchar(50) DEFAULT NULL, `did` int(10) DEFAULT NULL, PRIMARY KEY (`au_id`), KEY `Did` (`did`) ) ENGINE=InnoDB AUTO_INCREMENT=1 DEFAULT CHARSET=latin1 TRIGGER `binc_control` BEFORE INSERT ON `aut` FOR EACH ROW BEGIN SET NEW.AU_CONTROL = (SELECT COUNT(*)+1 FROM aut WHERE did = NEW.did); END;

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  • MySQLi - declaring variable after bind_param?

    - by Kerry
    This may be a completely dumb question, but I've seen a couple examples declaring the variables AFTER putting them in bind_param: http://devzone.zend.com/article/686 I've never seen this done before and all my programming knowledge says I should define them before hand. Is this a valid/preferred way?

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