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  • mysql :ordering table with day names

    - by Meko
    Hi.I am trying to get day of names with and correct order like Monday ,Tuesday.. But in my table I have records that after Monday comes Friday or I have Thursday between two Tuesday .I want to order them like Monday ,Monday ,Tuesday ,Tuesday, Wednesday so on .But I don`t want to group them. I used this query but it does not make order select Day_Name from mydb.schedule where Room_NO=(510) And Week_NO =(1) it outputs Monday Monday Tuesday Wednesday Wednesday Tuesday Thursday Thursday Thursday how can I correct it?

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  • mysql query if else statemnet?

    - by user253530
    I have this sql query: SELECT S.SEARCH, S.STATUS, C.TITLE AS CategoryName, E.SEARCH_ENGINES AS Engine, S.RESULTS, S.DATE, S.TOTAL_RESULTS AS Total, S.ID FROM PLD_SEARCHES AS S Join PLD_CATEGORY AS C ON C.ID = S.CATEGORY_ID Join PLD_SEARCH_ENGINES AS E ON S.SEARCH_ENGINES_ID = E.ID ORDER BY S.DATE ASC I want to identify if S.STATUS is either 1 or 0 and according to those values to return COMPLETE or PENDING in the query results

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  • What is wrong with this javascript?

    - by bala3569
    i use this javascript syntax for validating a checkbox... alert(document.getElementById("ctl00_ContentPlaceHolder1_Chkreg").checked); if (document.getElementById("ctl00_ContentPlaceHolder1_Chkreg").checked == false) { document.getElementById("ctl00_ContentPlaceHolder1_ErrorMsg").innerHTML = "please select the checkbox"; document.getElementById("ctl00_ContentPlaceHolder1_Chkreg").focus(); return false; } My alert showed me false but my if loop is not working... Any suggestion...

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  • Modifying MySQL Where Statement Based on Array

    - by Ryan
    Using an array like this: $data = array ( 'host' => 1, 'country' => 'fr', ) I would like to create a MySQL query that uses the values of the array to form its WHERE clause like: SELECT * FROM table WHERE host = 1 and country = 'fr' How can I generate this query string to use with MySQL?

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  • NHibernate (or other worth recommendation ORM), real life example?

    - by migajek
    I'd like to learn database applications in C# and I'm about to select some framework. I heard many recommendations of NHibernate, however I haven't decided yet. Anyway, I'd like to know if there's any real-life example (with sources) of NHibernate in C#, to learn best practices etc.? I know all of them are probably covered in the docs, but working example helps a lot understanding the proper development pattern.

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  • What VS 2010 Template to Chose?

    - by Dr Hydralisk
    I am just getting started with C++ and wanted to know does it really matter what template you chose in Visual Studio 2010 (for creating executables)? Like if I was creating a console application there is CLR Console Application, Win32 Console Application, and Win32 Project (description says it can be an application or a dll), what would I chose (or could i select Empty Project)?

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  • array with sessions, only prints one letter

    - by jolabero
    On login: $result = mysql_query("SELECT `id`, `username`, `email` FROM `users` WHERE `username` = '$username' AND `password` = '$passwd'"); $userdata = array('id','username','email'); $_SESSION['user'] = mysql_result($result, 0, $userdata); And when i want to print the users username echo $_SESSION['user']['username'] it only prints the first letter :/ whats wrong`?

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  • type casting in mysql

    - by muralikalpana
    i have passportno(varchar) in database. i am entering values like this 001,002,003. and i want to display like sorting order. now i wrote query like this "select * from passport_registration where status=1 ORDER BY passportno" then displaying output like this......077,088,099,100,1000,1001,1009,101,1010 i want to diplay sort order. how to do?

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  • Xcode: iPhone screenshots no longer work

    - by Chris Newman
    I can't be sure when this stopped working because I haven't used it for a while, but it was possibly since I upgraded to the 3.2 SDK. In Organizer, when I select the "Screenshots" tab and press "Capture", nothing happens. I've tried this with three different devices and I've restarted my Mac. What's happened, and how can I fix it?

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  • Any way to make this PostgreSQL count query any faster?

    - by Ben Dauphinee
    I'm running a case-insensitive search on a table with 7.2 million rows, and I was wondering if there was any way to make this query any faster? Currently, it takes approx 11.6 seconds to execute, with just one search parameter, and I'm worried that as soon as I add more than one, this query will become massively slow. SELECT count(*) FROM "exif_parse" WHERE (description ~* 'canon')

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  • Difference between dates when grouping in SQL

    - by CeejeeB
    I have a table of purchases containing a user_id and a date_of_purchase. I need to be able to select all the users who have made 2 purchases within 12 months of each other. The dates can be any point in time as long as they are less than 12 months apart. e.g. user_id date_of_purchase 123 01/Jan/2010 124 01/Aug/2010 123 01/Feb/2010 124 05/Aug/2008 In this example i want user_id 123

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  • pgSQL query error

    - by running4surival
    i tried using this query: "SELECT * FROM guests WHERE event_id=".$id." GROUP BY member_id;" and I'm getting this error: ERROR: column "guests.id" must appear in the GROUP BY clause or be used in an aggregate function can anyone explain how i can work around this?

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  • Determining an Oracle SQL MERGE statement result

    - by petejamd
    Follow up to this question This (similar version from old link) works in SQL Server 2008, however, Oracle is giving me trouble: MERGE INTO wdm_test USING ( select '10000000000000000000000000000000' Guid from DUAL ) val ON ( wdm_test.Guid = val.Guid ) WHEN MATCHED THEN UPDATE SET test_column = null WHEN NOT MATCHED THEN INSERT (Guid, test_column) VALUES ('10000000000000000000000000000000', null) OUTPUT $action; SQL Error: ORA-00933: SQL command not properly ended Does Oracle not support OUTPUT $action;? If not, is there an alternative?

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  • Django: GROUP BY two values

    - by AP257
    I would basically like to do the same as this question, but grouping by combinations of two values, rather than just one: SELECT player_type, team, COUNT(*) FROM players GROUP BY player_type, team; Does anyone know whether, and how, this is possible in Django? I'm using 1.2.

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  • sql - getting the id from a row based on a group by

    - by user85116
    Table A tableAID tableBID grade Table B tableBID name description Table A links to Table b from the tableBID found in both tables. If I want to find the row in Table A, which has the highest grade, for each row in Table B, I would write my query like this: select max(grade) from TableA group by tableBID However, I don't just want the grade, I want the grade plus id of that row.

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  • Calculating average (AVG) and grouping by week on large data set takes too long

    - by caioiglesias
    I'm getting average prices by week on 7 million rows, it's taking around 30 seconds to get the job done. This is the simple query: SELECT AVG(price) as price, yearWEEK(FROM_UNIXTIME(timelog)) as week from pricehistory where timelog > $range and product_id = $id GROUP BY week The only week that actually gets data changed and is worth averaging every time is always the last one, so this calculation for the whole period is a waste of resources. I just wanted to know if mysql has a tool to help out on this.

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  • not case sensitive query in mysql

    - by Mac Taylor
    hey guys i need to query my database and find results : mysql_query("select * from ".ALU_TABLE." where username like '%$q%' or name like '%$q%'"); if i have a name in my table such as Book and i enter book in search box it wont show the Book i need to query my database as not to be case sensitive.

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  • Ruby on Rails collection_select display attribute

    - by New2rails
    I'm new to Rails and am working with the collection_select method. I have two fields I'd like to display in my select box: first_name and last_name So far, I can only display one or the other, not both. Here's the code I'm working with: collection_select(:hour,:shopper_id,@shoppers,:id,"last_name") Thank you.

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