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  • MySQL: Update table column from subquery result

    - by Jhourlad Estrella
    On the Members table are columns "MemberID" and "PointsEarned". I want to update the PointsEarned column from the result of this query: SELECT m.MemberID, m.UserName, ( (SELECT COUNT(*) FROM EventsLog as e WHERE e.MemberID=m.MemberID AND e.EventsTypeID=2)*10 ) + ( (SELECT COUNT(*) FROM EventsLog as e WHERE e.MemberID=m.MemberID AND e.EventsTypeID=3)*3 ) + ( (SELECT COUNT(*) FROM ChatMessages as c WHERE c.MemberID=m.MemberID)*.1 ) as PointsEarned FROM Members as m Can anybody tell me how I should do it with a single query? Thanks!

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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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  • MySQL subquery and bracketing

    - by text
    Here are my tables respondents: field sample value respondentid : 1 age : 2 gender : male survey_questions: id : 1 question : Q1 answer : sample answer answers: respondentid : 1 question : Q1 answer : 1 --id of survey question I want to display all respondents who answered the certain survey, display all answers and total all the answer and group them according to the age bracket. I tried using this query: SELECT res.Age, res.Gender, answer.id, answer.respondentid, SUM(CASE WHEN res.Gender='Male' THEN 1 else 0 END) AS males, SUM(CASE WHEN res.Gender='Female' THEN 1 else 0 END) AS females, CASE WHEN res.Age < 1 THEN 'age1' WHEN res.Age BETWEEN 1 AND 4 THEN 'age2' WHEN res.Age BETWEEN 4 AND 9 THEN 'age3' WHEN res.Age BETWEEN 10 AND 14 THEN 'age4' WHEN res.Age BETWEEN 15 AND 19 THEN 'age5' WHEN res.Age BETWEEN 20 AND 29 THEN 'age6' WHEN res.Age BETWEEN 30 AND 39 THEN 'age7' WHEN res.Age BETWEEN 40 AND 49 THEN 'age8' ELSE 'age9' END AS ageband FROM Respondents AS res INNER JOIN Answers as answer ON answer.respondentid=res.respondentid INNER JOIN Questions as question ON answer.Answer=question.id WHERE answer.Question='Q1' GROUP BY ageband ORDER BY res.Age ASC I was able to get the data but the listing of all answers are not present. Do I have to subquery SELECT into my current SELECT statement to show the answers? I want to produce something like this: ex: # of Respondents is 3 ages: 2,3 and 6 Question: what are your favorite subjects? Ages 1-4: subject 1: 1 subject 2: 2 subject 3: 2 total respondents for ages 1-4 : 2 Ages 5-10: subject 1: 1 subject 2: 1 subject 3: 0 total respondents for ages 5-10 : 1

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  • MySQL Datefields: duplicate or calculate?

    - by Konerak
    We are using a table with a structure imposed upon us more than 10 years ago. We are allowed to add columns, but urged not to change existing columns. Certain columns are meant to represent dates, but are put in different format. Amongst others: * CHAR(6): YYMMDD * CHAR(6): DDMMYY * CHAR(8): YYYYMMDD * CHAR(8): DDMMYYYY * DATE * DATETIME Since we now would like to do some more complex queries, using advanced date functions, my manager proposed to d*uplicate those problem columns* to a proper FORMATTED_OLDCOLUMNNAME column using a DATE or DATETIME format. Is this the way to go? Couldn't we just use the STR_TO_DATE function each time we accessed the columns? To avoid every query having to copy-paste the function, I could still work with a view or a stored procedure, but duplicating data to avoid recalculation sounds wrong. Solutions I see (I guess I prefer 2.2.1) 1. Physically duplicate columns 1.1 In the same table 1.1.1 Added by each script that does a modification (INSERT/UPDATE/REPLACE/...) 1.1.2 Maintained by a trigger on each modification 1.2 In a separate table 1.2.1 Added by each script that does a modification (INSERT/UPDATE/REPLACE/...) 1.2.2 Maintained by a trigger on each modification 2. On-demand transformation 2.1 Each query has to perform the transformation 2.1.1 Using copy-paste in the source code 2.1.2 Using a library 2.1.3 Using a STORED PROCEDURE 2.2 A view performs the transformation 2.2.1 A separate table replacing the entire table 2.2.2 A separate table just adding the date-fields for the primary keys Am I right to say it's better to recalculate than to store? And would a view be a good solution?

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  • MySQL inner join different results

    - by Darryl at NetHosted
    I am trying to work out why the following two queries return different results: SELECT DISTINCT i.id, i.date FROM `tblinvoices` i INNER JOIN `tblinvoiceitems` it ON it.userid=i.userid INNER JOIN `tblcustomfieldsvalues` cf ON it.relid=cf.relid WHERE i.`tax` = 0 AND i.`date` BETWEEN '2012-07-01' AND '2012-09-31' and SELECT DISTINCT i.id, i.date FROM `tblinvoices` i WHERE i.`tax` = 0 AND i.`date` BETWEEN '2012-07-01' AND '2012-09-31' Obviously the difference is the inner join here, but I don't understand why the one with the inner join is returning less results than the one without it, I would have thought since I didn't do any cross table references they should return the same results. The final query I am working towards is SELECT DISTINCT i.id, i.date FROM `tblinvoices` i INNER JOIN `tblinvoiceitems` it ON it.userid=i.userid INNER JOIN `tblcustomfieldsvalues` cf ON it.relid=cf.relid WHERE cf.`fieldid` =5 AND cf.`value` REGEXP '[A-Za-z]' AND i.`tax` = 0 AND i.`date` BETWEEN '2012-07-01' AND '2012-09-31' But because of the different results that seem incorrect when I add the inner join (it removes some results that should be valid) it's not working at present, thanks.

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  • Can't connect to MySQL database hosted in CloudBees

    - by user3692698
    I have a free CloudBees account and created a free ClearDB database using their wizards. My trouble is when I use their connection information (whether I try to connect from my Java app, or an outside tool - SQLyog to be exact) I take the error: Access denied for user 'b51dbc5757d79f'@'%' to database 'mywiki. The username provided by CloudBees does not contain those extra characters that the error message is displaying which seems like it would be a problem, but I'm not sure there is anything I can do about that since everything is configured for me. The username I am given is: b51dbc5757d79f - which I can delete and rebuild after sharing here :)

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  • mysql split value from one field to two

    - by tsiger
    Hello, I 've got a table field (membername) which contains both the last name and the first name of users. Is it possible to split those into 2 fields (memberfirst - memberlast)? All the records have this format "Firstname Lastname" (without quotes and a space in between).

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  • mysql to xls sheet genration problemI(getting html code along with records ,unable get column names)

    - by pmms
    <?php if($_POST['Submit']=='Generatexml') { $tblname=$_GET['genratexml']; //mysql_connect("localhost","root",""); //mysql_select_db("hitnrunf_db"); global $obj_mysql; $result = mysql_query("SELECT * FROM tbl_js_login"); while($row = mysql_fetch_array($result)) { $csv_output .= "$row[fld_id],$row[fld_fname],$row[fld_lname]"; $csv_output .="\015\012"; } header("Content-type: application/vnd.ms-excel"); header("Content-disposition: csv; filename= Student_Data_". date("Y-m-d") . ".csv"); print $csv_output; exit; } include_once $path."includes/jobseeker_form.php"; ?> In the above we are getting html code along wtih id, firstname, lastname columns. we are unable to get the heading of the columns also How to remove Html code from xls file also need to get headers

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  • PHP & MySQL Pagination Update Help

    - by TaG
    Its been a while since I updated my pagination on my web page and I'm trying to add First and Last Links to my pagination as well as the ... when the search results are to long. For example I'm trying to achieve the following in the example below. Can some one help me fix my code so I can update my site. Thanks Previous First 1 2 3 4 5 6 7 ... 199 200 Last Next I currently have the following displayed using my code. Previous 1 2 3 4 5 6 7 Next Here is the part of my pagination code that displays the links. if ($pages > 1) { echo '<br /><p>'; $current_page = ($start/$display) + 1; if ($current_page != 1) { echo '<a href="index.php?s=' . ($start - $display) . '&p=' . $pages . '">Previous</a> '; } for ($i = 1; $i <= $pages; $i++) { if ($i != $current_page) { echo '<a href="index.php?s=' . (($display * ($i - 1))) . '&p=' . $pages . '">' . $i . '</a> '; } else { echo '<span>' . $i . '</span> '; } } if ($current_page != $pages) { echo '<a href="index.php?s=' . ($start + $display) . '&p=' . $pages . '">Next</a>'; } echo '</p>'; }

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  • mysql twitter/facebook like status feed

    - by barjonah
    Hi, I have two tables. One named status like this... user_id | status --------+----------- 1 | random status from user 1 2 | random status from user 2 3 | random message from user 3 4 | staus from user 4 1 | second status for user1 etc... and another named users_following like this... user_id | is_following --------+----------- 1 | 2 1 | 3 2 | 1 3 | 2 meaning that user 1 is following both users 2 and 3 etc... So, let's say I chose user 1. What is the best query (performance wise) to show the status updates of users that user 1 is following, in this case users 2 and 3 currently I have something like SELECT * from status WHERE user_id IN(SELECT is_following FROM users_following WHERE user_id='1') LIMIT 0,5 but I don't think this is good for performance if a user was following thousands+ of users

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  • mysql query to concat information from 3 tables - getting incorrect result count

    - by iPfaffy
    I have 3 tables in my database. ab_contacts id first_name last_name addressbook_id ab_addressbooks name id co_comments id link_id comment I'd like to create a query that will let me select all the contacts and comments related to them in a given addressbook. To select all the people in a given addressbook, I can use: select count(*) from ab_contacts where addressbook_id = '50'; This returns 8152 people. However, when I run my query: select ab_contacts.first_name, ab_contacts.last_name, ab_contacts.email, ab_addressbooks.name, co_comments.comments from ab_contacts JOIN ab_addressbooks ON (ab_contacts.addressbook_id = ab_addressbooks.id) JOIN co_comments ON (ab_contacts.id = co_comments.link_id) WHERE ab_contacts.addressbook_id = '50';` the format works, but I only get 1045 results. I'm sure there is something I am missing, but I cannot figure it out. Any help would be greatly appreciated.

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  • How to get this done in mysql?

    - by bala3569
    Consider i have a registartion table and there is field prefLocationId and it contains value like this 1,2,3,2,1,4 and so many.... And i have a table prefLocation which looks like this Id LocationName 1 Chennai 2 Mumbai 3 Kolkatta 4 Delhi and i want to select record of users and show values like Chennai,Mumbai,Kolkatta,Mumbai,Chennai,Delhi and so on...

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  • MySQL count statements error - operand should contain 1 column(s)

    - by jason
    I am trying to do multiple counts everyone was working accept the first sub select (list1) I get an error that reads "Operand should contain 1 column(s)" i'm guessing it has something to do with the AND, but i'm not sure how I would fix this one. Select Count(list0.ustatus) AS finished_count, (Select list1.ustatus, Count(*) From listofupdates list1 Where list1.ustarted != '0000-00-00 00:00:00' AND list1.ustatus != 1) AS start_count, (Select Count(list2.udifficulty) From listofupdates list2 Where list2.udifficulty = 2) AS recheck_count, (Select Count(list3.udifficulty) From listofupdates list3 Where list3.udifficulty = 4) AS question_count From listofupdates list0 Where list0.ustatus = 1

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  • MySQL customized join query using multiple tables

    - by itgeek
    I am searching one student from each class from one group. There are different class groups and every group has different classes and every class has multiple students. See below: Group1 --> Class1, Class2 etc Class1 --> GreenStudent1, GreenStudent2 etc Class2 --> RedStudent1, RedStudent2 etc ------------------------------------------------------ SELECT table1.id, table1.myname, table1.marks table2.studentid, table2.studentname FROM table1 INNER JOIN table3 ON table1.oldid = table3.id INNER JOIN table2 ON table2.studentid = table3.newid WHERE table1.classgroup = 'SCI79' GROUP BY table1.oldid ORDER BY table1.marks DESC There are different joins applied in the query. Above mentioned query giving me correct results but I need little modification in it. Current query returning me one student from each class. What I need? I need one student from each class but only that student who has MAXIMUM table1.marks So I should have one student from each class who has maximum number in their relevant classes. Can anyone suggest some solution or rewrite this query? Thanks :)

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  • Connect to web-service/API in MySQL?

    - by Jesse Figueroa
    I'm creating a sql based procedure which can Accept a table load the values one at a time send the variables to a remote API Record the response of the API Write the response to a table for viewing later I have successfully implemented 1,2, and 5. I am hoping there may be some way of choosing an address to contact and for SQL to listen too for a response. Please let me know if you have any suggestions!

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  • MySQL Query, Date Range From "Blacklist"

    - by erbaker
    I have 2 databases. One is properties and the other is dates. In dates I have associated the land_id and a date (In YYYYMMDD format) which means that the date is not available. I need to formulate a query that a user can specify a start and end date, and then choose a property for which dates are available (not in the date database). How do airline and hotel websites do this kind of logic? I was thinking about taking the date range and picking all days in between and doing a query where the dates do not match and ordering it by number of results, but I can see how that could easily turn into an intense query. CREATE TABLE IF NOT EXISTS `dates` ( `id` int(11) NOT NULL AUTO_INCREMENT, `land_id` int(11) NOT NULL, `date` varchar(255) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=44 ; -- -- Dumping data for table `dates` -- INSERT INTO `dates` (`id`, `land_id`, `date`) VALUES (43, 1, '20100526'), (39, 1, '20100522'), (40, 1, '20100523'), (41, 1, '20100521'), (42, 1, '20100525');

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  • how to do reverse fulltext search in MySQL?

    - by Shore
    By default it's like this: select * from main_table where match(col1,col2) against('search_item'); but what I want to fetch is the reverse, say,I've restored all the search_item(1000 records,for example), and I want to see which of them matches a specified row in main_table. Is that doable?

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  • MySQL Update query with left join and group by

    - by Rob
    I am trying to create an update query and making little progress in getting the right syntax. The following query is working: SELECT t.Index1, t.Index2, COUNT( m.EventType ) FROM Table t LEFT JOIN MEvents m ON (m.Index1 = t.Index1 AND m.Index2 = t.Index2 AND (m.EventType = 'A' OR m.EventType = 'B') ) WHERE (t.SpecialEventCount IS NULL) GROUP BY t.Index1, t.Index2 It creates a list of triplets Index1,Index2,EventCounts. It only does this for case where t.SpecialEventCount is NULL. The update query I am trying to write should set this SpecialEventCount to that count, i.e. COUNT(m.EventType) in the query above. This number could be 0 or any positive number (hence the left join). Index1 and Index2 together are unique in Table t and they are used to identify events in MEvent. How do I have to modify the select query to become an update query? I.e. something like UPDATE Table SET SpecialEventCount=COUNT(m.EventType)..... but I am confused what to put where and have failed with numerous different guesses.

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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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  • MySQL look for missing ratings

    - by Dustin
    Hopefully I can ask this without being confusing. I am a photographer and I am having some of our clients rate pictures that we have taken. We have hundreds of pictures in our portfolio that they may be rating. What I want to do is ask our clients to rate pictures again, but only show them the pictures they haven't yet rated. I currently have three tables: one that stores the actual ratings, one that stores the pictures (or location of each picture), and one that stores the information about the rater. I'm using codeigniter for my db management, if that helps at all. What I have so far is this: "SELECT * FROM ratings LEFT JOIN portfolio ON ratings.portfolioid = portfolio.portfolioid" This will give me a row for each rating, but won't show me where a rating is missing for a picture. Thanks in advance!

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  • php mysql error beginner

    - by Marcelo
    Hi, I'm trying to print some values on the screen from a table but I having a problem, I don't know much about string, vector and array but I think that my problem is related to them. I'm getting this on the screen Fatal error: Cannot use [] for reading ... My code $sql="SELECT * FROM $tbl_name"; $result=mysql_query($sql) or trigger_error(mysql_error().$sql); while($row = mysql_fetch_array($result)){ $DATA = $row[]; } //line with probelm mysql_close(); //html part <table> <? foreach($DATA as $row): ?> <tr> <td><?=$row['id']?></td> //more stuff </tr> <? endforeach ?> </table> What I'm trying to do is print somevalues form a database. But I'm getting this error. I'm sorry for any mistake in English, and thanks in advance for any help.

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  • My MYSQL & PHP while code is out of memory, when there is only one row

    - by Sam
    Hey all, why is this code throwing an out of memory error, when there is only 1 row in the database.. $request_db = mysql_query("SELECT * FROM requests WHERE haveplayed='0'") or die(mysql_error()); $request = mysql_fetch_array( $request_db ); echo "<table border=\"1\" align=\"center\">"; while ( $request['haveplayed'] == "0" ) { echo "<tr><td>"; echo $request['SongName']; echo "</td><td>"; echo "<tr><td>"; echo $request['Artist']; echo "</td><td>"; echo "<tr><td>"; echo $request['DedicatedTo']; echo "</td><td>"; } echo "</table>"; Thanks.

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