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  • Need help with many-to-many relationships....

    - by yuudachi
    I have a student and faculty table. The primary key for student is studendID (SID) and faculty's primary key is facultyID, naturally. Student has an advisor column and a requested advisor column, which are foreign key to faculty. That's simple enough, right? However, now I have to throw in dates. I want to be able to view who their advisor was for a certain quarter (such as 2009 Winter) and who they had requested. The result will be a table like this: Year | Term | SID | Current | Requested ------------------------------------------------ 2009 | Winter | 860123456 | 1 | NULL 2009 | Winter | 860445566 | 3 | NULL 2009 | Winter | 860369147 | 5 | 1 And then if I feel like it, I could also go ahead and view a different year and a different term. I am not sure how these new table(s) will look like. Will there be a year table with three columns that are Fall, Spring and Winter? And what will the Fall, Spring, Winter table have? I am new to the art of tables, so this is baffling me... Also, I feel I should clarify how the site works so far now. Admin can approve student requests, and what happens is that the student's current advisor gets overwritten with their request. However, I think I should not do that anymore, right?

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  • Enhancing an 'ORDER BY' clause to judge condition by more than 1 integer

    - by Yvonne
    Hi folks, I have some PHP code which allows me to sort a column into ascending and descending order (upon click of a table row title), which is good. It works perfectly for my D.O.B colum (with date/time field type), but not for a quantity column. For example, I have quantites of 10, 50, 100, 30 and another 100. The order seems to be only appreciating the 1st integer, so my sorting of the column ends up in this order: 10, 100, 100, 30, 50... and 50, 30, 100, 100, 10. This is obviously incorrect as 100 is bigger than 50, therefore both 100 values should appear at the end surely? It seems to me that 100 is only being taken into account as having the '1' value, then it appears before 10 because the system recognises it has another 0. Is this normal to happen? Is there any way I can solve this problem? Thanks for any help. P.S. I can show code if necessary, but would like to know if this is a common issue by default.

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  • 500 internal server error at form connection

    - by klox
    hi..all..i've a problem i can't connect to database what's wrong with my code?this is my code: $("#mod").change(function() { var barcode; barCode=$("#mod").val(); var data=barCode.split(" "); $("#mod").val(data[0]); $("#seri").val(data[1]); var str=data[0]; var matches=str.match(/(EE|[EJU]).*(D)/i); $.ajax({ type:"post", url:"process1.php", data:"value="+matches+"action=tunermatches", cache:false, async:false, success: function(res){ $('#rslt').replaceWith( "<div id='value'><h6>Tuner range is" + res + " .</h6></div>" ); } }); }); and this is my process file: switch(postVar('action')) { case 'tunermatches' : tunermatches(postVar('tuner')); break; function tunermatches($tuner)){ $Tuner=mysql_real_escape_string($tuner); $sql= "SELECT remark FROM settingdata WHERE itemname="Tuner_range" AND itemdata="$Tunermatches"; $res=mysql_query($sql); $dat=mysql_fetch_array($res,MYSQL_NUM); if($dat[0]>0) { echo $dat[0]; } mysql_close($dbc); }

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  • Accessing data entered into multiple Django forms and generating them onto a new URL

    - by pedjk
    I have a projects page where users can start up new projects. Each project has two forms. The two forms are: class ProjectForm(forms.Form): Title = forms.CharField(max_length=100, widget=_hfill) class SsdForm(forms.Form): Status = forms.ModelChoiceField(queryset=P.ProjectStatus.objects.all()) With their respective models as follows: class Project(DeleteFlagModel): Title = models.CharField(max_length=100) class Ssd(models.Model): Status = models.ForeignKey(ProjectStatus) Now when a user fills out these two forms, the data is saved into the database. What I want to do is access this data and generate it onto a new URL. So I want to get the "Title" and the "Status" from these two forms and then show them on a new page for that one project. I don't want the "Title" and "Status" from all the projects to show up, just for one project at a time. If this makes sense, how would I do this? I'm very new to Django and Python (though I've read the Django tutorials) so I need as much help as possible. Thanks in advance Edit: The ProjectStatus code is (under models): class ProjectStatus(models.Model): Name = models.CharField(max_length=30) def __unicode__(self): return self.Name

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  • getting sql records

    - by droidus
    when i run this code, it returns the topic fine... $query = mysql_query("SELECT topic FROM question WHERE id = '$id'"); if(mysql_num_rows($query) > 0) { $row = mysql_fetch_array($query) or die(mysql_error()); $topic = $row['topic']; } but when I change it to this, it doesn't run at all. why is this happening? $query = mysql_query("SELECT topic, lock FROM question WHERE id = '$id'"); if(mysql_num_rows($query) > 0) { $row = mysql_fetch_array($query) or die(mysql_error()); $topic = $row['topic']; $lockedThread = $row['lock']; echo "here: " . $lockedThread; }

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  • Wordpress inserting comments via wp_insert_comment()

    - by Cyber Junkie
    Hello all happy holidays! :) I'm trying to insert comments in my wordpress blog via the wp_insert_comment() function. It's for a plugin I'm trying to make. I have this code in my header for testing. It works every time I refresh the page. $agent = $_SERVER['HTTP_USER_AGENT']; $data = array( 'comment_post_ID' => 256, 'comment_author' => 'Dave', 'comment_author_email' => '[email protected]', 'comment_author_url' => 'http://www.someiste.com', 'comment_content' => 'Lorem ipsum dolor sit amet...', 'comment_author_IP' => '127.3.1.1', 'comment_agent' => $agent, 'comment_date' => date('Y-m-d H:i:s'), 'comment_date_gmt' => date('Y-m-d H:i:s'), 'comment_approved' => 1, ); $comment_id = wp_insert_comment($data); It successfully inserts comments into the database. The problem: Comments don't show via the Disqus comment system. I compared table rows and I noticed that user_agent differs. Normal comments use for example, Mozilla/5.0 (Windows; U; Windows NT 6.1; en-US; rv... and Disqus comments use Disqus/1.1(2.61):119598902 numbers are different for each comment. Does anyone know how to insert comments with wp_insert_comment() when Disqus is enabled?

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  • need help on my query.

    - by Dharmendra
    i have one table : nobel(yr, subject, winner) and i have this query : In which years was the Physics prize awarded but no Chemistry prize. this is what i tried : select distinct yr from nobel where subject='physics' and subject!='chemistry' but is not working where i am going wrong. see, i am not here to make my homework from someone. i am here to learn something. so, please give me suggetion.

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  • Define keys in temporary table creation

    - by imperium2335
    How do I define the keys for a temporary table that is being created from a SELECT statement? I have: CREATE temporary TABLE _temp_unique_parts_trading engine=memory AS (SELECT parts_trading.enquiryref, sellingcurrency, jobs.id AS jobID FROM parts_trading, jobs WHERE jobs.enquiryref = parts_trading.enquiryref GROUP BY parts_trading.enquiryref) But where do I define the keys?

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  • question with its query

    - by user329820
    Hi this is my homework and the question is this: List the average balance of customers by city and short zip code (the first five digits of thezip code). Only include customers residing in Washington State (‘WA’). also the Customer table has 5 columns(Name,Family,CustZip,CustCity,CustAVGBal) I wrote the query like below is this correct? SELECT CustCity,LEFT(CustZip,5) AS NewCustZip,CustAVGBal FROM Customer WHERE CustCity = 'WA' THANKS!!

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  • Normalise this Table?

    - by Abs
    Hello all, I am creating a social bookmarking app. I am having a re-thought of the DB design in the middle of development. Should I normalise the bookmarks table and remove the tag columns that I have into a separate table. I have 10 tags per bookmark and therefore 10 columns per record (per bookmark). It seems to me that breaking the table into two would just mean I would have to do a join but the way I currently have it, its a straight select - but the table doesn't feel right...? Thanks all

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  • How to handle customers with multiple addresses in CakePHP

    - by Ryan
    I'm putting together a system to track customer orders. Each order will have three addresses; a Main contact address, a billing address and a shipping address. I do not want to have columns in my orders table for the three addresses, I'd like to reference them from a separate table and have some way to enumerate the entry so I can determine if the addressing is main, shipping or billing. Does it make sense to create a column in the address table for AddressType and enumerate that or create another table - AddressTypes - that defines the address enumeration and link to that table? I have found other questions that touch on this topic and that is where I've taken my model. The problem I'm having is taking that into the cakePHP convention. I've been struggling to internalize the direction OneToMany relationships are formed - the way the documentation states feels backwards to me. Any help would be appreciated, Thanks!

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  • Unknown Column?

    - by Kenny
    ok im trying to get mutual friends between these Two users, user1 and user92 This is the sql that is successful in displaying them SELECT IF(user_a = 1 OR user_a = 92, user_b, user_a) friend FROM friendship WHERE (user_a = 1 OR user_a = 92) OR (user_b = 1 OR user_b = 92) GROUP BY 1 HAVING COUNT(*) > 1 THis is how it looks friend 61 72 73 74 75 76 77 78 79 80 81 So now i want to select all users after the number 72, and i try to do it with this sql but its not working? It gives me the error, "unknown coulum name friend in where clause" SELECT IF(user_a = 1 OR user_a = 92, user_b, user_a) friend FROM friendship WHERE friend > 72 and (user_a = 1 OR user_a = 92) OR (user_b = 1 OR user_b = 92) GROUP BY 1 HAVING COUNT(*) > 1 what am i doing wrong? or what is the correct way?? thx

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  • SQL: Find the max record per group

    - by user319088
    I have one table, which has three fields and data. Name , Top , Total cat , 1 , 10 dog , 2 , 7 cat , 3 , 20 horse , 4 , 4 cat , 5 , 10 dog , 6 , 9 I want to select the record which has highest value of Total for each Name, so my result should be like this: Name , Top , Total cat , 3 , 20 horse , 4 , 4 Dog , 6 , 9 I tried group by name order by total, but it give top most record of group by result. Can anyone guide me, please?

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  • Select statement that combines similar rows with certain ids?

    - by vegatron
    hi I have a warehouse_products table which defines how many products in the warehouses so lets say I have 20 records/rows in the table, some rows may contain the same product id but in a different warehouse I need to create select statement that give every product one row, and in this row I must have the quantity in warehouse A and warehouse B .. so in the end I will get for example 10 rows that contain all the data

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  • Resetting AUTO_INCREMENT on myISAM without rebuilding the table

    - by Artem
    Please help I am in major trouble with our production database. I had accidentally inserted a key with a very large value into an autoincrement column, and now I can't seem to change this value without a huge rebuild time. "ALTER TABLE tracks_copy AUTO_INCREMENT = 661482981" Is super-slow. How can I fix this in production? I can't get this to work either (has no effect): myisamchk tracks.MYI --set-auto-increment=661482982 Any ideas? Basically, no matter what I do I get an overflow: SHOW CREATE TABLE tracks CREATE TABLE tracks ( ... ) ENGINE=MYISAM AUTO_INCREMENT=2147483648 DEFAULT CHARSET=latin1

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  • How to compare filename of uploaded file and string

    - by user225269
    I use this code to upload image files in xammp server: <?php if ((($_FILES["file"]["type"] == "image/gif") || ($_FILES["file"]["type"] == "image/jpeg") || ($_FILES["file"]["type"] == "image/pjpeg")) && ($_FILES["file"]["size"] < 100000)) { if ($_FILES["file"]["error"] > 0) { echo "Return Code: " . $_FILES["file"]["error"] . "<br />"; } else { echo "Upload: " . $_FILES["file"]["name"] . "<br />"; echo "Type: " . $_FILES["file"]["type"] . "<br />"; echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />"; echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />"; if (file_exists("upload/" . $_FILES["file"]["name"])) { echo $_FILES["file"]["name"] . " already exists. "; } else { move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $_FILES["file"]["name"]); echo "Stored in: " . "upload/" . $_FILES["file"]["name"]; } } } else { echo "Invalid file, File must be less than 100Kb in size with .jpg, .jpeg, or .gif file extension"; } ?> What do I do to compare the file name of the uploaded files with the text inputted by the user? My goal is to be able to compare the user input(ID number) and the file name of the image file which should also be an ID number. So that I will be able to display the image that corresponds with the ID Number provided. What do I need to do?Please give me an idea on how can I achieve this. Thanks

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  • Zend_Table_Db and Zend_Paginator num rows

    - by Uffo
    I have the following query: $this->select() ->where("`name` LIKE ?",'%'.mysql_escape_string($name).'%') Now I have the Zend_Paginator code: $paginator = new Zend_Paginator( // $d is an instance of Zend_Db_Select new Zend_Paginator_Adapter_DbSelect($d) ); $paginator->getAdapter()->setRowCount(200); $paginator->setItemCountPerPage(15) ->setPageRange(10) ->setCurrentPageNumber($pag); $this->view->data = $paginator; As you see I'm passing the data to the view using $this->view->data = $paginator Before I didn't had $paginator->getAdapter()->setRowCount(200);I could determinate If I have any data or not, what I mean with data, if the query has some results, so If the query has some results I show the to the user, if not, I need to show them a message(No results!) But in this moment I don't know how can I determinate this, since count($paginator) doesn't work anymore because of $paginator->getAdapter()->setRowCount(200);and I'm using this because it taks about 7 sec for Zend_Paginator to count the page numbers. So how can I find If my query has any results?

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  • how to validate username and password in vb6?

    - by srikanth
    i have created a database in mysql5.0. i want to display the data from it. it has table named login. it has 2 columns username and password. in form i have 2 text fields username and password i just want to validate input with database values and display message box. connection from vb to database is established successfully. but its not validating input. its giving error as 'object required'. please any body help i'm new to vb. i'm using vb6 and mysql5.0 thank you

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  • User has many computers, computers have many attributes in different tables, best way to JOIN?

    - by krismeld
    I have a table for users: USERS: ID | NAME | ---------------- 1 | JOHN | 2 | STEVE | a table for computers: COMPUTERS: ID | USER_ID | ------------------ 13 | 1 | 14 | 1 | a table for processors: PROCESSORS: ID | NAME | --------------------------- 27 | PROCESSOR TYPE 1 | 28 | PROCESSOR TYPE 2 | and a table for harddrives: HARDDRIVES: ID | NAME | ---------------------------| 35 | HARDDRIVE TYPE 25 | 36 | HARDDRIVE TYPE 90 | Each computer can have many attributes from the different attributes tables (processors, harddrives etc), so I have intersection tables like this, to link the attributes to the computers: COMPUTER_PROCESSORS: C_ID | P_ID | --------------| 13 | 27 | 13 | 28 | 14 | 27 | COMPUTER_HARDDRIVES: C_ID | H_ID | --------------| 13 | 35 | So user JOHN, with id 1 owns computer 13 and 14. Computer 13 has processor 27 and 28, and computer 13 has harddrive 35. Computer 14 has processor 27 and no harddrive. Given a user's id, I would like to retrieve a list of that user's computers with each computers attributes. I have figured out a query that gives me a somewhat of a result: SELECT computers.id, processors.id AS p_id, processors.name AS p_name, harddrives.id AS h_id, harddrives.name AS h_name, FROM computers JOIN computer_processors ON (computer_processors.c_id = computers.id) JOIN processors ON (processors.id = computer_processors.p_id) JOIN computer_harddrives ON (computer_harddrives.c_id = computers.id) JOIN harddrives ON (harddrives.id = computer_harddrives.h_id) WHERE computers.user_id = 1 Result: ID | P_ID | P_NAME | H_ID | H_NAME | ----------------------------------------------------------- 13 | 27 | PROCESSOR TYPE 1 | 35 | HARDDRIVE TYPE 25 | 13 | 28 | PROCESSOR TYPE 2 | 35 | HARDDRIVE TYPE 25 | But this has several problems... Computer 14 doesnt show up, because it has no harddrive. Can I somehow make an OUTER JOIN to make sure that all computers show up, even if there a some attributes they don't have? Computer 13 shows up twice, with the same harddrive listet for both. When more attributes are added to a computer (like 3 blocks of ram), the number of rows returned for that computer gets pretty big, and it makes it had to sort the result out in application code. Can I somehow make a query, that groups the two returned rows together? Or a query that returns NULL in the h_name column in the second row, so that all values returned are unique? EDIT: What I would like to return is something like this: ID | P_ID | P_NAME | H_ID | H_NAME | ----------------------------------------------------------- 13 | 27 | PROCESSOR TYPE 1 | 35 | HARDDRIVE TYPE 25 | 13 | 28 | PROCESSOR TYPE 2 | 35 | NULL | 14 | 27 | PROCESSOR TYPE 1 | NULL | NULL | Or whatever result that make it easy to turn it into an array like this [13] => [P_NAME] => [0] => PROCESSOR TYPE 1 [1] => PROCESSOR TYPE 2 [H_NAME] => [0] => HARDDRIVE TYPE 25 [14] => [P_NAME] => [0] => PROCESSOR TYPE 1

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  • How to retrieve column total when rows are paginated?

    - by Rick
    Hey guys I have a column "price" in a table and I used a pagination script to generate the data displayed. Now the pagination is working perfectly however I am trying to have a final row in my HTML table to show the total of all the price. So I wrote a script to do just that with a foreach loop and it sort of works where it does give me the total of all the price summed up together however it is the sum of all the rows, even the ones that are on following pages. How can I retrieve just the sum of the rows displayed within the pagination? Thank you! Here is the query.. SELECT purchase_log.id, purchase_log.date_purchased, purchase_log.total_cost, purchase_log.payment_status, cart_contents.product_name, members.first_name, members.last_name, members.email FROM purchase_log LEFT JOIN cart_contents ON purchase_log.id = cart_contents.purchase_id LEFT JOIN members ON purchase_log.member_id = members.id GROUP BY id ORDER BY id DESC LIMIT 0,30";

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