Search Results

Search found 17036 results on 682 pages for 'mysql administrator'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 371/682 | < Previous Page | 367 368 369 370 371 372 373 374 375 376 377 378  | Next Page >

  • 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?

    Read the article

  • 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

    Read the article

  • 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

    Read the article

  • SQL LEFT JOIN help

    - by Stolz
    My scenario: There are 3 tables for storing tv show information; season, episode and episode_translation. My data: There are 3 seasons, with 3 episodes each one, but there is only translation for one episode. My objetive: I want to get a list of all the seasons and episodes for a show. If there is a translation available in a specified language, show it, otherwise show null. My attempt to get serie 1 information in language 1: SELECT season_number AS season,number AS episode,name FROM season NATURAL JOIN episode NATURAL LEFT JOIN episode_trans WHERE id_serie=1 AND id_lang=1 ORDER BY season_number,number result: +--------+---------+--------------------------------+ | season | episode | name | +--------+---------+--------------------------------+ | 3 | 3 | Episode translated into lang 1 | +--------+---------+--------------------------------+ expected result +-----------------+--------------------------------+ | season | episode| name | +-----------------+--------------------------------+ | 1 | 1 | NULL | | 1 | 2 | NULL | | 1 | 3 | NULL | | 2 | 1 | NULL | | 2 | 2 | NULL | | 2 | 3 | NULL | | 3 | 1 | NULL | | 3 | 2 | NULL | | 3 | 3 | Episode translated into lang 1 | +--------+--------+--------------------------------+ Full DB dump http://pastebin.com/Y8yXNHrH

    Read the article

  • 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.

    Read the article

  • How to do left joins with least-n-per-group query?

    - by Nate
    I'm trying to get a somewhat complicated query working and am not having any luck whatsoever. Suppose I have the following tables: cart_items: +--------------------------------------------+ | item_id | cart_id | movie_name | quantity | +--------------------------------------------+ | 0 | 0 | braveheart | 4 | | 1 | 0 | braveheart | 9 | | . | . | . | . | | . | . | . | . | | . | . | . | . | | . | . | . | . | +--------------------------------------------+ movies: +------------------------------+ | movie_id | movie_name | ... | +------------------------------+ | 0 | braveheart | . | | . | . | . | | . | . | . | | . | . | . | | . | . | . | +------------------------------+ pricing: +-----------------------------------------+ | id | movie_name | quantity | price_per | +-----------------------------------------+ | 0 | braveheart | 1 | 1.99 | | 1 | braveheart | 2 | 1.50 | | 2 | braveheart | 4 | 1.25 | | 3 | braveheart | 8 | 1.00 | | . | . | . | . | | . | . | . | . | | . | . | . | . | | . | . | . | . | | . | . | . | . | +-----------------------------------------+ I need to join the data from the tables, but with the added complexity that I need to get appropriate price_per from the pricing table. Only one price should be returned for each cart_item, and that should be the lowest price from the pricing table where the quantity for the cart item is at least the quantity in the pricing table. So, the query should return for each item in cart_items the following: +---------------------------------------------+ | item_id | movie_name | quantity | price_per | +---------------------------------------------+ Example 1: Variable passed to the query: cart_id = 0. Return: +---------------------------------------------+ | item_id | movie_name | quantity | price_per | +---------------------------------------------+ | 0 | braveheart | 4 | 1.25 | | 1 | braveheart | 9 | 1.00 | +---------------------------------------------+ Note that this is a minimalist example and that additional data will be pulled from the tables mentioned (particularly the movies table). How could this query be composed? I have tried using left joins and subqueries, but the difficult part is getting the price and nothing I have tried has worked. Thanks for your help. EDIT: I think this is similar to what I have working with my "real" tables: SELECT t1.item_id, t2.movie_name, t1.quantity FROM cart_items t1 LEFT JOIN movies t2 ON t2.movie_name = t1.movie_name WHERE t1.cart_id = 0 Assuming I wrote that correctly (I quickly tried to "port over" my real query), then the output would currently be: +---------------------------------+ | item_id | movie_name | quantity | +---------------------------------+ | 0 | braveheart | 4 | | 1 | braveheart | 9 | +---------------------------------+ The trouble I'm having is joining the price at a certain quantity for a movie. I simply cannot figure out how to do it.

    Read the article

  • 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

    Read the article

  • Sql Query to get total rows and total rows matching specific condition

    - by mrNepal
    OK, Here is what my table looks like ------------------------------------------------ id type ----------------------------------------------- 1 a 2 b 3 a 4 c 5 c 7 a 8 a ------------------------------------------------ Now, I need a query that can give me this output... ----------------------------------------------------------------- count(*) | count(type=a) | count(type=b) | count(type=c) ----------------------------------------------------------------- 8 4 1 3 ------------------------------------------------------------------ I only know to get the total set using count(*), but how to do the remaining

    Read the article

  • 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

    Read the article

  • Concatinating Multiple Rows in SQL

    - by Dave C
    Hello, I have a table structure that looks like this: ID String ----------- 1 A 1 Test 1 String 2 Dear 2 Person I need the final output to look like this: ID FullString -------------------- 1 A, Test, String 2 Dear, Person I am really lost on how to approach this... I looked on a couple examples online but they seemed to be VERY complex... this seems like it should be a real easy problem to solve in sql. Thank you for all assistance!

    Read the article

  • node.js: looping, email each matching row from query but it it emails the same user by # number of matches?

    - by udonsoup16
    I have a node.js email server that works fine however I noticed a problem. If the query string found 4 rows, it would send four emails but only to the first result instead of to each found email address. var querystring1 = 'SELECT `date_created`,`first_name`,`last_name`,`email` FROM `users` WHERE TIMESTAMPDIFF(SECOND,date_created, NOW()) <=298 AND `email`!="" LIMIT 0,5'; connection.query(querystring1, function (err, row) { if (err) throw err; for (var i in row) { // Email content; change text to html to have html emails. row[i].column name will pull relevant database info var sendit = { to: row[i].email, from: '******@******', subject: 'Hi ' + row[i].first_name + ', Thanks for joining ******!', html: {path:__dirname + '/templates/welcome.html'} }; // Send emails transporter.sendMail(sendit,function(error,response){ if (error) { console.log(error); }else{ console.log("Message sent1: " + row[i].first_name);} transporter.close();} ) }}); .... How do I have it loop through each found row and send a custom email using row data individual row data?

    Read the article

  • 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?

    Read the article

  • Oracle access from iOS

    - by Michael Lowman
    I'm developing an iPad app that needs read-only access to an Oracle database. Is there any way to do this? As far as I can see, the only options are using OCI, which requires a prebuilt binary in the form of the instant client (and not built for ARM), or OJDBC drivers. Both of these seem to be out of the question. In my research I have discovered that libmysqlclient compiles for arm with minimal tuning. This is a stretch, but is there any possible way to use this to my advantage? I have seen this product providing odbc connectivity through the use of a Windows gateway machine using the ODBC client libraries, but this solution really isn't an option for me at the present time. Any ideas?

    Read the article

  • unknown column in where clause

    - by ranzy
    $result = mysql_query("SELECT * FROM Volunteers WHERE Volunteers.eventID = " . $var); $sql = mysql_query("SELECT * FROM Members WHERE Members.pid = " . $temp); I am also doing or die(mysql_error()) at the end of both statements if that matter. My problem is that the first statement executes perfectly but in that table I store an attribute called pid. So the second statement is supposed to take that and return the row where it equals that pid so I can get the name. I get an error that says unknown column in 'a2' in 'where clause' where a2 the pid attribute returned from the first statement. Thanks for any help!

    Read the article

  • COUNT issue across multiple tables

    - by Kim
    I am trying to count across 2 tables and I dont see whats wrong with my query yet I get a wrong result. User 2 does not exist in table_two, so the zero is correct. SELECT t1.creator_user_id, COUNT(t1.creator_user_id), COUNT(t2.user_id) FROM table_one AS t1 LEFT JOIN table_two AS t2 ON t2.user_id = t1.creator_user_id GROUP BY t1.creator_user_id, t2.user_id Actual result 1 192 192 2 9 0 Expected result 1 16 12 2 9 0 The result indicate a missing group by condition, but I already got both fields used. Where am I wrong ? Also, can I sum up all users that doesnt exist in table_two for t1 ? Like user 3 exists 21 times in t1, then the results would be: 1 16 12 (users with > 0 in t2 will need their own row) 2 30 0 (user 2=9 + user 3=21 => 30) Its okay for the user Id to be wrong for sum of t1 for all users with 0 in t2. If not possible, then I'll just do two queries.

    Read the article

  • 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

    Read the article

  • Correct Sql Script for Formula

    - by Madan Madan
    Can anyone help me write SQL script for the following formula? If DEP = 1 If DROP 1 PLV = 334.86 * exp(0.3541 * ACTIVE_DAYS) + 0.25 * DROP + 20 * DEP Else If DROP < 0 PLV = DROP + 70 * ACTIVE_DAYS Else PLV = 0.25 * DROP + 70 * ACTIVE_DAYS The SQL script which I have is the following SELECT IF(dep=1, if(dep=1, (334.86 * exp(0.3541 * act_days)) + (0.25 * 'drop') + (20 * dep), if('drop'<0, 'drop' + (70 * act_days), (0.25 * 'drop') + (70 * act_days))),'0') as PLV But the above query is not right as something is missing where the formula says Else PLV = 0.26 * DROP Thanks,

    Read the article

  • How add column order in careers table???

    - by Mahran Elneel
    this table i want to create and how assign job to first position??? job_id dynamic Jobs Title text Job Description text Order combo box to choose after what job or at first position in the website i create this table and cannot choose first job to view in my website

    Read the article

  • sql statement question. Need to query 3 tables in one go!

    - by Stefan
    Hey there, I have an sql database. In this database is 3 tables I need to query. The first table has all the item info called item and the other two tables has data for votes and comments called userComment and the third for votes called userItem I currently have a function which uses this sql query to get the latest more popular (in terms of both votes and comments): $sql = "SELECT itemID, COUNT(*) AS cnt FROM ( SELECT `itemID` FROM `userItem` WHERE FROM_UNIXTIME( `time` ) >= NOW() - INTERVAL 1 DAY UNION ALL SELECT `itemID` FROM `userComment` WHERE FROM_UNIXTIME( `time` ) >= NOW() - INTERVAL 1 DAY AND `itemID` > 0 ) q GROUP BY `itemID` ORDER BY cnt DESC"; I know how to change this for either by votes alone or comments.... HOWEVER - I need to query the database to only return the itemID's of the ones which have specific conditions in only the item table these are WHERE categoryID = 'xx' AND typeID = 'xx' If the sql ninja could please help me on this one? Do I have to first return the results from the above query and the for each in the array fetched then check each against the item table and see if it fits the conditions to build a new array - or is that overkill? Thanks, Stefan

    Read the article

  • storing image_id after uploading image in article table

    - by Bader
    According to this question i successed to create upload image , but now i need to store the image_id to another table called articles , i do not know if this is correct , but i tried to select the image_id from table image like this $select_image=mysql_query("select image_id from image where image_name = $fileName") or die(mysql_error()); and fetch the result to my article insert query like this $fetch=mysql_fetch_array($select_image); $qeuery=mysql_query("insert into articles (article_name,article_category,article_subcategory,article_body,article_summary,article_tags,article_photo,article_timedate) values ('$article_title','$CategoryID','$ProductID','$article_body','$article_summary','$fetch[image_id]','$time')") or die ('Error, Query Faild'.mysql_error()); is this correct ? the mysql_error keeps saying " Unknown column 'Penguins.jpg' in 'where clause'"

    Read the article

  • Having a Link Only Appear If a Logged-In User Appears on a Dynamic List

    - by John
    Hello, For the function below, I would like the link <div class="footervote"><a href="http://www...com/.../footervote.php">Vote</a></div> to only appear if the logged in user currently appears on editorlist.php. (I. e. if the loginid in the function corresponds to any of the usernames that currently appear in editorlist.php.) Appearing on editorlist.php is something that is dynamic. How can I do this? Thanks in advance, John function show_userbox() { // retrieve the session information $u = $_SESSION['username']; $uid = $_SESSION['loginid']; // display the user box echo '<div id="userbox"> <div class="username">'.$u.'</div> <div class="submit"><a href="http://www...com/.../submit.php">Submit an item.</a></div> <div class="changepassword"><a href="http://www...com/.../changepassword.php">Change Password</a></div> <div class="logout"><a href="http://www...com/.../logout.php">Logout</a></div> <div class="footervote"><a href="http://www...com/.../footervote.php">Vote</a></div> </div>'; } On editorlist.php: $sqlStr = "SELECT l.loginid, l.username, l.created, DATEDIFF(NOW(), l.created) AS days, COALESCE(s.total, 0) AS countSubmissions, COALESCE(c.total, 0) AS countComments, COALESCE(s.total, 0) * 10 + COALESCE(c.total, 0) AS totalScore, DATEDIFF(NOW(), l.created) + COALESCE(s.total, 0) * 10 + COALESCE(c.total, 0) AS totalScore2 FROM login l LEFT JOIN ( SELECT loginid, COUNT(1) AS total FROM submission GROUP BY loginid ) s ON l.loginid = s.loginid LEFT JOIN ( SELECT loginid, COUNT(1) AS total FROM comment GROUP BY loginid ) c ON l.loginid = c.loginid GROUP BY l.loginid ORDER BY totalScore2 DESC LIMIT 10"; $result = mysql_query($sqlStr); $arr = array(); echo "<table class=\"samplesrec1edit\">"; while ($row = mysql_fetch_array($result)) { echo '<tr>'; echo '<td class="sitename1edit1"><a href="http://www...com/.../members/index.php?profile='.$row["username"].'">'.stripslashes($row["username"]).'</a></td>'; echo '<td class="sitename1edit2">'.($row["countSubmissions"]).'</td>'; echo '<td class="sitename1edit2">'.($row["countComments"]).'</td>'; echo '<td class="sitename1edit2">'.($row["days"]).'</td>'; echo '<td class="sitename1edit2">'.($row["totalScore2"]).'</td>'; echo '</tr>'; } echo "</table>";

    Read the article

  • passing a scalar query result to coalesce

    - by Fakrudeen
    How can I pass the result from a scalar [single row, single value] query to coalesce? I am trying to pick the priority as (the biggest priority so far in the table) + 1. [0 if it is the first row.] create trigger priority_SuperRuleSamples before insert on SuperRuleSamples FOR EACH ROW SET NEW.Priority=coalesce(NEW.Priority, coalesce( select Priority from SuperRuleSamples order by Priority desc limit 1, -1 )+1 )

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 367 368 369 370 371 372 373 374 375 376 377 378  | Next Page >