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  • Combine SQL statement

    - by ninumedia
    I have 3 tables (follows, postings, users) follows has 2 fields - profile_id , following_id postings has 3 fields - post_id, profile_id, content users has 3 fields - profile_id, first_name, last_name I have a follows.profile_id value of 1 that I want to match against. When I run the SQL statement below I get the 1st step in obtaining the correct data. However, I now want to match the postings.profile_id of this resulting set against the users table so each of the names (first and last name) are displayed as well for all the listed postings. Thank you for your help! :) Ex: SELECT * FROM follows JOIN postings ON follows.following_id = postings.profile_id WHERE follows.profile_id = 1

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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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  • How to add condition on multiple-join table

    - by Jean-Philippe
    Hi, I have those two tables: client: id (int) #PK name (varchar) client_category: id (int) #PK client_id (int) category (int) Let's say I have those datas: client: {(1, "JP"), (2, "Simon")} client_category: {(1, 1, 1), (2, 1, 2), (3, 1, 3), (4,2,2)} tl;dr client #1 has category 1, 2, 3 and client #2 has only category 2 I am trying to build a query that would allow me to search multiple categories. For example, I would like to search every clients that has at least category 1 and 2 (would return client #1). How can I achieve that? Thanks!

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  • How to set a filter for an estimated maximum price

    - by David
    I cannot figure out how to set an estimated maximum price for a collection of records. What I want to avoid is to simply use SQL MAX, because maybe there are records with exorbitant prices. For example, in the "computers-hardware" category of OLX (http://www.olx.com/computers-hardware-cat-240) the filter for maximum price is estimately set to $1400, but sorting by price, the first items are above $10000 Maybe they calculated the average and then estimated some maximum price... what do you think? And what about the stepping? How would you calculate it?

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  • Printing All Entries in A PHP Table

    - by mgunawan
    I'm trying to insert a php excerpt with SQL (I understand this is outdated, but am trying to grasp the syntax first) into my HTML page, and I've got the following table: ID Name Element1 Element2 0 John John's 1st John's 2nd 1 Bill Bill's 1st Bill's 2nd 2 Steve Steven's 1st Steve's 2nd I'm trying to get the for loop that will essentially print out the following in my html page Name: Name where ID=0 Element1: Element1 where ID=0 Element2: Element2 where ID=0 Name: Name where ID=1 Element1: Element1 where ID=1 Element2: Element2 where ID=1 and so forth. Basically, I am trying to make this process automated so that whenever a new record is added into the table, the HTML page will automatically update with a new "profile". Thank you for your help!

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  • PHP Multiple User Login Form - Navigation to Different Pages Based on Login Credentials

    - by Zulu Irminger
    I am trying to create a login page that will send the user to a different index.php page based on their login credentials. For example, should a user with the "IT Technician" role log in, they will be sent to "index.php", and if a user with the "Student" role log in, they will be sent to the "student/index.php" page. I can't see what's wrong with my code, but it's not working... I'm getting the "wrong login credentials" message every time I press the login button. My code for the user login page is here: <?php session_start(); if (isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'IT Technician')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'Student')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/student/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> And the form that the data is pulled from is shown here: <form id="LoginForm" name="LoginForm" method="post" action="http://www.zuluirminger.com/SchoolAdmin/user_login.php"> User Name:<br /> <input type="text" name="username" id="username" size="50" /><br /> <br /> Password:<br /> <input type="password" name="password" id="password" size="50" /><br /> <br /> Log in as: <select name="role" id="role"> <option value="">...</option> <option value="Head">Head</option> <option value="Deputy Head">Deputy Head</option> <option value="IT Technician">IT Technician</option> <option value="Pastoral Care">Pastoral Care</option> <option value="Bursar">Bursar</option> <option value="Secretary">Secretary</option> <option value="Housemaster">Housemaster</option> <option value="Teacher">Teacher</option> <option value="Tutor">Tutor</option> <option value="Sanatorium Staff">Sanatorium Staff</option> <option value="Kitchen Staff">Kitchen Staff</option> <option value="Parent">Parent</option> <option value="Student">Student</option> </select><br /> <br /> <input type="submit" name = "button" id="button" value="Log In" onclick="javascript:return validateLoginForm();" /> </h3> </form> Once logged in (and should the correct page be loaded, the validation code I have at the top of the script looks like this: <?php session_start(); if (!isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/user_login.php"); exit(); } $managerID = preg_replace('#[^0-9]#i', '', $_SESSION["id"]); $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["manager"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if ($existCount == 0) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> Just so you're aware, the database table has the following fields: id, username, password and role. Any help would be greatly appreciated! Many thanks, Zulu

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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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  • 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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  • Keeping some choices in the Table for the Field of Type Dropdown

    - by Mercy
    Hi, i am having a Table named Attributes which has id form_id label size sequence_no Type 1 1 Name 200 1 Text 2 1 Age 150 2 Number 3 1 Address 300 3 Textarea 4 1 Gender 200 4 Dropdown I am having the doubt how can i keep the Choices of the Field of type "Dropdown" in the Table Eg. For Gender the choices will Male , Female.. Please give me the suggestions...

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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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  • I can't insert data into my database

    - by Ken
    I don't know why, but my data doesn't go into my database 'users' with the table 'data'. <html> <body> <?php date_default_timezone_set("America/Los_Angeles"); include("mainmenu.php"); $con = mysql_connect("localhost", "root", "g00dfor@boy"); if(!$con){ die(mysql_error()); } $usrname = $_POST['usrname']; $fname = $_POST['fname']; $lname = $_POST['lname']; $password = $_POST['password']; $email = $_POST['email']; mysql_select_db(`users`, $con) or die(mysql_error()); mysql_query(INSERT INTO `users`.`data` (`id`, `usrname`, `fname`, `lname`, `email`, `password`) VALUES (NULL, '$usrname', '$fname', '$lname', '$email', 'password')) or die(mysql_error()); mysql_close($con) echo("Thank you for registering!"); ?> </body> </html> All i get is a blank page.

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  • 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

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  • PHP MySQLi isn't letting me alter a table (adding a new column)

    - by asdasd
    Well thats pretty much it. This is my query: $query = 'ALTER TABLE permissions ADD '.$name.' INT NOT NULL DEFAULT \'0\''; Where $name is already checked to exist with only lower case alpha letters, and not more than 20 length. Im just starting this out with very simple names. The next 4 lines of code after that one are: if($stmt = $db -> prepare($query)) { $success = $stmt -> execute(); $stmt -> close(); if(!$success) echo 'ERROR: Unsuccessful query: ',$db->error,PHP_EOL; } And I get back, every time ERROR: Unsuccessful query: And no error message. Is there a way to get more error messages so I can see what is failing? I can add new columns through phpmyadmin, but that really doesnt help me at all. The $db is fine, i do lots of stuff before and after this one section. It is only adding new column to the table that fails. side question: prepare() rejected my query every time when i tried to make those 2 variables, the $name and the 0 value as ? ? prepared statement values. Thats why they are in the real query and not bound later. If i could change that too I would like that.

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  • Passing dynamic string in hyperlink as parameter in jsp

    - by user3660263
    I am trying to pass a dynamic string builder variable in jsp I am generating a string through code. String Builder variable has some value but i am not able to pass it in at run time.It doesn't get the value. CODE FOR VARIABLE <% StringBuilder sb=new StringBuilder(""); if(request.getAttribute("Brand")!=null) { String Brand[]=(String[])request.getAttribute("Brand"); for(String brand:Brand) { sb.append("Brand="); sb.append(brand); sb.append("&&"); } } if(request.getAttribute("Flavour")!=null) { String Flavour[]=(String[])request.getAttribute("Flavour"); for(String flavour:Flavour) { sb.append(flavour); sb.append("&&"); } sb.trimToSize(); pageContext.setAttribute("sb", sb); } out.print("this is string"+sb); %> CODE FOR HYPERLINK <a href="Filter_Products?${sb}page=${currentPage + 1}" style="color: white;text-decoration: none;">Next</a></td>

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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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  • 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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  • 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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  • 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

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  • Database Design for One to One relationships

    - by Greelmo
    I'm trying to finalize my design of the data model for my project, and am having difficulty figuring out which way to go with it. I have a table of users, and an undetermined number of attributes that apply to that user. The attributes are in almost every case optional, so null values are allowed. Each of these attributes are one to one for the user. Should I put them on the same table, and keep adding columns when attributes are added (making the user table quite wide), or should I put each attribute on a separate table with a foreign key to the user table. I have decided against using the EAV model. Thanks!

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