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• PHP - MySQL - Select runs indefinitely

  - by John

  I have three tables listings: id, pid, beds, baths, etc, etc, etc, db locations: id, pid, zip, lat, lon, etc, etc, etc, db images id, pid, height, width, raw, etc, etc, db id, pid & db are indexed. db just references the mls provider a particular item came from. in images the raw column holds raw image data there are about 15k rows in listings/locations, and about 120k rows in images so there are multiple rows that have the same pid. when i do "select pid from listings" or "select pid from locations" the query completes successfully in about 100ms. when i do "select pid from images" it just hangs in sqlyog and never completes... i was thinking since the raw column contains alot of information that it might be trying to select that too, but my query doesn't try to select that so I can't imagine why it's taking so long... any idea why this is happening??

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• mysql join 3 tables and count

  - by air

  Please look at this image here is 3 tables , and out i want is uid from table1 industry from table 3 of same uid count of fid from table 2 of same uid like in the sample example output will be 2 records Thanks

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• mysql joining three specific tables

  - by sam lim

  Here what i would like to pull date from this three table. Table users i have three columns uid, username , data(text) Table users_order i have three columns uid, orders_id , users_email Table order_products i have three columns orders_id, product_id, product_name I would like to use product_id as the ref/search to pull the user info from those three tables. If product_id = 5 The query will display uid; username; users_email; orders_id; product_name; data (text) how would i right the sql query for this situation. Thanks,

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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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• Average of a Sum in Mysql query

  - by chupeman

  I am having some problems creating a query that gives me the average of a sum. I read a few examples here in stackoverflow and couldn't do it. Can anyone help me to understand how to do this please? This is the data I have: Basically I need the average transaction value by cashier. I can't run a basic avg because it will take all rows but each transaction can have multiple rows. At the end I want to have: Cashier| Average| 131 | 44.31 |(Which comes from the sum divided by 3 transactions not 5 rows) 130 | 33.15 | etc. This is the query I have to SUM the transactions but don't know how or where to include the AVG function. SELECT `products`.`Transaction_x0020_Number`, Sum(`products`.`Sales_x0020_Value`) AS `SUM of Sales_x0020_Value`, `products`.`Cashier` FROM `products` GROUP BY `products`.`Transaction_x0020_Number`, `products`.`Date`, `products`.`Cashier` HAVING (`products`.`Date` ={d'2010-06-04'}) Any help is appreciated.

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• Mysql query question

  - by brux

  I have 2 tables: Customer: customerid - int, pri-key,auto fname - varchar sname -varchar housenum - varchar street -varchar Items: itemid - int,pri-key,auto type - varchar collectiondate - date releasedate - date customerid - int I need a query which will get me all items that have a releasedate 3 days prior to (and including) the current date. i.e The query should return customerid,fname,sname,street,housenum,type,releasedate for all items which have releasedate within (and including)3 days prior today thanks in advance

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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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• Casting a Calculated Column in a MySQL view.

  - by Chris Brent

  I have a view that contains a calculated column. Is there are a way to cast it as a CHAR or VARCHAR rather than a VARBINARY ? Obviously, I have tried using CAST(... as CHAR) but it gives an error. Here is a simple replicable example. CREATE VIEW view_example AS SELECT concat_ws('_', lpad(9, 3,'0'), lpad(1,3,'0'), date_format(now(),'%Y%m%d%H%i%S')) AS calculated_field_id; This is how my view is created: describe view_example; +---------------------+---------------+------+-----+---------+-------+ | Field | Type | Null | Key | Default | Extra | +---------------------+---------------+------+-----+---------+-------+ | calculated_field_id | varbinary(27) | YES | | NULL | | +---------------------+---------------+------+-----+---------+-------+ select version(); +-----------------------+ | version() | +-----------------------+ | 5.0.51a-community-log | +-----------------------+

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• counting sub rows in mysql

  - by moustafa

  i have 2 table ok catgories and artilces i have this structure catgories web > design > photoshop > layers web > design > photoshop > effects and each one is a catgory and layers catgories has 100 article and effects catgories has 50 article now i want when count the articles 'web' catgory it show 150 article how i can do this give me an example

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• mySQL query not returning correct results!

  - by Pete Herbert Penito

  Hi! This query that I have is returning therapists whose 'therapistTable.activated' is equal to false as well as those set to true! so it's basically selecting all of the db, any advice would be appreciated! ` $query = "SELECT therapistTable.* FROM therapistTable WHERE therapistTable.activated = 'true' ORDER BY therapistTable.distance "; `

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• MySQL: optimization of table (indexing, foreign key) with no primary keys

  - by Haradzieniec

  Each member has 0 or more orders. Each order contains at least 1 item. memberid - varchar, not integer - that's OK (please do not mention that's not very good, I can't change it). So, thera 3 tables: members, orders and order_items. Orders and order_items are below: CREATE TABLE `orders` ( `orderid` INT(11) UNSIGNED NOT NULL AUTO_INCREMENT, `memberid` VARCHAR( 20 ), `Time` TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP , `info` VARCHAR( 3200 ) NULL , PRIMARY KEY (orderid) , FOREIGN KEY (memberid) REFERENCES members(memberid) ) ENGINE = InnoDB; CREATE TABLE `order_items` ( `orderid` INT(11) UNSIGNED NOT NULL, `item_number_in_cart` tinyint(1) NOT NULL , --- 5 items in cart= 5 rows `price` DECIMAL (6,2) NOT NULL, FOREIGN KEY (orderid) REFERENCES orders(orderid) ) ENGINE = InnoDB; So, order_items table looks like: orderid - item_number_in_cart - price: ... 1000456 - 1 - 24.99 1000456 - 2 - 39.99 1000456 - 3 - 4.99 1000456 - 4 - 17.97 1000457 - 1 - 20.00 1000458 - 1 - 99.99 1000459 - 1 - 2.99 1000459 - 2 - 69.99 1000460 - 1 - 4.99 ... As you see, order_items table has no primary keys (and I think there is no sense to create an auto_increment id for this table, because once we want to extract data, we always extract it as WHERE orderid='1000456' order by item_number_in_card asc - the whole block, id woudn't be helpful in queries). Once data is inserted into order_items, it's not UPDATEd, just SELECTed. The questions are: I think it's a good idea to put index on item_number_in_cart. Could anybody please confirm that? Is there anything else I have to do with order_items to increase the performance, or that looks pretty good? I could miss something because I'm a newbie. Thank you in advance.

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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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• effective counter for unique number of visits in PHP & MySQL

  - by Adnan

  Hello, I am creating a counter for unique number of visits on a post, so what I have until now is a table for storing data like this; cvp_post_id | cvp_ip | cvp_user_id In cases a registered user visits a post, for the first time a record is inserted with cpv_post_id and cvp_user_id, so for his next visit I query the table and if the record is available I do not count him as a new visitor. In cases of an anonymous user the same happens but now the cvp_ip and cpv_post_id are used. My concerns is that I do a query every time anyone visits a post for checking if there has been a visit, what would be a more effective way for doing this?

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• How to output items in Order in MySQL?

  - by RailsRor

  I have a column called "menu_order" which has no default value. When I select the contents of this column using the following select statement: SELECT * FROM categories ORDER BY menu_order ASC It lists the category items that have nothing as their menu order first and then the one's that have 1's and 2's and 3's. Is there any way to prevent SQL from take nothing to come before numbers when I am trying to list things in order? So for example, if I have: cat_name | menu_order ---------------------- Lunch | 1 Dinner | And I perform my query, the output should be: Lunch Dinner Not: Dinner Lunch

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• Optimize SELECT DISTINCT CONCAT query in MySQL

  - by L. Cosio

  Hello! I'm running this query: SELECT DISTINCT CONCAT(ALFA_CLAVE, FECHA_NACI) FROM listado GROUP BY ALFA_CLAVE HAVING count(CONCAT(ALFA_CLAVE, FECHA_NACI)) > 1 Is there any way to optimize it? Queries are taking 2-3 hours on a table with 850,000 rows. Adding an index to ALFA_CLAVE and FECHA_NACI would work? Thanks in advanced

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• MySql Alter Syntax error with mulitple FK

  - by acidzombie24

  If i do the first one i have no problem. When i do addition i get a syntax error. What is wrong with the syntax? The error says syntax error near [entire 2nd line] alter table `ban_Status` add FOREIGN KEY (`banned_user`) REFERENCES `user_data`(`id`) alter table `ban_Status` add FOREIGN KEY (`banned_user`) REFERENCES `user_data`(`id`), FOREIGN KEY (`banning_user`) REFERENCES `user_data`(`id`), FOREIGN KEY (`unban_user`) REFERENCES `user_data`(`id`)

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• Retrieving specific tuples using Mysql

  - by Narayanan

  Hi, I have some problems retrieving specific tuples. I am actually a student trying to build a Room management system. I have two tables: Room(roomID,hotelname,rate) and Reservation(resID,arriveDate,departDate,roomID). I am not sure how to retrieve the rooms that are available between 2 specific dates. This was the query that i used. SELECT Room.roomID,hotelname,rate FROM Room LEFT JOIN Reservation on ( Room.roomID=Reservation.resID and arriveDate >='2010-02-16' and departDate <='2010-02-20' ) GROUP BY roomID,hotelname,rate HAVING count(*)=0;' but it returns an empty set. Can any1 be kind enough to tell me what mistake i am doing??

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• How can I update a field in a MySQL database table by addition in MySQL database in a single query

  - by undefined

  I have a table that stores a value that will be added to over time. When I want to add to the value I would like to do so in a single query rather than - Get oldValue from database newValue = oldValue + X update row with newValue $query1 = "SELECT value FROM table WHERE id = thisID"; $result1 = mysql_query($query1); while($row=mysql_fetch_array($result)) { $oldValue = $row['value']; } $newValue = $oldValue + x $query1 = "UPDATE table SET value = $newValue WHERE id = thisID"; Can this be done in a single query?

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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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• mysql rollback when auto commit is on

  - by devang

  hi, by mistake i fired a update query and all the records in the table in dat field got updated nd also cannot rollback as auto_commit is on..is der ny other way in which i can retreive the records.plz help its urgent.. Thanks in advance

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• MySQL - How do I insert an additional where clause into this full-text search (updated)

  - by Steven

  I want to add a WHERE clause to a full text search query (to limit to past 24 hours), but wherever I insert it I get Low Level Error. Is it possible to add the clause and if so, how? Here is the code WITHOUT the where clause: $query = "SELECT *, MATCH (story_title) AGAINST ('$query' IN BOOLEAN MODE) AS Relevance FROM stories WHERE MATCH (story_title) AGAINST ('+$query' IN BOOLEAN MODE) HAVING Relevance > 0.2 ORDER BY Relevance DESC, story_time DESC;

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• MySQL Queries using Doctrine & CodeIgniter

  - by 01010011

  Hi, How do I write plane SQL queries using Doctrine connection object and display the results? For example, how do I perform: SELECT * FROM table_name WHERE column_name LIKE '%anything_similar_to_this%'; using Doctrine something like this (this example does not work) $search_key = 'search_for_this'; $conn = Doctrine_Manager::connection(); $conn->execute('SELECT * FROM table_name WHERE column_name LIKE ?)', $search_key); echo $conn;

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• MySql insert data in multiple table at a time

  - by Sanjay

  hi, i have a query regarding to insert data in multiple table.. i have a two tables. one is item table and second is field table. i want to insert data in both table with one query at a time. any idea about it.

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• Mysql Query is not working, why?

  - by Furkan Kadioglu

  Im using this example: www.jtable.org Now i download to jtable php version. And i edited script. Jtable simple version is working but my edition is doesnt working. I can do list, but i cant do add a row; this codes having problem. But php doesnt giving any error. else if($_GET["action"] == "create") { //Insert record into database $result = mysql_query("INSERT INTO veriler(bolge, sehir, firma, adres, tel, web) VALUES('" . $_POST["bolge"] . "', '" . $_POST["sehir"] . "', '" . $_POST["firma"] . "', '" . $_POST["adres"] . "', '" . $_POST["tel"] . "', '" . $_POST["web"] . "'"); //Get last inserted record (to return to jTable) $result = mysql_query("SELECT * FROM veriler WHERE id = LAST_INSERT_ID();"); $row = mysql_fetch_array($result); //Return result to jTable $jTableResult = array(); $jTableResult['Result'] = "OK"; $jTableResult['Record'] = $row; print json_encode($jTableResult); } if you doesnt understand, please ask me question. Where is the problem?

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• How to insert form data into MySQL database table

  - by Richard

  So I have this registration script: The HTML: <form action="register.php" method="POST"> <label>Username:</label> <input type="text" name="username" /><br /> <label>Password:</label> <input type="text" name="password" /><br /> <label>Gender:</label> <select name="gender"> <optgroup label="genderset"> <option value="Male">Male</option> <option value="Female">Female</option> <option value="Hermaphrodite">Hermaphrodite</option> <option value="Not Sure!!!">Not Sure!!!</option> </optgroup> </select><br /> <input type="submit" value="Register" /> </form> The PHP/SQL: <?php $username = $_POST['username']; $password = $_POST['password']; $gender = $_POST['gender']; mysql_query("INSERT INTO registration_info (username, password, gender) VALUES ('$username', '$password', '$gender') ") ?> The problem is, the username and password gets inserted into the "registration_info" table just fine. But the Gender input from the select drop down menu doesn't. Can some one tell me how to fix this, thanks.

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