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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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  • Display only latest three results from PHP and MySQL

    - by nogggin1
    <?php $result = @mysql_query('SELECT Article FROM news WHERE ID = (SELECT MAX(ID) FROM News)'); if (!$result) { die('<p>Error performing query: ' . mysql_error() . '</p>'); } while ( $row = mysql_fetch_array($result) ) { echo('<p>' . $row['Article'] . '</p>'); } ?> basically i need to tweak this so that it shows the latest 3 results instead of just the latest one, also i need the order to be: newest 2nd newest 3rd newest any help would be greatly appreciated!

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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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  • 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 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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  • 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 select one field from table WHERE condition is in multiple rows

    - by Alex
    Tried to find the answer, but still couldn't.. The table is as follows: id, keyword, value 1 display 15.6 1 harddrive 320 1 ram 3 So what i need is something like this.. Select an id from this table where (keyword="display" and value="15.6") AND (keyword="harddrive" and value="320") There's also a possibility that there will be 3 or 4 such keyword conditions which should result into returning one id (one row) It seems there's something to deal with UNION but i didn't use it before so i can't figure it out Thanks in advance

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  • Find all those columns which have only null values, in a MySQL table

    - by Robin v. G.
    The situation is as follows: I have a substantial number of tables, with each a substantial number of columns. I need to deal with this old and to-be-deprecated database for a new system, and I'm looking for a way to eliminate all columns that have - apparently - never been in use. I wanna do this by filtering out all columns that have a value on any given row, leaving me with a set of columns where the value is NULL in all rows. Of course I could manually sort every column descending, but that'd take too long as I'm dealing with loads of tables and columns. I estimate it to be 400 tables with up to 50 (!) columns per table. Is there any way I can get this information from the information_schema? EDIT: Here's an example: column_a column_b column_c column_d NULL NULL NULL 1 NULL 1 NULL 1 NULL 1 NULL NULL NULL NULL NULL NULL The output should be 'column_a' and 'column_c', for being the only columns without any filled in values.

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  • mysql update enums

    - by FFish
    I have a field with enums: 'preview','active','closed' When I query like this: $query = "UPDATE albums SET album_active = preview WHERE album_id = 3"; $result = mysql_query($query); if (!$result) die('Invalid query: ' . mysql_error()); I get : Invalid query: Unknown column 'preview' in 'field list

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  • Why isn't this simple MySQL statement working?

    - by Clark
    I am trying to match a user inputted search term against two tables: posts and galleries. The problem is the union all clause isn't working. Is there something wrong with my code? $query = mysql_query(" SELECT * FROM posts WHERE title LIKE '%$searchTerm%' OR author LIKE '%$searchTerm%' OR location LIKE '%$searchTerm%' OR excerpt LIKE '%$searchTerm%' OR content LIKE '%$searchTerm%' UNION ALL SELECT * FROM galleries WHERE title LIKE '%$searchTerm%' ");

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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 php problem: no error message despite error_reporting(E_ALL) line

    - by herrturtur
    index.php <html> <head> <title>Josh's Online Playground</title> </head> <body> <form method="POST" action="action.php"> <table> <tr> <td>"data for stuff"</td> <td><input type="text" ?></td> </tr> <tr> <td><input type="submit"></td> </tr> </table> </form> </body> </html> action.php <?php error_reporting(E_ALL); ini_sit("display_errors", 1); $mysqli = new mysqli('localhost', 'root', 'password', 'website'); $result = $mysqli->query("insert into stuff (data) values (' .$_POST['data'] ."'); echo $mysqli->error(); if($result = $mysqli->query("select data from stuff")){ echo 'There are '.$result->num_rows.' results.'; } while ($row = $result->fetch_object()){ echo 'stuff' . $row->data; } ?> Despite the first two lines in action.php, I get no error or warning messages. Instead I get a blank page after clicking the submit button. Do I have to do something differently to insert data?

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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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  • 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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  • php array into mysql

    - by mckenzie
    Hello, $sql_where = ''; $exclude = '30,35,36,122,123,124,125'; if($exclude != '') { $exclude_forums = explode(',', $exclude); foreach ($exclude_forums as $id) { if ($id > 0) { $sql_where = ' AND forum_id <> ' . trim($id); } } } $sql = 'SELECT topic_title, forum_id, topic_id, topic_type, topic_last_poster_id, topic_last_poster_name, topic_last_poster_colour, topic_last_post_time FROM ' . TOPICS_TABLE . ' WHERE topic_status <> 2 AND topic_approved = 1 ' . $sql_where . ' ORDER BY topic_time DESC'; The above code i use to exclude the id of forum to be displayed on sql queries. Why doesn't it work and still display it? Any solution

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  • MySQL foreign key constraint disappearing

    - by Bramjam
    This is my table: /* oefenreeks leerplan */ CREATE TABLE leerplan_oefenreeks ( leerplan_oefenreeks_id INT PRIMARY KEY AUTO_INCREMENT NOT NULL, leerplan_id INT NOT NULL, oefenreeks_id INT NOT NULL, plaats INT NOT NULL ); /* fk */ ALTER TABLE leerplan_oefenreeks ADD CONSTRAINT fk_leerp_oefenr_leerplan FOREIGN KEY(leerplan_id) REFERENCES leerplan (leerplan_id) ON DELETE CASCADE; ALTER TABLE leerplan_oefenreeks ADD CONSTRAINT fk_leerp_oefenr_oefenreeks FOREIGN KEY(oefenreeks_id) REFERENCES oefenreeks (oefenreeks_id) ON DELETE CASCADE; /* when I execute the nexline, my fk_leerp_oefenr_leerplan constraint vanishes/disappears*/ ALTER TABLE leerplan_oefenreeks ADD CONSTRAINT un_leerp_oefenr UNIQUE(leerplan_id, oefenreeks_id); ALTER TABLE leerplan_oefenreeks ADD CONSTRAINT un_leerp_oefenr_plaats UNIQUE(leerplan_id, plaats); When I go and check only 3 constraints exist. fk_leerp_oefenr_leerplan disappears. I don't understand why this happens.

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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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  • multiple-to-one relationship mysql, submissions

    - by Yulia
    Hello, I have the following problem. Basically I have a form with an option to submit up to 3 images. Right now, after each submission it creates 3 records for album table and 3 records for images. I need it to be one record for album and 3 for images, plus to link images to the album. I hope it all makes sense... Here is my structure. TABLE `albums` ( `id` int(11) NOT NULL auto_increment, `title` varchar(50) NOT NULL, `fullname` varchar(40) NOT NULL, `email` varchar(100) NOT NULL, `created_at` datetime NOT NULL, `theme_id` int(11) NOT NULL, `description` int(11) NOT NULL, `vote_cache` int(11) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=20 ; TABLE `images` ( `id` int(11) NOT NULL auto_increment, `album_id` int(11) NOT NULL, `name` varchar(30) NOT NULL, and my code function create_album($params) { db_connect(); $query = sprintf("INSERT INTO albums set albums.title = '%s', albums.email = '%s', albums.discuss_url = '%s', albums.theme_id = '%s', albums.fullname = '%s', albums.description = '%s', created_at = NOW()", mysql_real_escape_string($params['title']), mysql_real_escape_string($params['email']), mysql_real_escape_string($params['theme_id']), mysql_real_escape_string($params['fullname']), mysql_real_escape_string($params['description']) ); $result = mysql_query($query); if(!$result) { return false; } $album_id = mysql_insert_id(); return $album_id; } if(!is_uploaded_file($_FILES['userfile']['tmp_name'][$i])) { $warning = 'No file uploaded'; } elseif is_valid_file_size($_FILES['userfile']['size'][$i])) { $_POST['album']['theme_id'] = $theme['id']; create_album($_POST['album']); mysql_query("INSERT INTO images(name) VALUES('$newName')"); copy($_FILES['userfile']['tmp_name'][$i], './photos/'.$original_dir.'/' .$newName.'.jpg');

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  • MySQL LEFT JOIN, GROUP BY and ORDER BY not working as required

    - by Simon
    I have a table 'products' => ('product_id', 'name', 'description') and a table 'product_price' => ('product_price_id', 'product_id', 'price', 'date_updated') I want to perform a query something like SELECT `p`.*, `pp`.`price` FROM `products` `p` LEFT JOIN `product_price` `pp` ON `pp`.`product_id` = `p`.`product_id` GROUP BY `p`.`product_id` ORDER BY `pp`.`date_updated` DESC As you can probably guess the price changes often and I need to pull out the latest one. The trouble is I cannot work out how to order the LEFT JOINed table. I tried using some of the GROUP BY functions like MAX() but that would only pull out the column not the row. 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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