Search Results

Search found 63877 results on 2556 pages for 'mysql error 1452'.

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 212/2556 | < Previous Page | 208 209 210 211 212 213 214 215 216 217 218 219  | Next Page >

  • Best way to limit results in MySQL with user subcategories

    - by JM4
    I am trying to essentially solve for the following: 1) Find all users in the system who ONLY have programID 1. 2) Find all users in the system who have programID 1 AND any other active program. My tables structures (in very simple terms are as follows): users userID | Name ================ 1 | John Smith 2 | Lewis Black 3 | Mickey Mantle 4 | Babe Ruth 5 | Tommy Bahama plans ID | userID | plan | status --------------------------- 1 | 1 | 1 | 1 2 | 1 | 2 | 1 3 | 1 | 3 | 1 4 | 2 | 1 | 1 5 | 2 | 3 | 1 6 | 3 | 1 | 0 7 | 3 | 2 | 1 8 | 3 | 3 | 1 9 | 3 | 4 | 1 10 | 4 | 2 | 1 11 | 4 | 4 | 1 12 | 5 | 1 | 1 I know I can easily find all members with a specific plan with something like the following: SELECT * FROM users a JOIN plans b ON (a.userID = b.userID) WHERE b.plan = 1 AND b.status = 1 but this will only tell me which users have an 'active' plan 1. How can I tell who ONLY has plan 1 (in this case only userID 5) and how to tell who has plan 1 AND any other active plan? Update: This is not to get a count, I will actually need the original member information, including all the plans they have so a COUNT(*) response may not be what I'm trying to achieve.

    Read the article

  • MySQL: Select pages that are not tagged?

    - by lauthiamkok
    Hi, I have a db with two tables like these below, page table pg_id title 1 a 2 b 3 c 4 d tagged table tagged_id pg_id 1 1 2 4 I want to select the pages which are tagged, I tried with this query below but doesn't work, SELECT * FROM root_pages LEFT JOIN root_tagged ON ( root_tagged.pg_id = root_pages.pg_id ) WHERE root_pages.pg_id != root_tagged.pg_id It returns zero - Showing rows 0 - 1 (2 total, Query took 0.0021 sec) But I want it to return pg_id title 2 b 3 c My query must have been wrong?? How can I return the pages which are not tagged correctly? Thanks.

    Read the article

  • An array of MySQL results...

    - by Michael Falk
    What am I doing wrong here? I am attempting to return a json object and I can't seem to get past the array... I've built hundreds of regular array and returned them as a json object but I am having a hard time wrapping my head around this one. $rows = array(); $post_array = array(); $i = 0; $result = mysql_query(" SELECT * FROM forum_posts WHERE permalink = '$permalink' AND LOWER(raw_text) LIKE '%$str%' " ); while($row = mysql_fetch_assoc($result)) { $post_array[$i] = $rows[ "id" => htmlentities($row["id"]), "post_content" => htmlentities($row["content"]), "author" => $row["author"], "last_updated" => $row["last_updated"], "author_id" => $row["author_id"], "editing_author" => $row["editing_author"], "date" => $outputQuoteDate ]; $i++; }

    Read the article

  • Entering Content Into A MySQL Database Via A Form

    - by ThatMacLad
    I've been working on creating a form that submits content into my database but I decided that rather than using a drop down menu to select the date I'd rather use a textfield. I was wondering what changes I will need to make to my table creation file. <?php mysql_connect ('localhost', 'root', 'root') ; mysql_select_db ('tmlblog'); $sql = "CREATE TABLE php_blog ( id int(20) NOT NULL auto_increment, timestamp int(20) NOT NULL, title varchar(255) NOT NULL, entry longtext NOT NULL, PRIMARY KEY (id) )"; $result = mysql_query($sql) or print ("Can't create the table 'php_blog' in the database.<br />" . $sql . "<br />" . mysql_error()); mysql_close(); if ($result != false) { echo "Table 'php_blog' was successfully created."; } ?> It's the timestamp that I need to edit to enter in via a textfield. The Title and Entry are currently being entered via that method anyway.

    Read the article

  • mysql update too slow when joining multiple tables

    - by user293487
    Hi, I have two tables. they looks like as follows: Id (int) Tags char(128). the column Tags in table A does not have value. It is empty. The column Tags in table B has value. What I want to copy the Tags in table B to corresponding place of table A. the mapping is based on Id. My sql query is: update A INNER JOIN B set A.Tags = B.Tags where A.Id = B.Id There are about 2,000,000 rows in table A, and 50,000 rows in table B. The update seems very slow.... Could anyone tell me how to make it run faster?

    Read the article

  • Generate MySQL data dump in SQL from PHP

    - by Álvaro G. Vicario
    I'm writing a PHP script to generate SQL dumps from my database for version control purposes. It already dumps the data structure by means of running the appropriate SHOW CREATE .... query. Now I want to dump data itself but I'm unsure about the best method. My requirements are: I need a record per row Rows must be sorted by primary key SQL must be valid and exact no matter the data type (integers, strings, binary data...) Dumps should be identical when data has not changed I can detect and run mysqldump as external command but that adds an extra system requirement and I need to parse the output in order to remove headers and footers with dump information I don't need (such as server version or dump date). I'd love to keep my script as simple as I can so it can be hold in an standalone file. What are my alternatives?

    Read the article

  • Best database (mysql) structure for this case:

    - by robert
    we have three types of data (tables): Book (id,name,author...) ( about 3 million of rows) Category (id,name) ( about 2000 rows) Location (id,name) ( about 10000 rows) A Book must have at least 1 type of Category (up to 3) AND a Book must have only one Location. I need to correlate this data to get this query faster: Select Books where Category = 'cat_id' AND Location = 'loc_id' Select Books where match(name) against ('name of book') AND Location = 'loc_id' Please I need some help. Thanks

    Read the article

  • 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');

    Read the article

  • 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.

    Read the article

  • Help with PHP MySQL join

    - by kester martinez
    Please help me to understand proper join syntax. I have table named inventory which has: trans_id trans_items items -> item_id trans_user employees -> person_id trans_date trans_comment trans_inventory As you can see above, trans_items is a foreign key in items table, and trans_user is a foreign key in employees table. Now what I want to do is to display in HTML the inventory table, but instead of displaying the item id, I want the ITEM NAME to be displayed. Here is what I have done. Please note I'm using CodeIgniter. public function getData(array $inputs) { $this->db->select('trans_items, trans_user, trans_date, trans_inventory, trans_comment'); $this->db->from('inventory'); $this->db->order_by('trans_date desc'); return $this->db->get()->result_array(); }

    Read the article

  • MySQL - Extracting Time in correct format

    - by mouthpiec
    Hi, I have a table with a coloumn of type "time", and the values in this coloumn are stored as follows: 20:45:00, 18:00:00, etc. Now when displaying the result, I am not getting the minutes, but just 00. I am using the following to get the time: SELECT TIME_FORMAT(time, '%h:%m') as time FROM ......... etc

    Read the article

  • MySQL SELECT WHERE returning empty with long numbers, although they are there

    - by brybam
    Alright, so basically the most simple query ever... I've done this a million times... SELECT * FROM purchased_items WHERE uid = '$uid' if $uid == 123 It works fine and returns all data in rows where uid is 123 if $uid == 351565051447743 It returns empty... I'm positive 351565051447743 is a possible uid in some rows, i literally copied and pasted it into the table. $uid is a string, and is being passed as a string. This is something i've done a million times, and i've never had this simple query not work. Any ideas why this is not working?

    Read the article

  • 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,

    Read the article

  • Export a MYSQL column to a plain txt file with no headings

    - by Kohl Sharples
    So what I'm trying to do is write a script or CRON job (Linux- CentOS) to export the usernames listed in my wordpress database to a simple .txt file with just on username per line. So with the picture, I would like the .txt file to read like this: Sir_Fluffulus NunjaX007 (Except with all the username in the user_login column.) See screenshot at: http://i.stack.imgur.com/lZQai.png I have found how to export the entire table to a CVS file, but that contains about 10+ fields (Columns) that I DO NOT what to show up in this text file. Can anyone point me in the right direction on how to do this? If it helps, this is going to be for exporting users that have signed up on our website (Wordpress) to a whitelist.txt file for Minecraft. Thanks!

    Read the article

  • [mysql] having a number in

    - by Wiika
    Hi all, ID Name Hidden 1 Mika 1,4,2 2 Loca 0 3 Nosta 4 4 Like 2 can someone give me a query that will return as a result rows ID 1 & 3 something like this SELECT * FROM table WHERE Hidden HAVING(4) Thanks, I Appreciate

    Read the article

  • Removing groups of similar records in MySQL query

    - by user1182155
    I'm trying to wrap my head around this... (it may be simple, been a long day!) I have a database with sometimes multiple similar records... ie. Apples 2008-09-03 Apples 2012-01-01 Apples 2013-10-24 Oranges 2012-01-04 What I need to do is do a query that will show only records that haven't been updated today... So in this case, since Apples has an entry that was updated today, none of the records for the Apples should appear in the results. Oranges should be the only record it returns. I have a query similar to this... SELECT fruit FROM fruitnames where date < CURDATE() Which works to remove the record that was updated today... But it keeps the other records for Apples (obviously)... How would I remove those results as well?

    Read the article

  • 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%' ");

    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 | 208 209 210 211 212 213 214 215 216 217 218 219  | Next Page >