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  • Help needed to construct a SQL query

    - by song202y
    Need your help to get the list of suggested friends (who aren't friends of the current user but are friends of 2 or more of the current user's friends). The primary ordering should put people at the same school at the top, and the secondary ordering should put people with more common friends (that is, the number of people who are friends of that person and the current user) near the top. Users: user_id PK, user_name Profiles: user_id PK, school_name, ... Friendships: id PK, user_id FK, friend_id FK Thank you in advance. Joe

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  • MySQL join problem

    - by snaken
    Whats wrong with this SQL? It should return results but returns nothing SELECT `pid` FROM `products` LEFT JOIN `prods_to_features` ON (`ptf_pid` = `pid`) WHERE (`ptf_id` = '66' OR `ptf_id` = '67') AND (`ptf_id` = '76') Is it not possible to have the 2nd where clause for the table that has been used in the left join?

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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 Update Statement + File Upload

    - by Jason Sweet
    Greetings! Been staring at this all day and can't seem to figure out why my update statement fails to update the field 'image_filename': $fileName = $_FILES['image_filename']; if($fileName["name"] <> ""){ $imageFile = $fileName['name']; $destination = "../../../../assets/resources/images/".$fileName['name']; move_uploaded_file($fileName['name'], $destination); } $updateSQL = sprintf("UPDATE content SET image_filename='$imageFile' WHERE id=%s", GetSQLValueString($_POST['resource_id'], "int")); mysql_select_db($database_conn_talent, $conn_talent); $Result1 = mysql_query($updateSQL, $conn_talent) or die(mysql_error()); Can a SQL pro tell me what I"m missing? Much thanks in advance for your feedback!

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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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  • 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: Use CASE/ELSE value as join parameter

    - by DRJ
    I'm trying to join the NAME and PHOTO from USERS table to the TRANSACTIONS table based on who is the payer or payee. It keeps telling me can't find the table this -- What am I doing wrong? SELECT name,photo,amount,comment, ( CASE payer_id WHEN 72823 THEN payee_id ELSE payer_id END ) AS this FROM transactions RIGHT JOIN users ON (users.id=this) WHERE payee_id=72823 OR payer_id=72823

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  • MySql returning multiple rows from stored procedure / function

    - by pistacchio
    Hi, I need to make a stored procedure or function that returns a set of rows. I've noted that in a stored procedure i can SELECT * FROM table with success. If i fetch rows in a loop and SELECT something, something_other FROM table once per loop execution, I only get one single result. What I need to do is looping, doing some calculations and returning a rowset. What's the best way to do this? A temporary table? Stored functions? Any help appreciated.

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  • % confuses python raw sql query

    - by Jonathan
    Following this SO question, I'm trying to "truncate" all tables related to a certain django application using the following raw sql commands in python: cursor.execute("set foreign_key_checks = 0") cursor.execute("select concat('truncate table ',table_schema,'.',table_name,';') as sql_stmt from information_schema.tables where table_schema = 'my_db' and table_type = 'base table' AND table_name LIKE 'some_prefix%'") for sql in [sql[0] for sql in cursor.fetchall()]: cursor.execute(sql) cursor.execute("set foreign_key_checks = 1") Alas I receive the following error: C:\dev\my_project>my_script.py Traceback (most recent call last): File "C:\dev\my_project\my_script.py", line 295, in <module> cursor.execute(r"select concat('truncate table ',table_schema,'.',table_name,';') as sql_stmt from information_schema.tables where table_schema = 'my_db' and table_type = 'base table' AND table_name LIKE 'some_prefix%'") File "C:\Python26\lib\site-packages\django\db\backends\util.py", line 18, in execute sql = self.db.ops.last_executed_query(self.cursor, sql, params) File "C:\Python26\lib\site-packages\django\db\backends\__init__.py", line 216, in last_executed_query return smart_unicode(sql) % u_params TypeError: not enough arguments for format string Is the % in the LIKE making trouble? How can I workaround it?

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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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  • Foreign keys and NULL in mySQL

    - by Industrial
    Hi everyone, Can I have a column in my values table (value) referenced as a foreign key to knownValues table, and let it be NULL whenever needed, like in the example: Table: values product type value freevalue 0 1 NULL 100 1 2 NULL 25 3 3 1 NULL Table: types id name prefix 0 length cm 1 weight kg 2 fruit NULL Table: knownValues id Type name 0 2 banana Note: The types in the table values & knownValues are of course referenced into the types table. Thanks!

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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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  • MYSQL updating data from other table

    - by atif089
    Hi, I have two tables like of this structure content (content_id, content_type, user_id, time, comment_count) comments (comment_id, content_id, userid, comment, comment_time) What I wold like to do is update the comments_count field with sum of comments i.e COUNT(content_id) from the comments table. I am not able to figure out the right syntax Thanks

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  • Quality questionnaire php mysql graphipcs

    - by Marcelo
    Hi, i'm making a questionnaire about a service quality, its contains the options (poor, regular, good, very good). It's contains 6 questions (radio button) and a suggestion box (textbox). In the table of the database i created 6 rows for questions, 1 for suggestion and 1 for date (a friend of mine tole me to use this but i didn't get why). q1) I'm going to atribute a value form 1 to 4 to the radio buttons options, and i'd like to sum every answer for each question, and then divide by the numbers of user that answered that question and give the mean. how am i supposed to to that? I'd also like generate reports of the month, of the year. q2) not only about the questionnaire but for registration too. I need all the fields to be completed, no blank options, if he don't complete all of fields it'll not be submitted and there will be a warning message to the user. q3) about the field type, i'd like it to be the same class that is in the database, i'm having a "problem". Ex: Name(varchar) : 1234(int), in the field 'name' of the table of the database 1234 will be shown as name, and i don't want this, i want only the type that i declared in the construction of the table. q4) i'd also like to know if it's possible to create pizza graphics, about the percentage of each question, is this possible? q5) I'm using phpmyadmin and some of my id's are auto_increment, but 'cause of my tests they at a high number, i'd like to restart to 0 the ids number, is this possible? Thanks for the attention.

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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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  • If inside Where mysql

    - by Barno
    Can I do an if inside Where? or something that allows me to do the checks only if the field is not null (path=null) SELECT IF(path IS NOT NULL, concat("/uploads/attachments/",path, "/thumbnails/" , nome), "/uploads/attachments/default/thumbnails/avatar.png") as avatar_mittente FROM prof_foto   WHERE profilo_id = 15  -- only if path != "/uploads/attachments/default/thumbnails/avatar.png" AND foto_eliminata = 0 AND foto_profilo = 1

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  • Seeking assistance with Escaping Data for MySQL queries

    - by JM4
    Please don't send me a link to php.net referencing mysql_real_escape_string as the only response. I have read through the page and while I understand the general concepts, I am having some trouble based on how my INSERT statement is currently built. Today, I am using the following: $sql = "INSERT INTO tablename VALUES ('', '$_SESSION['Member1FirstName'], '$_SESSION['Member1LastName'], '$_SESSION['Member1ID'], '$_SESSION['Member2FirstName'], '$_SESSION['Member2LastName'], '$_SESSION['Member2ID'] .... and the list goes on for 20+ members with some other values entered. It seems most people in the examples already have all their data stored in an array. On my site, I accept form inputs, action="" is set to self, php validation takes place and if validation passes, data is stored into SESSION variables on page 2 then redirected to the next page in the process (page 3) (approximately 8-10 pages in the whole process). thanks!

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