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  • Efficient way to build a MySQL update query in Python

    - by ensnare
    I have a class variable called attributes which lists the instance variables I want to update in a database: attributes = ['id', 'first_name', 'last_name', 'name', 'name_url', 'email', 'password', 'password_salt', 'picture_id'] Each of the class attributes are updated upon instantiation. I would like to loop through each of the attributes and build a MySQL update query in the form of: UPDATE members SET id = self._id, first_name = self._first name ... Thanks.

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  • Recommended approach for error handling with PHP and MYSQL

    - by iama
    I am trying to capture database (MYSQL) errors in my PHP web application. Currently, I see that there are functions like mysqli_error(), mysqli_errno() for capturing the last occurred error. However, this still requires me to check for error occurrence using repeated if/else statements in my php code. You may check my code below to see what I mean. Is there a better approach to doing this? (or) Should I write my own code to raise exceptions and catch them in one single place? What is the recommended approach? Also, does PDO raise exceptions? Thanks. function db_userexists($name, $pwd, &$dbErr) { $bUserExists = false; $uid = 0; $dbErr = ''; $db = new mysqli(SERVER, USER, PASSWORD, DB); if (!mysqli_connect_errno()) { $query = "select uid from user where uname = ? and pwd = ?"; $stmt = $db->prepare($query); if ($stmt) { if ($stmt->bind_param("ss", $name, $pwd)) { if ($stmt->bind_result($uid)) { if ($stmt->execute()) { if ($stmt->fetch()) { if ($uid) $bUserExists = true; } } } } if (!$bUserExists) $dbErr = $db->error(); $stmt->close(); } if (!$bUserExists) $dbErr = $db->error(); $db->close(); } else { $dbErr = mysqli_connect_error(); } return $bUserExists; }

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  • mysql category tree search

    - by ffffff
    I have the following schema on MySQL 5.1 CREATE TABLE `mytest` ( `category` varchar(32) , `item_name` varchar(255) KEY `key1` (`category`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1; category column is filled with like that [:parent_parent_cat_id][:parent_cat_id][:leaf_cat_id] "10000200003000" if you can search all of the under categories :parent_parent_category_id SELECT * FROM mytest WHERE category LIKE "10000%"; it's using index key1; but How to use index when I wanna search :parent_cat_id? SELECT * FROM mytest WHERE category LIKE "%20000%"; Do you have a better solutions?

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  • MYSQL and Array with PHP for create Tag Cloud

    - by asilloo
    Hi, I'm trying to make a Tag cloud for every user in own page, I'm using PHP5 and Mysql, My table named "tags" and I want to make a array but in short way. The table like below, The array can be like for user1 array={[car,1],[cat,null],[pen,1],[dvd,1],[cd,null]} Username totaltag tag1 tag2 tag3 tag4 tag5 admin 5 car cat pen dvd cd user1 1 1 1 user2 1 2 12 1 user3 3 2 10 1

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  • Php + mysql transactions examples

    - by Donator
    I really haven't found normal example of php file where mysql transactions are being used. Can you show me simple example of that? And one more question. I've already created a lot of programming and didn't use transaction, maybe I can put any php function or smth to header.php that if one mysql_query fails, then others too? Thank you.

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  • mysql get last auto increment value

    - by Mick
    Hi I have a field in mySql table called jobnumber which auto increments with each new entry. What I want to do is when a user views my form, get the value of the next 'jobnumber' . But I want to do this before any entry is made to the table. IE so when a user looks at a form it will display something like 'this is job number 6954' I have tried $rr = mysql_insert_id() but this only work after I have made an entry Can anyone help please thanks

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  • Unique text field in MySQL and php

    - by Jonathan
    I've created a salt using; md5(rand(0,10000000)); (there is probably a better way?) There doesn't seem to be possible to make a text field unique in MYSQL. So how do I check if the salt has already been used for a previous user? Or should I generate the salt based on the current date/time? as it is impossible for 2 users to register at exactly the same time correct?

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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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  • easy mysql query question

    - by Ahmet vardar
    Hi, here is the "msg" table on mysql sent_to customer msg read ------- -------- ------ ----- 45 3 bla 0 34 4 bla 1 34 6 bla 0 45 3 bla 0 56 7 bla 1 45 8 bla 0 for example user whose id number is 45 logs in, i want him to see this, you have 2 unread msg to your "number 3" customer you have 1 unread msg to your "number 8" customer like a news feed what query should i use for this ? thx

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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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  • ORDERING A MYSQL QUERY WITH JOINS AND GROUPS

    - by Oliver
    I have this mysql query: SELECT * FROM Customer c JOIN eHRDemographic ehD ON ehD.CxID = c.CustomerID JOIN CustPrimaryWeight cpW ON cpW.CxID = c.CustomerID WHERE c.CustomerID =22703 GROUP BY c.CustomerID ORDER BY cpW.CustPrimaryWeightID DESC This doesn't really work correctly as the CustPrimaryWeight table has multiple enteries and it's simply joining the first entry and not the more recent one as the ORDER statement doesn't seem to do anything. Any ideas?

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  • Which MySql line is faster:

    - by Camran
    I have a classified_id variable which matches one document in a MySql table. I am currently fetching the information about that one record like this: SELECT * FROM table WHERE table.classified_id = $classified_id I wonder if there is a faster approach, for example like this: SELECT 1 FROM table WHERE table.classified_id = $classified_id Wont the last one only select 1 record, which is exactly what I need, so that it doesn't have to scan the entire table but instead stops searching for records after 1 is found? Or am I dreaming this? Thanks

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  • mysql command for update

    - by Mac Taylor
    hey guys i need to add a special text to all rows in my mysql table , how to add some text to the end of all rows' content in a table just for one field i used this code : UPDATE `blogs` SET `title`= `title`+ 'mytext'; but didnt work for me

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  • MySQL Datefields: duplicate or calculate?

    - by Konerak
    We are using a table with a structure imposed upon us more than 10 years ago. We are allowed to add columns, but urged not to change existing columns. Certain columns are meant to represent dates, but are put in different format. Amongst others: * CHAR(6): YYMMDD * CHAR(6): DDMMYY * CHAR(8): YYYYMMDD * CHAR(8): DDMMYYYY * DATE * DATETIME Since we now would like to do some more complex queries, using advanced date functions, my manager proposed to d*uplicate those problem columns* to a proper FORMATTED_OLDCOLUMNNAME column using a DATE or DATETIME format. Is this the way to go? Couldn't we just use the STR_TO_DATE function each time we accessed the columns? To avoid every query having to copy-paste the function, I could still work with a view or a stored procedure, but duplicating data to avoid recalculation sounds wrong. Solutions I see (I guess I prefer 2.2.1) 1. Physically duplicate columns 1.1 In the same table 1.1.1 Added by each script that does a modification (INSERT/UPDATE/REPLACE/...) 1.1.2 Maintained by a trigger on each modification 1.2 In a separate table 1.2.1 Added by each script that does a modification (INSERT/UPDATE/REPLACE/...) 1.2.2 Maintained by a trigger on each modification 2. On-demand transformation 2.1 Each query has to perform the transformation 2.1.1 Using copy-paste in the source code 2.1.2 Using a library 2.1.3 Using a STORED PROCEDURE 2.2 A view performs the transformation 2.2.1 A separate table replacing the entire table 2.2.2 A separate table just adding the date-fields for the primary keys Am I right to say it's better to recalculate than to store? And would a view be a good solution?

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  • help with delete where not in query

    - by kralco626
    I have a lookup table (##lookup). I know it's bad design because I'm duplicating data, but it speeds up my queries tremendously. I have a query that populates this table insert into ##lookup select distinct col1,col2,... from table1...join...etc... I would like to simulate this behavior: delete from ##lookup insert into ##lookup select distinct col1,col2,... from table1...join...etc... This would clearly update the table correctly. But this is a lot of inserting and deleting. It messes with my indexes and locks up the table for selecting from. This table could also be updated by something like: delete from ##lookup where not in (select distinct col1,col2,... from table1...join...etc...) insert into ##lookup (select distinct col1,col2,... from table1...join...etc...) except if it is already in the table The second way may take longer, but I can say "with no lock" and I will be able to select from the table. Any ideas on how to write the query the second way?

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  • mysql insert with auto-increment column

    - by czuroski
    Hello, I am trying to insert data into a mysql table from a csv file. I am using the infile sql command, but I am having trouble because the first column of the table is an id that is set as an auto increment field. what do I have to set my first column value to in order to get this to work, or can I do it at all? Thanks

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  • MySQL connection timeout

    - by NikolayGS
    I'm running program at apache tomcat server, that should be on permanently, but every morning(the client part isn't accessible at night) i receive error message (in apache tomcat console) that MySQL server is off. So is there any way to prevent this? Thanks in advance!

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  • MySQL friends table

    - by asmo
    I have a MySQL DB in which I store data about each user. I would like to add a list of friends for each user. Should I create a table of friends for each user in the DB or is there a better way?

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