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  • my sql insert query not working

    - by Piyush
    I am inserting userId.It is displaying correct but inserting 0 in spite of actual userId. mycode- If(! empty($userIDToCheck) || $userIDToCheck != '' ) { echo $userIDToCheck; $sql = "INSERT INTOpnpdb.ruser(userid) VALUES ('$userIDToCheck');"; mysql_query($sql)or die(mysql_error()); echo "Done"; } Output : pi203713 Done But is database it is inserting "0"???

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  • Confused as to why my PHP include isn't working

    - by Sam
    I had a prototype of my website working correctly, meaning it connected to the database correctly. This was done with just one file called "connect.php" which had mysql_connect() and such inside it. I then separated the connect information into to separate files, one containing the account information (account.php) and one containing the connect function (connect.php), with correct information (I triple checked) and it isn't connecting properly. All I can think of is that I'm not including it the right way. This is what I have in a file: <?php include('account.php'); include('connect.php'); include('functions.php'); ..... ?>

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  • Query to add a column depending of outcome of there columns

    - by Tam
    I have a user table 'users' that has fields like: id first_name last_name ... and have another table that determines relationships: user_id friend_id user_accepted friend_accepted .... I would like to generate a query that selects all the users but also add another field/column say 'network_status' that depends on the values of user_accepted and fiend_accepted. For example, if user_accepted is true friend_accepted is false I want the 'network_status' field to say 'request sent'. Can I possibly do this in one query? (I would prefer not to user if/else inside the query but if that's the only way so be it)

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  • Use where clause with Like in codeigniter

    - by user2524013
    I am working on a project. I am implementing the Search functionality in my System. I will have to show the search record from two tables base on the current use login. I have tried the following code: function searchActivity($limit,$offset,$keyword1,$keyword2,$recruiter_id) { $q=$this->db->select('*')->from('tbl_activity')->limit($limit,$offset); $this->db->join('tbl_job', 'tbl_job.job_id = tbl_activity.job_id_fk', 'left outer'); $this->db->order_by("activity_id", "ASC"); $this->db->like('job_title',$keyword1,'both'); $this->db->or_like('job_title',$keyword2,'both'); $this->db->or_like('activity_subject',$keyword1,'both'); $this->db->or_like('activity_subject',$keyword2,'both'); $this->db->or_like('activity_details',$keyword1,'both'); $this->db->or_like('activity_details',$keyword2,'both'); $this->db->where('tbl_activity.recruiter_id_fk',$recruiter_id); $ret['rows']=$q->get()->result(); return $ret; } I want to show search results based on the current user id, which is currently store in $recruiter. Thanks in advance.

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  • PHP classes, parse syntax errors when using 'var' to declare variables

    - by jon
    I am a C# guy trying to translate some of my OOP understanding over to php. I'm trying to make my first class object, and are hitting a few hitches. Here is the beginning of the class: <?php require("Database/UserDB.php"); class User { private var $uid; private var $username; private var $password; private var $realname; private var $email; private var $address; private var $phone; private var $projectArray; public function _construct($username) { $userArray = UserDB::GetUserArray($username); $uid = $userArray['uid']; $username = $userArray['username']; $realname = $userArray['realname']; $email = $userArray['email']; $phone = $userArray['phone']; $i = 1; $projectArray = UserDB::GetUserProjects($this->GetID()); while($projectArray[$i] != null) { $projectArray[$i] = new Project($projectArray[$i]); } UserDB.php is where I have all my static functions interacting with the Database for this User Class. I am getting errors using when I use var, and I'm getting confused. I know I don't HAVE to use var, or declare the variables at all, but I feel it is a better practice to do so. the error is "unexpected T_VAR, expecting T_VARIABLE" When I simply remove var from the declarations it works. Why is this?

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  • SQL Query That Should Return Least two days record

    - by Aryans
    I have a table "abc" where i store timestamp having multiple records let suppose 1334034000 Date:10-April-2012 1334126289 Date:11-April-2012 1334291399 Date:13-April-2012 I want to build a sql query where I can find at first attempt the records having last two day values and so second time the next two days . . . Example: Select *,dayofmonth(FROM_UNIXTIME(i_created)) from notes where dayofmonth(FROM_UNIXTIME(i_created)) > dayofmonth(FROM_UNIXTIME(i_created)) -2 order by dayofmonth(FROM_UNIXTIME(i_created)) this query returns all the records date wise but we need very most two days record. Please suggest accordingly. Thanks in advance

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  • Unknown column '' in 'field list'

    - by Rixius
    I am running a sql in PHP query that is dieing with the mysql_error() of Unknown column '' in 'field list' The query: SELECT `standard` AS fee FROM `corporation_state_fee` WHERE `stateid` = '8' LIMIT 1 when I run the query in PHPmyadmin, it return the information without flagging the error

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  • In SQL, a Join is actually an Intersection? And it is also a linkage or a "Sideway Union"?

    - by Jian Lin
    I always thought of a Join in SQL as some kind of linkage between two tables. For example, select e.name, d.name from employees e, departments d where employees.deptID = departments.deptID In this case, it is linking two tables, to show each employee with a department name instead of a department ID. And kind of like a "linkage" or "Union" sideway". But, after learning about inner join vs outer join, it shows that a Join (Inner join) is actually an intersection. For example, when one table has the ID 1, 2, 7, 8, while another table has the ID 7 and 8 only, the way we get the intersection is: select * from t1, t2 where t1.ID = t2.ID to get the two records of "7 and 8". So it is actually an intersection. So we have the "Intersection" of 2 tables. Compare this with the "Union" operation on 2 tables. Can a Join be thought of as an "Intersection"? But what about the "linking" or "sideway union" aspect of it?

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  • Locking DB w/ Large Reads (Ruby-on-Rails/Heroku)

    - by Splashlin
    Currently I have a Web API running on Heroku that is constantly writing information we're collecting from other data sources (currently theres about half a GB of data and it's growing very quickly). We're looking to add a reporting system on top of the current database that we can use to extract useful information out of the DB. The problem is that when we're running reports we're locking the DB and any other sites communicating with the DB are timing out. Does anyone have any solutions on how to solve this type of issue? Amazon RDS seems to have some interesting stuff with database replication but I don't know if that will solve my problems. Any advice would be greatly appreciated. Thanks

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  • Database design for business numbers

    - by Rob Morris
    I'm in need of some help, I need to store the information below into a database, what would the relational database structure be for this: Then I need to create a dropdown for the insurance company followed by another dropdown depending on what the first dropdown selected value was, then once both selects have been chosen display the relevant telephone number. I guess i need to query the database, then display the dropdowns using javascript(jquery) or Ajax?

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  • Random Alpha numeric generator

    - by AAA
    Hi, I want to give our users in the database a unique alpha-numeric id. I am using the code below, will this always generate a unique id? Below is the old and updated version of the code: New php: // Generate Guid function NewGuid() { $s = strtoupper(md5(uniqid("something",true))); $guidText = substr($s,0,8) . '-' . substr($s,8,4) . '-' . substr($s,12,4). '-' . substr($s,16,4). '-' . substr($s,20); return $guidText; } // End Generate Guid $Guid = NewGuid(); echo $Guid; echo "<br><br><br>"; Old PHP: // Generate Guid function NewGuid() { $s = strtoupper(md5(uniqid("something",true))); $guidText = substr($s,0,8) . '-' . substr($s,8,4) . '-' . substr($s,12,4). '-' . substr($s,16,4). '-' . substr($s,20); return $guidText; } // End Generate Guid $Guid = NewGuid(); echo $Guid; echo "<br><br><br>"; Will this first code guarantee uniqueness?

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  • Inventory count in CakePHP

    - by metrobalderas
    We are developing an inventory tracking system. Basically we've got an order table in which orders are placed. When an order is payed, the status changes from 0 to 1. This table has multiple children in another table order_items. This is the main structure. CREATE TABLE order( id INT UNSIGNED AUTO_INCREMENT PRIMARY KEY, user_id INT UNSIGNED, status INT(1), total INT UNSIGNED ); CREATE TABLE order_items( id INT UNSIGNED AUTO_INCREMENT PRIMARY KEY, order_id INT UNSIGNED, article_id INT UNSIGNED, size enum('s', 'm', 'l', 'xl'), quantity INT UNSIGNED ); Now, we've got a stocks table with similar architecture for the acquisitions. This is the structure. CREATE TABLE stock( id INT UNSIGNED AUTO_INCREMENT PRIMARY KEY, article_id INT UNSIGNED ); CREATE TABLE stock_items( id INT UNSIGNED AUTO_INCREMENT PRIMARY KEY, stock_id INT UNSIGNED, size enum('s', 'm', 'l', 'xl'), quantity INT(2) ); The main difference is that stocks has no status field. What we are looking for is a way to sum each article size from stock_items, then sum each article size from order_items where Order.status = 1 and substract both these items to find our current inventory. This is the table we want to get from a single query: Size | Stocks | Sales | Available s | 10 | 3 | 7 m | 15 | 13 | 2 l | 7 | 4 | 3 Initially we thought abouth using complex find conditions, but perhaps that's the wrong approach. Also, since it's not a direct join, it turns out to be quite hard. This is the code we have to retrieve the stock's total for each item. function stocks_total($id){ $find = $this->StockItem->find('all', array( 'conditions' => array( 'StockItem.stock_id' => $this->find('list', array('conditions' => array('Stock.article_id' => $id))) ), 'fields' => array_merge( array( 'SUM(StockItem.cantidad) as total' ), array_keys($this->StockItem->_schema) ), 'group' => 'StockItem.size', 'order' => 'FIELD(StockItem.size, \'s\', \'m\' ,\'l\' ,\'xl\') ASC' )); return $find; } Thanks.

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  • I cant get a field on report from a view

    - by felipedz
    When I get a field, this work good. But, when get a field from a 'VIEW', is a problem because the code of a VIEW is: CREATE OR REPLACE VIEW tabla_clientes AS SELECT id_cliente,nombre, CONCAT('$ ',FORMAT(monto_a_favor,0), '???'), CONCAT('$ ',FORMAT(calcular_monto_por_cobrar_cliente(id_cliente),0)) FROM cliente; When I compile this. Appears errors from the name of fields. Description | Object ---------------------------------------------------------------------------- Syntax error, insert ";" to complete BlockStatements | ${CONCAT('$ ',FORMAT(monto_a_favor,0)} Syntax error on tokens, delete these tokens | ${CONCAT('$ ',FORMAT(monto_a_favor,0)} Syntax error on token ",", delete this token | ${CONCAT('$ ',FORMAT(monto_a_favor,0)} If I change the name at this field appears other error.

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  • Does UNIQ constraint mean also an index on that field(s)?

    - by Gremo
    As title, should i defined a separate index on email column (for searching purposes) or the index is "automatically" added along with UNIQ_EMAIL_USER constraint? CREATE TABLE IF NOT EXISTS `customer` ( `id` int(11) NOT NULL AUTO_INCREMENT, `user_id` int(11) NOT NULL, `first` varchar(255) NOT NULL, `last` varchar(255) NOT NULL, `slug` varchar(255) NOT NULL, `email` varchar(255) NOT NULL, `created_at` datetime NOT NULL, `updated_at` datetime NOT NULL, PRIMARY KEY (`id`), UNIQUE KEY `UNIQ_SLUG` (`slug`), UNIQUE KEY `UNIQ_EMAIL_USER` (`email`,`user_id`), KEY `IDX_USER` (`user_id`) ) ENGINE=InnoDB;

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  • Converting SQL with subselect in select to HQL

    - by UltraVi01
    I have the following SQL that I am having problems converting to HQL. A NPE is getting thrown -- which I think has something to do with the SUM function. Also, I'd like to sort on the subselect alias -- is this possible? SQL (subselect): SELECT q.title, q.author_id, (SELECT IFNULL(SUM(IF(vote_up=true,1,-1)), 0) FROM vote WHERE question_id = q.id) AS votecount FROM question q ORDER BY votecount DESC HQL (not working) SELECT q, (SELECT COALESCE(SUM(IF(v.voteUp=true,1,-1)), 0) FROM Vote v WHERE v.question = q) AS votecount FROM Question AS q LEFT JOIN q.author u LEFT JOIN u.blockedUsers ub WHERE q.dateCreated BETWEEN :week AND :now AND u.id NOT IN ( SELECT ub.blocked FROM UserBlock AS ub WHERE ub.blocker =:loggedInUser ) AND (u.blockedUsers IS EMPTY OR ub.blocked !=:loggedInUser) ORDER BY votecount DESC

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  • Array "tree" creation from db table

    - by Tural Teyyuboglu
    Trying to create array tree for db driven navigation. Getting following errur: array_key_exists() expects exactly 2 parameters, 1 given on line if (!array_key_exists($tree[$parent]['children'][$id])) Function looks like that $tree = array(); $sql = "SELECT id, parent, name FROM menu WHERE parent ... etc.... "; $results = mysql_query($sql) or die(mysql_error()); while(list($id, $parent, $name) = mysql_fetch_assoc($results)) { $tree[$id] = array('name' => $name, 'children' => array(), 'parent' => $parent); if (!array_key_exists($tree[$parent]['children'][$id])) { $tree[$parent]['children'][$id] = $id; } } Db structure How can I fix that? Whats wrong in this function?

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  • PHP-How to Pass Multiple Value In Form Field

    - by Tall boY
    hi i have a php based sorting method with drop down menu to sort no of rows, it is working fine. i have another sorting links to sort id & title, it is also working fine. but together they are not working fine. what happens is that when i sort(say by title) using links, result gets sorted by title, then if i sort rows using drop down menu rows get sorted but result gets back to default of id sort. sorting codes for id & tite is if ($orderby == 'title' && $sortby == 'asc') {echo " <li id='scurrent'><a href='?rpp=$rowsperpage&order=title&sort=asc'>title-asc:</a></li> ";} else {echo " <li><a href='?rpp=$rowsperpage&order=title&sort=asc'>title-asc:</a></li> ";} if ($orderby == 'title' && $sortby == 'desc') {echo " <li id='scurrent'><a href='?rpp=$rowsperpage&order=title&sort=desc'>title-desc:</a></li> ";} else {echo " <li><a href='?rpp=$rowsperpage&order=title&sort=desc'>title-desc:</a></li> ";} if ($orderby == 'id' && $sortby == 'asc') {echo " <li id='scurrent'><a href='?rpp=$rowsperpage&order=id&sort=asc'>id-asc:</a></li> ";} else {echo " <li><a href='?rpp=$rowsperpage&order=id&sort=asc'>id-asc:</a></li> ";} if ($orderby == 'id' && $sortby == 'desc') {echo " <li id='scurrent'><a href='?rpp=$rowsperpage&order=id&sort=desc'>id-desc:</a></li> ";} else {echo " <li><a href='?rpp=$rowsperpage&order=id&sort=desc'>id-desc:</a></li> ";} ?> sorting codes for rows is <form action="is-test.php" method="get"> <select name="rpp" onchange="this.form.submit()"> <option value="10" <?php if ($rowsperpage == 10) echo 'selected="selected"' ?>>10</option> <option value="20" <?php if ($rowsperpage == 20) echo 'selected="selected"' ?>>20</option> <option value="30" <?php if ($rowsperpage == 30) echo 'selected="selected"' ?>>30</option> </select> </form> this method passes only rows per page(rpp) into url. i want it to pass order, sort& rpp. is there a way around to pass multiple values in form fields like this. <form action="is-test.php" method="get"> <select name="rpp, order, sort" onchange="this.form.submit()"> <option value="10, $orderby, $sortby" <?php if ($rowsperpage == 10) echo 'selected="selected"' ?>>10</option> <option value="20, $orderby, $sortby" <?php if ($rowsperpage == 20) echo 'selected="selected"' ?>>20</option> <option value="30, $orderby, $sortby" <?php if ($rowsperpage == 30) echo 'selected="selected"' ?>>30</option> </select> </form> this may seem silly but it just to give you an idea of what i am trying to implement,(i am very new to php) please suggest any way to make this work. thanks

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  • Where to place web server root?

    - by nefo_x
    Hello everybody, I've just made an upgrade and now partly thinking on web-server directory structure for local workstation for web-development on linux platform. Running multiple hosts and different projects required. Where is it better to put all the server's docroots? /var/www? /srv? /www? I plan to make it as separate partition - could it be good for backups? :) I'm looking forward to your thoughts on this.

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  • Stop Submit With empty and error Input Values Using PHP

    - by user3615781
    I am using the following code for sending data to database, but it sends the data even the values of the fields are incorrect or empty. So can anyone help me solve this by using php? Here is my code: <?php //Connecting to sql db $connect = mysqli_connect("localhost","root","","form"); /* check connection */ if (!$connect) { die('Connect Error: ' . mysqli_connect_error()); } //Sending data to sql db $result = mysqli_query($connect,"INSERT INTO students(name,email,website,comment,gender) VALUES('$name','$email','$website','$comment','$gender')"); if (!$result) { die('Query Error:'. mysqli_error($connect)); } ?>

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  • How do I all the registered users on a day report

    - by Nadal
    I have a table called users where I have two columns: name and created_at. created_at column column is of type datetime and it stores the datetime when this user was created. I need to know the number of users created for a given date range. Let's say I ask give me user report between 1-nov-2010 and 30-nov-2010 . I need something like this 1-nov-2010: 2 2-nov-2010: 5 The problem I am running into is that created_at data has value upto second. How do I check if a created_at date falls within a given date. Any help in solving this problem is appreciated. I am using mysql5.

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  • php query cant figure out the problem

    - by Eat
    I dont get an error but it does not return me the row values as well. How can I address the problem? <?php //DATABASE $dbConn = mysql_connect($host,$username,$password); mysql_select_db($database,$dbConn); $SQL = mysql_query("SELECT N.*, C.CatID FROM News N INNER JOIN Categories C ON N.CatID = C.CatID WHERE N.Active = 1 ORDER BY DateEntered DESC"); while ( $Result = mysql_fetch_array($SQL) or die(mysql_error())) { $CatID[] = $Result[CatID]; $NewsName[] = $Result[NewsName]; $NewsShortDesc[] = $Result[NewsShortDesc]; } // mysql_free_result($Result); ?> <div class="toparticle"> <span class="section"><?=$CatID[0] ?> </span> <span class="headline"><?=$NewsName[0] ?></span> <p><?=$NewsShortDesc[0] ?></p> </div>

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