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  • How do I show the print with AJAX/jQuery?

    - by Doug
    So I'm trying to understand this whole AJAX/jQuery thing. Right now, when I run this PHP script alone, I would have to wait and watch the wheel spin until it's done with the loop and then it will load. while ( $row = mysql_fetch_array($res) ) { postcode_to_storm( $row['Test'] ); $dom = new DOMDocument(); @$dom->loadHTML($result); $xPath = new DOMXPath($dom); $failInvite = 'Rejected'; $findFalse = strpos($result, $failInvite); if ( $findFalse == true ) { $array[$i] = $row['Test']; $i++; echo $array[$i]}; } } Now, how do I use AJAX/jQuery to show echo $array[$i]}; everytime it is invoked instead of waiting for the whole process to complete?

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  • need help to construct query

    - by Learner
    i have the following result and i would like to construct the select query from the following result in java, Please help me how to go about , tablename columnname size order employee name 25 1 employee sex 25 2 employee contactNumber 50 3 employee salary 25 4 address street 25 5 address country 25 6 from this i would like to construct query like select T1.name, T1.sex,T1.contactNumber, T1.salaryT2.street, T2.contry from tablename1[employee] T1, tablename2[address] T2 how to construt the above query in java, here table name can be N also the columname can be also N. Please help me to achieve the above. Thanks and Regards

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  • How to UNION ALL two SELECT statements?

    - by Lisa
    I have 2 tables, one looks like this: TABLE ONE id | Last Name | First Name | Username | Password | Secret Question and another that looks like this: TABLE TWO id | Hobby | Country | I want to combine a Select statement that grabs data from both tables and output the results. The following code: $select = mysql_query(" SELECT * FROM table_one WHERE Username = 'Bob' UNION ALL SELECT * FROM table_two WHERE Hobby = 'Baseball' "); while ($return = mysql_fetch_assoc($select)) { $userName = $return['Username']; $hobby = $return['Hobby']; } echo "$userName likes $hobby"; results in a The used SELECT statements have a different number of columns error, what am I doing wrong?

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  • Finding comma separated values with a colon delimiter

    - by iconMatrix
    I am setting values in my database for tourneyID,Selected,Paid,Entered,date then separating each selection with a colon So I have a string that may look like this 187,S,,,09-21-2013:141,S,,,06-21-2013:144,S,,,05-24-2013 but it also could look like this 145,S,,,07-12-2013:142,S,,,05-24-2013:187,S,,,09-21-2013 and some times is looks like this 87,S,,,07-11-2013:125,S,,,06-14-2013 I am trying to find this sequence: 187,S,,,09-21-2013 I have data stored like that because I paid a programmer to code it for me. Now, as I learn, I see it was not the best solution, but it is what I have till I learn more and it is working. My problem is when using LIKE it returns both the 187 and 87 values $getTeams = mysql_query("SELECT * FROM teams WHERE (team_tourney_vector LIKE '%$tid,S,P,,$tourney_start_date%' OR team_tourney_vector LIKE '%$tid,S,,,$tourney_start_date%') AND division='$division'"); I tried this using FIND_IN_SET() but it would only return the the team id for this string 187,S,,,09-21-2013:141,S,,,06-21-2013:144,S,,,05-24-2013 and does not find the team id for this string 145,S,,,07-12-2013:142,S,,,05-24-2013:187,S,,,09-21-2013 SELECT * FROM teams WHERE FIND_IN_SET('187',team_tourney_vector) AND (team_tourney_vector LIKE '%S,,,09-21-2013%') Any thoughts on how to achieve this?

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  • Xampp error on windows

    - by Deepak Kumar
    My problem is when i use xampp i see many error and when i use my web it has no error Notice: Undefined index: action in C:\xampp\htdocs\xyz\index.php on line 3 Notice: Undefined index: usNick in C:\xampp\htdocs\xyz\config.php on line 11 Notice: Use of undefined constant setname - assumed 'setname' in C:\xampp\htdocs\xyz\config.php on line 31 Notice: Use of undefined constant setname - assumed 'setname' in C:\xampp\htdocs\xyz\config.php on line 31 Notice: Undefined index: usNick in C:\xampp\htdocs\xyz\config.php on line 34 Notice: A session had already been started - ignoring session_start() in C:\xampp\htdocs\xyz\data.php on line 2 Notice: Undefined index: r in C:\xampp\htdocs\xyz\data.php on line 4 Notice: Undefined index: ucNick in C:\xampp\htdocs\xyz\data.php on line 8 I have tried many time changing things in Setting, Security, Privileges etc but nothing changed, I want to know if im missing something out Thanks

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  • Query returning related assets

    - by GMo
    I have 2 tables, one is an assets table which holds digital assets (e.g. article, images etc), the 2nd table is an asset_links table which maps 1-1 relationships between assets contained within the assets table. Here are the table definitions: Asset +---------------+--------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +---------------+--------------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | source | varchar(255) | YES | | NULL | | | title | varchar(255) | YES | | NULL | | | date_created | datetime | YES | | NULL | | | date_embargo | datetime | YES | | NULL | | | date_expires | datetime | YES | | NULL | | | date_updated | datetime | YES | | NULL | | | keywords | varchar(255) | YES | | NULL | | | status | int(11) | YES | | NULL | | | priority | int(11) | YES | | NULL | | | fk_site | int(11) | YES | MUL | NULL | | | resource_type | varchar(255) | YES | | NULL | | | resource_id | int(11) | YES | | NULL | | | fk_user | int(11) | YES | MUL | NULL | | +---------------+--------------+------+-----+---------+----------------+ Asset_links +-----------+---------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +-----------+---------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | asset_id1 | int(11) | YES | | NULL | | | asset_id2 | int(11) | YES | | NULL | | +-----------+---------+------+-----+---------+----------------+ In the asset_links table there are the following rows: 1 - 3, 1 - 4, 2 - 10, 2 - 56 I am looking to write one query which will return all assets which satisfy any asset search criteria and within the same query return all of the linked asset data for linked assets for that asset. e.g. The query returning assets 1 and 2 would return : Asset 1 attributes - Asset 3 attributes - Asset 4 attributes Asset 2 attributes - Asset 10 attributes - Asset 56 attributes What is the best way to write the query?

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  • Session is working in Localhost Properly but not Online (Cpanel)

    - by nando pandi
    Hello guys Sorry for my stupid question regarding to my yesterday question its not solved yet even the advice you have given but still not working. i have removed all of spaces but still showing the problem for me. it's working perfect in localhost but not in CPANEL. Here is the errors which give: Warning: session_start() [function.session-start]: Cannot send session cookie - headers already sent by (output started at /home/scalepro/public_html/Admin Panel/Remote Employee/main.php:1) in /home/scalepro/public_html/Admin Panel/Remote Employee/main.php on line 1 Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at /home/scalepro/public_html/Admin Panel/Remote Employee/main.php:1) in /home/scalepro/public_html/Admin Panel/Remote Employee/main.php on line 1 Warning: Cannot modify header information - headers already sent by (output started at /home/scalepro/public_html/Admin Panel/Remote Employee/main.php:1) in /home/scalepro/public_html/Admin Panel/Remote Employee/main.php on line 13 Warning: Unknown: Your script possibly relies on a session side-effect which existed until PHP 4.2.3. Please be advised that the session extension does not consider global variables as a source of data, unless register_globals is enabled. You can disable this functionality and this warning by setting session.bug_compat_42 or session.bug_compat_warn to off, respectively in Unknown on line 0 ANY ONE PLEASE ??? Here is my code: <?php session_start(); require_once('../../Admin Panel/db.php'); if(isset($_POST['email']) && !empty($_POST['email']) && isset($_POST['password']) && !empty($_POST['password'])) { $email = $_POST['email']; $password = $_POST['password']; $query="SELECT RemoteEmployeeFullName, RemoteEmployeeEmail, RemoteEmployeePassword FROM remoteemployees WHERE RemoteEmployeeEmail='".$email."' AND RemoteEmployeePassword='".$password."'"; $queryrun=$connection->query($query); if($queryrun->num_rows > 0) { $_SESSION['email']=$RemoteEmployeeFullName; header("Location: /home/scalepro/public_html/Admin Panel/Remote Employee/REPLists.php"); } else { echo 'Email: <b>'.$email. '</b> or Password <b>'. $password.'</b> Is Not Typed Correctly Try Again Please!.'; header( "refresh:5;url= /home/scalepro/public_html/spd/myaccount.php" ); } } else { header( "refresh:5;url= /home/scalepro/public_html/spd/myaccount.php" ); } ?> if the condition gets true this will be redirected to a page by the name of REPLists.php here is the page. <?php session_start(); require_once('../../Admin Panel/db.php'); ?> <html> <head> <style> .wrapper { width:1250px; height:auto; border:solid 1px #000; margin:0 auto; padding:5px; border-radius:5px; -webkit-border-radius:5px; -moz-border-radius:5px; -ms-border-radius:5px; } .wrapper .header { width:1250px; height:20px; border-bottom:solid 1px #f0eeee; margin:auto 0; margin-bottom:12px; } .wrapper .header div { text-decoration:none; color:#F60; } .wrapper .header div a { text-decoration:none; color:#F60; } .wrapper .Labelcon { width:1250px; height:29px; border-bottom:solid 1px #ccc; } .wrapper .Labelcon .Label { width:125px; height:20px; float:left; text-align:center; border-left:1px solid #f0eeee; font:Verdana, Geneva, sans-serif; font-size:14.3px; font-weight:bold; } .wrapper .Valuecon { width:1250px; height:29px; border-bottom:solid 1px #ccc; color:#F60; text-decoration:none; } .wrapper .Valuecon .Value { width:125px; height:20px; float:left; text-align:center; border-left:1px solid #f0eeee; font-size:14px; } </style> </head> <body> <div class="wrapper"> <div class="header"> <div style="float:left;"><font color="#000000">Email: </font> <?php if(isset($_SESSION['email'])) { echo $_SESSION['email']; } ?> </div> <div style="float:right;"> <a href="#">My Profile</a> | <a href="logout.php">Logout</a></div> </div> <div class="Labelcon"> <div class="Label">Property ID</div> <div class="Label">Property Type</div> <div class="Label">Property Deal Type</div> <div class="Label">Property Owner</div> <div class="Label">Proposted Price</div> </div> <?php if(!isset($_SESSION['email'])) { header('Location:../../spd/myaccount.php'); } else { $query = "SELECT properties.PropertyID, properties.PropertyType, properties.PropertyDealType, properties.Status, properties.PropostedPrice, remoteemployees.RemoteEmployeeFullName, propertyowners.PropertyOwnerName, propertydealers.PropertyDealerName FROM remoteemployees, propertyowners, propertydealers, properties WHERE properties.PropertyOwnerID=propertyowners.PropertyOwnerID AND properties.PropertyDealerID=propertydealers.PropertyDealerID AND remoteemployees.RemoteEmployeeID=properties.RemoteEmployeeID ORDER BY properties.PropertyID "; $query_run = $connection->query($query); if( $connection->error ) exit( $connection->error ); while($row=$query_run->fetch_assoc()) { ?> <div class="Valuecon"> <div class="Value"><?php echo $row['PropertyID'] ?></div> <div class="Value"><?php echo $row['PropertyType'] ?></div> <div class="Value"><?php echo $row['PropertyDealType']?></div> <div class="Value"><?php echo $row['PropertyOwnerName'] ?></div> <div class="Value"><?php echo $row['PropostedPrice'];?></div> </div> <?php } }?> </div> </body> </html>

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  • Cannot add or update a child row: a foreign key constraint fails

    - by myaccount
    // Getting the id of the restaurant to which we are uploading the pictures $restaurant_id = intval($_GET['restaurant-id']); if(isset($_POST['submit'])) { $tmp_files = $_FILES['rest_pics']['tmp_name']; $target_files = $_FILES['rest_pics']['name']; $tmp_target = array_combine($tmp_files, $target_files); $upload_dir = $rest_pics_path; foreach($tmp_target as $tmp_file => $target_file) { if(move_uploaded_file($tmp_file, $upload_dir."/".$target_file)) { $sql = sprintf(" INSERT INTO rest_pics (branch_id, pic_name) VALUES ('%s', '%s')" , mysql_real_escape_string($restaurant_id) , mysql_real_escape_string(basename($target_file))); $result = mysql_query($sql) or die(mysql_error()); } I get the next error: Cannot add or update a child row: a foreign key constraint fails (rest_v2.rest_pics, CONSTRAINT rest_pics_ibfk_1 FOREIGN KEY (branch_id) REFERENCES rest_branches (branch_id) ON DELETE CASCADE ON UPDATE CASCADE However, this error totally disappears and everything goes well when I put directly the restaurant id (14 for example) instead of $restaurant_id variable in the sql query. The URL am getting the id from is: http://localhost/rest_v2/public_html/admin/add-delete-pics.php?restaurant-id=2 Any help please?

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  • Having to insert a record, then update the same record warrants 1:1 relationship design?

    - by dianovich
    Let's say an Order has many Line items and we're storing the total cost of an order (based on the sum of prices on order lines) in the orders table. -------------- orders -------------- id ref total_cost -------------- -------------- lines -------------- id order_id price -------------- In a simple application, the order and line are created during the same step of the checkout process. So this means INSERT INTO orders .... -- Get ID of inserted order record INSERT into lines VALUES(null, order_id, ...), ... where we get the order ID after creating the order record. The problem I'm having is trying to figure out the best way to store the total cost of an order. I don't want to have to create an order create lines on an order calculate cost on order based on lines then update record created in 1. in orders table This would mean a nullable total_cost field on orders for starters... My solution thus far is to have an order_totals table with a 1:1 relationship to the orders table. But I think it's redundant. Ideally, since everything required to calculate total costs (lines on an order) is in the database, I would work out the value every time I need it, but this is very expensive. What are your thoughts?

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  • Calculate distances and sort them

    - by Emir
    Hi guys, I wrote a function that can calculate the distance between two addresses using the Google Maps API. The addresses are obtained from the database. What I want to do is calculate the distance using the function I wrote and sort the places according to the distance. Just like "Locate Store Near You" feature in online stores. I'm going to specify what I want to do with an example: So, lets say we have 10 addresses in database. And we have a variable $currentlocation. And I have a function called calcdist(), so that I can calculate the distances between 10 addresses and $currentlocation, and sort them. Here is how I do it: $query = mysql_query("SELECT name, address FROM table"); while ($write = mysql_fetch_array($query)) { $distance = array(calcdist($currentlocation, $write["address"])); sort($distance); for ($i=0; $i<1; $i++) { echo "<tr><td><strong>".$distance[$i]." kms</strong></td><td>".$write['name']."</td></tr>"; } } But this doesn't work very well. It doesn't sort the numbers. Another challenge: How can I do this in an efficient way? Imagine there are infinite numbers of addresses; how can I sort these addresses and page them?

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  • Symfony 1.4: Deleting a sfGuardUser

    - by Tom
    Hi, I'm having some trouble with the following... I have a sfGuardUser table set up normally, and it has a one-to-one relationship with a Profile table, which contains some additional user info. When a user wants to delete themselves from the site, I'd like to retain their info in the Profile table for various purposes BUT delete the sfGuardUser in order to keep that table cleaner/shorter (not just set it to inactive). I was under the impression that I could set the FK in the Profile table to NULL and then delete the sfGuardUser, but it seems the FK-constraint fails. Indeed, it seems I can't delete either and the queries fail: If I try to delete the sfGuardUser, the Profile table will have an invalid FK If I try to delete a Profile, the sfGuardUser will have an invalid FK Other than leaving outdated sfGuardUsers and Profiles in these tables, or having to use a cascaded delete to get rid of both, can anyone tell me if there's any other way around this? Thank you.

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  • Why should I abstract my data layer?

    - by Gazillion
    OOP principles were difficult for me to grasp because for some reason I could never apply them to web development. As I developed more and more projects I started understanding how some parts of my code could use certain design patterns to make them easier to read, reuse, and maintain so I started to use it more and more. The one thing I still can't quite comprehend is why I should abstract my data layer. Basically if I need to print a list of items stored in my DB to the browser I do something along the lines of: $sql = 'SELECT * FROM table WHERE type = "type1"';' $result = mysql_query($sql); while($row = mysql_fetch_assoc($result)) { echo '<li>'.$row['name'].'</li>'; } I'm reading all these How-Tos or articles preaching about the greatness of PDO but I don't understand why. I don't seem to be saving any LoCs and I don't see how it would be more reusable because all the functions that I call above just seem to be encapsulated in a class but do the exact same thing. The only advantage I'm seeing to PDO are prepared statements. I'm not saying data abstraction is a bad thing, I'm asking these questions because I'm trying to design my current classes correctly and they need to connect to a DB so I figured I'd do this the right way. Maybe I'm just reading bad articles on the subject :) I would really appreciate any advice, links, or concrete real-life examples on the subject!

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  • Referral System PHP

    - by Liam
    I have a membership based website and im planning on implementing a referral system. My website is credit based, the idea is that if User X refers User Y, then User X gets 100 bonus credits. Has anybody built a referral system before and if so what obstacles should I bear in mind? I've had a snoop round SO tonight but couldn't find any suitable answers. My theory is to give each user a random string which is generated and stored in the DB when they sign up, The user will then be presented with a URL incl. that string which when they pass to somebody (User Z), User Z is then sent to a page, the page then uses the GET method to gather the Random string and update the DB Row they currently occupy, does this sound feasible or could it easily be breached? Thanks

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  • Confusion using MYSQLI

    - by user1020069
    I just started using mysqli API for PHP. Apparently, every time an object of the class MYSQLI is instantiated, it can setup a connection to the database as it connects to the server unlike mysql_connect, which connects to the server first and then you are required to specify the database to query. Now this is a good problem if the db exists, in my case, the db does not exist on the first ever connection to the server/execution of the problem, hence I must connect without specifying the database, which is fine, since the msyqli constructor does not make this database mandatory. My challenge is essentially, how do I check if the database exists before attempting the first connection. The only way to really do this would be to establish a conection to the server and then use the result of the following query to gauge if the database exists: SELECT COUNT(*) AS `exists` FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMATA.SCHEMA_NAME="dbname" ; If this returns true, then the database exists, but now the challenge is how do I get the mysqli object to query this database rather than having to prefix the name of the database in the query. Thanks much

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  • Unique constraint with nullable column

    - by Álvaro G. Vicario
    I have a table that holds nested categories. I want to avoid duplicate names on same-level items (i.e., categories with same parent). I've come with this: CREATE TABLE `category` ( `category_id` int(10) unsigned NOT NULL AUTO_INCREMENT, `category_name` varchar(100) NOT NULL, `parent_id` int(10) unsigned DEFAULT NULL, PRIMARY KEY (`category_id`), UNIQUE KEY `category_name_UNIQUE` (`category_name`,`parent_id`), KEY `fk_category_category1` (`parent_id`,`category_id`), CONSTRAINT `fk_category_category1` FOREIGN KEY (`parent_id`) REFERENCES `category` (`category_id`) ON DELETE SET NULL ON UPDATE CASCADE ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_spanish_ci Unluckily, category_name_UNIQUE does not enforce my rule for root level categories (those where parent_id is NULL). Is there a reasonable workaround?

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  • Database related Questions

    - by alokpatil
    I am planning to make a railway reservation project... I am maintaining following tables.. trainTable (trainId,trainName,trainFrom,trainTo,trainDate,trainNoOfBoogies)...PK(trainId) Boogie (trainId,boogieId,boogieName,boogieNoOfseats)...CompositeKey(trainId,boogieId)... Seats (trainId,boogieId,seatId,seatStatus,seatType)...CompositeKey(trainId,boogieId,seatId)... user (userId,name...personal details) userBooking (userId,trainId,boogieId,seatId)...Is this good design reply me please...

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  • group by with 3 diffrent

    - by NN
    I have 2 table and I wanna a query with 3 column result in on of them 2 column with view count and title name and in the other 1 column with type_ and i wanna to grouping type_ with max(view count) and show the them title but i didn't have any idea about grouping expression. i think we can solve in by using sub query but i don't know which column use in group by. 2 table join with this expression class pk=resource key i exam this query: SELECT t.title,j.type_ FROM tags asset t,journal article j where type_ in (select type_ from journal article,tags asset where class pk=resource key group by type_) but the answer was wrong

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  • Difficults on sql query

    - by João Madureira Pires
    I have the following tables: TableA (id, tableB_id, tableC_id) TableB (id, expirationDate) TableC (id, expirationDate) I want to retrieve all the results from TableA ordered by tableB.expirationDate and tableC.expirationDate. thanks

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  • How much does an InnoDB table benefit from having fixed-length rows?

    - by Philip Eve
    I know that dependent on the database storage engine in use, a performance benefit can be found if all of the rows in the table can be guaranteed to be the same length (by avoiding nullable columns and not using any VARCHAR, TEXT or BLOB columns). I'm not clear on how far this applies to InnoDB, with its funny table arrangements. Let's give an example: I have the following table CREATE TABLE `PlayerGameRcd` ( `User` SMALLINT UNSIGNED NOT NULL, `Game` MEDIUMINT UNSIGNED NOT NULL, `GameResult` ENUM('Quit', 'Kicked by Vote', 'Kicked by Admin', 'Kicked by System', 'Finished 5th', 'Finished 4th', 'Finished 3rd', 'Finished 2nd', 'Finished 1st', 'Game Aborted', 'Playing', 'Hide' ) NOT NULL DEFAULT 'Playing', `Inherited` TINYINT NOT NULL, `GameCounts` TINYINT NOT NULL, `Colour` TINYINT UNSIGNED NOT NULL, `Score` SMALLINT UNSIGNED NOT NULL DEFAULT 0, `NumLongTurns` TINYINT UNSIGNED NOT NULL DEFAULT 0, `Notes` MEDIUMTEXT, `CurrentOccupant` TINYINT UNSIGNED NOT NULL DEFAULT 0, PRIMARY KEY (`Game`, `User`), UNIQUE KEY `PGR_multi_uk` (`Game`, `CurrentOccupant`, `Colour`), INDEX `Stats_ind_PGR` (`GameCounts`, `GameResult`, `Score`, `User`), INDEX `GameList_ind_PGR` (`User`, `CurrentOccupant`, `Game`, `Colour`), CONSTRAINT `Constr_PlayerGameRcd_User_fk` FOREIGN KEY `User_fk` (`User`) REFERENCES `User` (`UserID`) ON DELETE CASCADE ON UPDATE CASCADE, CONSTRAINT `Constr_PlayerGameRcd_Game_fk` FOREIGN KEY `Game_fk` (`Game`) REFERENCES `Game` (`GameID`) ON DELETE CASCADE ON UPDATE CASCADE ) ENGINE=INNODB CHARACTER SET utf8 COLLATE utf8_general_ci The only column that is nullable is Notes, which is MEDIUMTEXT. This table presently has 33097 rows (which I appreciate is small as yet). Of these rows, only 61 have values in Notes. How much of an improvement might I see from, say, adding a new table to store the Notes column in and performing LEFT JOINs when necessary?

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  • How to access website CMS with only access to database

    - by user1741615
    I have a website that uses an "in-house" cms and I don't know the login details. The platform itself doesn't have the "reset your password" functionality. I do have access to ftp and phmyadmin and I found the SQL table containing the user details, but of course the password is MD5 encryption. I tried manually creating a user in php my admin and filling in a password encrypted in MD5 (used a md service online for that), but it still doesn't work. Does anybody know other tricks I can use? Thanks.

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