Search Results

Search found 17036 results on 682 pages for 'mysql administrator'.

Page 357/682 | < Previous Page | 353 354 355 356 357 358 359 360 361 362 363 364  | Next Page >

  • Need help in Select query..

    - by Parth
    If I have four rows for a field with different values with other fields similar and then other four rows with same condition, as given below here u can see there different rows for insert with only difference in the "newvalue" and "field" excluding "id" for the table_name=jos_menu, operation=INSERT and live=0 now here what select query should be used to get only single row from the table on every change of table_name...??

    Read the article

  • AIR: sync gui with data-base?

    - by John Isaacks
    I am going to be building an AIR application that shows a list (about 1-25 rows of data) from a data-base. The data-base is on the web. I want the list to be as accurate as possible, meaning as soon as the data-base data changes, the list displayed in the app should update asap. I do not know of anyway that the air application could be notified when there is a change, I am thinking I am going to have to poll the data-base at certain intervals to keep an up to date list. So my question is, first is there any way to NOT have to keep checking the data-base? or if I do keep have to keep checking the data-base what is a reasonable interval to do that at? Thanks.

    Read the article

  • PHP auto refresh page without losing user input

    - by Tony
    I'm working on a PHP collaboration software project. I have a page that shows the latest updates from other users who are adding content to the database, but also has a form input to allow the user to enter text. I am currently using this code to refresh the page automatically every 30 seconds: header('Refresh: 30'); The problem is that the header code refreshes the entire page, and not just what is pulled from the database. Is there any PHP code that will just pull any new data from the database without refreshing the entire page? If someone could point me in the right direction I'd appreciate it.

    Read the article

  • Can a binary tree or tree be always represented in a Database as 1 table and self-referencing?

    - by Jian Lin
    I didn't feel this rule before, but it seems that a binary tree or any tree (each node can have many children but children cannot point back to any parent), then this data structure can be represented as 1 table in a database, with each row having an ID for itself and a parentID that points back to the parent node. That is in fact the classical Employee - Manager diagram: one boss can have many people under him... and each person can have n people under him, etc. This is a tree structure and is represented in database books as a common example as a single table Employee.

    Read the article

  • PHP - Select from database the same query

    - by How to PHP
    I created a table that contains the name of the user and his job, and created PHP page that shows me all the users that works doctor, I entered doctor into a variable then I selected from the table where Jobs equal to $doctor, that is great, but I need it to get the same Jobs into a table in the page and the other same jobs into a table in the same page. this is my code that shows only the users works doctor in one table, <html> <h1>Doctors</h1> </html> <?php mysql_connect('localhost','root',''); mysql_select_db('data'); $doctor='doctor'; $query= mysql_query("SELECT * FROM `users` WHERE `job` = '$doctor'")or die(mysql_error()); while ($arr = mysql_fetch_array($query)) $name= $arr['name']; echo $name; } ?> That shows me doctors when I put doctor in a variable I want to show all same Jobs in a table. Is there is a way to do this? Thanks :)

    Read the article

  • ob_start() is partially capturing data

    - by AAA
    I am using the following code: PHP: // Generate Guid function NewGuid() { $s = strtoupper(uniqid(rand(),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(); $alphabet = '123456789abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ'; function base_encode($num, $alphabet) { $base_count = strlen($alphabet); $encoded = ''; while ($num >= $base_count) { $div = $num/$base_count; $mod = ($num-($base_count*intval($div))); $encoded = $alphabet[$mod] . $encoded; $num = intval($div); } if ($num) $encoded = $alphabet[$num] . $encoded; return $encoded; } function base_decode($num, $alphabet) { $decoded = 0; $multi = 1; while (strlen($num) > 0) { $digit = $num[strlen($num)-1]; $decoded += $multi * strpos($alphabet, $digit); $multi = $multi * strlen($alphabet); $num = substr($num, 0, -1); } return $decoded; } ob_start(); echo base_encode($Guid, $alphabet); //should output: bUKpk $theid = ob_get_contents(); ob_get_clean(); The problem: When i echo $theid, it shows the complete entry, but as it is being inserted into the database, only the first entry in the sequence gets inserted, for example for the entry buKPK, only 'b' is being inserted not the rest.

    Read the article

  • How to display thumbnails with category by category

    - by shin
    http://pastie.org/962429 I have the above SQL. There are three categories, webdesign, logos and prints. I want to display them as following html. http://pastie.org/962432 Basically pulling thumbnail image and name from database category by category. I am not really sure how to do it. I will appreciate any inputs or help. Thanks in advance.

    Read the article

  • Remove duplicate records/objects uniquely identified by multiple attributes

    - by keruilin
    I have a model called HeroStatus with the following attributes: id user_id recordable_type hero_type (can be NULL!) recordable_id created_at There are over 100 hero_statuses, and a user can have many hero_statuses, but can't have the same hero_status more than once. A user's hero_status is uniquely identified by the combination of recordable_type + hero_type + recordable_id. What I'm trying to say essentially is that there can't be a duplicate hero_status for a specific user. Unfortunately, I didn't have a validation in place to assure this, so I got some duplicate hero_statuses for users after I made some code changes. For example: user_id = 18 recordable_type = 'Evil' hero_type = 'Halitosis' recordable_id = 1 created_at = '2010-05-03 18:30:30' user_id = 18 recordable_type = 'Evil' hero_type = 'Halitosis' recordable_id = 1 created_at = '2009-03-03 15:30:00' user_id = 18 recordable_type = 'Good' hero_type = 'Hugs' recordable_id = 1 created_at = '2009-02-03 12:30:00' user_id = 18 recordable_type = 'Good' hero_type = NULL recordable_id = 2 created_at = '2009-012-03 08:30:00' (Last two are not a dups obviously. First two are.) So what I want to do is get rid of the duplicate hero_status. Which one? The one with the most-recent date. I have three questions: How do I remove the duplicates using a SQL-only approach? How do I remove the duplicates using a pure Ruby solution? Something similar to this: http://stackoverflow.com/questions/2790004/removing-duplicate-objects. How do I put a validation in place to prevent duplicate entries in the future?

    Read the article

  • Unknown column even thoug it exits

    - by george
    I have SELECT servisler.geo_location, servisler.ADRES_MERKEZ, servisler.ADRES_ILCE, servisler.ADRES_IL, servisler.FIRMA_UNVANI, servisler.ADRES_ISTEL, servisler.YETKILI_ADISOYADI, urun_gruplari.GRUP_ADI FROM servisler INNER JOIN urun_gruplari ON kullanici_cihaz.URUN_GRUP_NO= urun_gruplari.RECNO INNER JOIN kullanici ON kullanici.SERVIS_RECNO = servisler.RECNO INNER JOIN kullanici_cihaz ON kullanici.RECNO = kullanici_cihaz.KUL_RECNO AND kullanici_cihaz.URUN_GRUP_NO = urun_gruplari.RECNO where kullanici.kullanici = 'MAR.EDI.003' but it says [Err] 1054 - Unknown column 'kullanici_cihaz.URUN_GRUP_NO' in 'on clause' enen though the column exits. What is its problem? schema Server version: 5.1.33-community-log

    Read the article

  • select query from mysql_num_rows

    - by Andi Nugroho
    i want create multiple search where statement $where_search is a multiple condition from post form. but stil error when iam using this code ".where_search." in where condition with mysql_num_rows for paging $tampil2 = mysql_query("SELECT * FROM bb where ".$where_search." and kd_kelompok='2' and kd_komoditi='11' and nm_sebutan IS NOT NULL " ); this is the complete code. $where_search = "kd_pok='2' and kd_komoditi='11' "; if (isset($_POST['lakpus'])) { if (empty($_POST['lakpus'])) { } else { if (empty($where_search)) { $where_search .= "lakpus = '$lakpus' "; } else { $where_search .= "AND lakpus = '$lakpus' "; } } } if (isset($_POST['kd_por'])) { $kd_por = $_POST['kd_por'] ; if (empty($_POST['kd_por'])) { } else { if (empty($where_search)) { $where_search .= "kd_por = '$kd_por' "; } else { $where_search .= "AND tab1.kd_por = '$kd_por' "; } } } $max=15; $tampil2 = mysql_query("SELECT * FROM bb where ".$where_search." and kd_kelompok='2' and kd_komoditi='11' and nm_sebutan IS NOT NULL " ); $jml = mysql_num_rows($tampil2); $jmlhal = ceil($jml/$max);

    Read the article

  • how to set a status

    - by ejah85
    hello guys..here i've a problem where i want to set the status whether it is approved or reject.. the condition are if admin select the registration number and driver name, that means the status is approve otherwise, if admin fill up the reason, that means the request is reject.. here is the code to set status if ($reason =='null'){ $query2 = "UPDATE usage SET status ='APPROVED' WHERE '$bookingno'=bookingno"; $result2 = @mysql_query($query2); } elseif (($regno =='null')&&($d_name =='null')) { $query3 = "UPDATE usage SET status ='REJECT' WHERE '$bookingno'=bookingno"; $result3 = @mysql_query($query3); } when i save the data, the status field are not updates..

    Read the article

  • Wordpress - Total User Count who only have posts

    - by knightrider
    I want to display total number of user who only have posts at Wordpress. I can get all users by this query <?php $user_count = $wpdb->get_var("SELECT COUNT(*) FROM $wpdb->users;"); echo $user_count ?> But for the user count only with posts, i think i might need to join another table, does anyone have snippets ? Thanks.

    Read the article

  • WordPress on other parts of my site

    - by SHiNKiROU
    I have a WordPress installation on my site, and I want to display WP posts on other parts of my site (that is outside the WP installation). How do I do that with PHP? I tried to search this type of question on Stack Overflow, Google and WP official site but I didn't find anything.

    Read the article

  • php - upload script mkdir saying file already exists when same directory even though different filename

    - by neeko
    my upload script says my file already exists when i try upload even though different filename <?php // Start a session for error reporting session_start(); ?> <?php // Check, if username session is NOT set then this page will jump to login page if (!isset($_SESSION['username'])) { header('Location: index.html'); } // Call our connection file include('config.php'); // Check to see if the type of file uploaded is a valid image type function is_valid_type($file) { // This is an array that holds all the valid image MIME types $valid_types = array("image/jpg", "image/JPG", "image/jpeg", "image/bmp", "image/gif", "image/png"); if (in_array($file['type'], $valid_types)) return 1; return 0; } // Just a short function that prints out the contents of an array in a manner that's easy to read // I used this function during debugging but it serves no purpose at run time for this example function showContents($array) { echo "<pre>"; print_r($array); echo "</pre>"; } // Set some constants // Grab the User ID we sent from our form $user_id = $_SESSION['username']; $category = $_POST['category']; // This variable is the path to the image folder where all the images are going to be stored // Note that there is a trailing forward slash $TARGET_PATH = "img/users/$category/$user_id/"; mkdir($TARGET_PATH, 0755, true); // Get our POSTed variables $fname = $_POST['fname']; $lname = $_POST['lname']; $contact = $_POST['contact']; $price = $_POST['price']; $image = $_FILES['image']; // Build our target path full string. This is where the file will be moved do // i.e. images/picture.jpg $TARGET_PATH .= $image['name']; // Make sure all the fields from the form have inputs if ( $fname == "" || $lname == "" || $image['name'] == "" ) { $_SESSION['error'] = "All fields are required"; header("Location: error.php"); exit; } // Check to make sure that our file is actually an image // You check the file type instead of the extension because the extension can easily be faked if (!is_valid_type($image)) { $_SESSION['error'] = "You must upload a jpeg, gif, or bmp"; header("Location: error.php"); exit; } // Here we check to see if a file with that name already exists // You could get past filename problems by appending a timestamp to the filename and then continuing if (file_exists($TARGET_PATH)) { $_SESSION['error'] = "A file with that name already exists"; header("Location: error.php"); exit; } // Lets attempt to move the file from its temporary directory to its new home if (move_uploaded_file($image['tmp_name'], $TARGET_PATH)) { // NOTE: This is where a lot of people make mistakes. // We are *not* putting the image into the database; we are putting a reference to the file's location on the server $imagename = $image['name']; $sql = "insert into people (price, contact, category, username, fname, lname, expire, filename) values (:price, :contact, :category, :user_id, :fname, :lname, now() + INTERVAL 1 MONTH, :imagename)"; $q = $conn->prepare($sql) or die("failed!"); $q->bindParam(':price', $price, PDO::PARAM_STR); $q->bindParam(':contact', $contact, PDO::PARAM_STR); $q->bindParam(':category', $category, PDO::PARAM_STR); $q->bindParam(':user_id', $user_id, PDO::PARAM_STR); $q->bindParam(':fname', $fname, PDO::PARAM_STR); $q->bindParam(':lname', $lname, PDO::PARAM_STR); $q->bindParam(':imagename', $imagename, PDO::PARAM_STR); $q->execute(); $sql1 = "UPDATE people SET firstname = (SELECT firstname FROM user WHERE username=:user_id1) WHERE username=:user_id2"; $q = $conn->prepare($sql1) or die("failed!"); $q->bindParam(':user_id1', $user_id, PDO::PARAM_STR); $q->bindParam(':user_id2', $user_id, PDO::PARAM_STR); $q->execute(); $sql2 = "UPDATE people SET surname = (SELECT surname FROM user WHERE username=:user_id1) WHERE username=:user_id2"; $q = $conn->prepare($sql2) or die("failed!"); $q->bindParam(':user_id1', $user_id, PDO::PARAM_STR); $q->bindParam(':user_id2', $user_id, PDO::PARAM_STR); $q->execute(); header("Location: search.php"); exit; } else { // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to // Make sure you chmod the directory to be writeable $_SESSION['error'] = "Could not upload file. Check read/write persmissions on the directory"; header("Location: error.php"); exit; } ?>

    Read the article

  • how to specify a BIGINT in a rails scaffold?

    - by webdestroya
    I am trying to create a model in ruby that uses a BIGINT datatype (as opposed to the INT done by :integer). I have search all over Google, but all I seem to find is "run an SQL statement to alter the table to a BIGINT" - This seems a bit hack-ish to me, so I wanted to know if there was a way to specify a bigint in the ruby system like :big_int or something Any ideas?

    Read the article

  • Need help with SQL Query

    - by StackOverflowNewbie
    Say I have 2 tables: Person - Id - Name PersonAttribute - Id - PersonId - Name - Value Further, let's say that each person had 2 attributes (say, gender and age). A sample record would be like this: Person->Id = 1 Person->Name = 'John Doe' PersonAttribute->Id = 1 PersonAttribute->PersonId = 1 PersonAttribute->Name = 'Gender' PersonAttribute->Value = 'Male' PersonAttribute->Id = 2 PersonAttribute->PersonId = 1 PersonAttribute->Name = 'Age' PersonAttribute->Value = '30' Question: how do I query this such that I get a result like this: 'John Doe', 'Male', '30'

    Read the article

  • Data Modeling Help - Do I add another table, change existing table's usage, or something else?

    - by StackOverflowNewbie
    Assume I have the following tables and relationships: Person - Id (PK) - Name A Person can have 0 or more pets: Pet - Id (PK) - PersonId (FK) - Name A person can have 0 or more attributes (e.g. age, height, weight): PersonAttribute _ Id (PK) - PersonId (FK) - Name - Value PROBLEM: I need to represent pet attributes, too. As it turns out, these pet attributes are, in most cases, identical to the attributes of a person (e.g. a pet can have an age, height, and weight too). How do I represent pet attributes? Do I create a PetAttribute table? PetAttribute Id (PK) PetId (FK) Name Value Do I change PersonAttribute to GenericAttribute and have 2 foreign keys in it - one connecting to Person, the other connecting to Pet? GenericAttribute Id (PK) PersonId (FK) PetId (FK) Name Value NOTE: if PersonId is set, then PetId is not set. If PetId is set, PersonId is not set. Do something else?

    Read the article

  • getting notice like undefined index

    - by user2533308
    $result = mysql_query("SELECT * FROM customers WHERE loginid='$_POST[login]' AND accpassword='$_POST[password]'"); if(mysql_num_rows($result) == 1) { while($recarr = mysql_fetch_array($result)) { $_SESSION[customerid] = $recarr[customerid]; $_SESSION[ifsccode] = $recarr[ifsccode]; $_SESSION[customername] = $recarr[firstname]. " ". $recarr[lastname]; $_SESSION[loginid] = $recarr[loginid]; $_SESSION[accstatus] = $recarr[accstatus]; $_SESSION[accopendate] = $recarr[accopendate]; $_SESSION[lastlogin] = $recarr[lastlogin]; } $_SESSION["loginid"] =$_POST["login"]; header("Location: accountalerts.php"); } else { $logininfo = "Invalid Username or password entered"; } Notice: Undefined index:login and Notice: Undefined index:password try to help me out getting error message in second line

    Read the article

  • Zend Framework Error:Invalid parameter number: no parameters were bound'

    - by roast_soul
    I'm using the Zend Frameworker 1.12. According to the help file, I used the Zend_Db_Statement to execute my sql. Below is my php code: $sql = "delete from options where id=?"; $stmt = new Zend_Db_Statement_Mysqli($this->getAdapter(), $sql); return $stmt->execute(array('1')); But the error is exception 'PDOException' with message 'SQLSTATE[HY093]: Invalid parameter number: no parameters were bound' in D:\Zend\workspaces\DefaultWorkspace.metadata.plugins\org.zend.php.framework.resource\resources\ZendFramework-1\library\Zend\Db\Statement\Mysqli.php:209 Stack trace: ......... ......... I googled for days, but nothing works. Any one know how to fix it?

    Read the article

  • Prevent two users from editing the same data

    - by Industrial
    Hi everyone, I have seen a feature in different web applications including Wordpress (not sure?) that warns a user if he/she opens an article/post/page/whatever from the database, while someone else is editing the same data simultaneously. I would like to implement the same feature in my own application and I have given this a bit of thought. Is the following example a good practice on how to do this? It goes a little something like this: 1) User A enters a the editing page for the mysterious article X. The database tableEvents is queried to make sure that no one else is editing the same page for the moment, which no one is by then. A token is then randomly being generated and is inserted into a database table called Events. 1) User B also want's to make updates to the article X. Now since our User A already is editing the article, the Events table is queried and looks like this: | timestamp | owner | Origin | token | ------------------------------------------------------------ | 1273226321 | User A | article-x | uniqueid## | 2) The timestamp is being checked. If it's valid and less than say 100 seconds old, a message appears and the user cannot make any changes to the requested article X: Warning: User A is currently working with this article. In the meantime, editing cannot be done. Please do something else with your life. 3) If User A decides to go on and save his changes, the token is posted along with all other data to update the database, and toggles a query to delete the row with token uniqueid##. If he decides to do something else instead of committing his changes, the article X will still be available for editing in 100 seconds for User B Let me know what you think about this approach! Wish everyone a great weekend!

    Read the article

  • [WEB] Local/Dev/Live deployment - best workflow

    - by Adam Kiss
    Hello, situation We our little company with 3 people, each has a localhost webserver and most projects (previous and current) are on one PC network shared disk. We have virtual server, where some of our clients' sites and our site. Our standard workflow is: Coder PC ? Programmer localhost ? dev domain (client.company.com) ? live version (client.com) It often happens, that there are two or three guys working on same projects at the same time - one is on dev version, two are on localhost. When finished, we try to synchronize the files on dev version and ideally not to mess (thanks ILMV:]) up any files, which *knock knock * doesn't happen often. And then one of us deploys dev version on live webserver. question we are looking for a way to simplify this workflow while updating websites - ideally some sort of diff uploader or VCS probably (Git/SVN/VCS/...), but we are not completely sure where to begin or what way would be ideal, therefore I ask you, fellow stackoverflowers for your experience with website / application deployment and recommended workflow. We probably will also need to use Mac in process, so if it won't be a problem, that would be even better. Thank you

    Read the article

  • How to get Joomla users data into a json array

    - by sami
    $sql = "SELECT * FROM `jos_users` LIMIT 0, 30 "; $response = array(); $posts = array(); $result=mysql_query($sql); while($row=mysql_fetch_array($result)) { $id=$row['id']; $id=$row['name']; $posts[] = array('id'=> $title, 'name'=> $name); } $response['jos_users'] = $posts; $fp = fopen('results.json', 'w'); fwrite($fp, json_encode($response)); fclose($fp); I want to fetch the user id and name to the json file.i thought id did wrong code.can anyone correct it ?

    Read the article

< Previous Page | 353 354 355 356 357 358 359 360 361 362 363 364  | Next Page >