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  • 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..

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  • 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?

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  • counting twice in a query, once using restrictions

    - by Andrew Heath
    Given the following tables: Table1 [class] [child] math boy1 math boy2 math boy3 art boy1 Table2 [child] [glasses] boy1 yes boy2 yes boy3 no If I want to query for number of children per class, I'd do this: SELECT class, COUNT(child) FROM Table1 GROUP BY class and if I wanted to query for number of children per class wearing glasses, I'd do this: SELECT Table1.class, COUNT(table1.child) FROM Table1 LEFT JOIN Table2 ON Table1.child=Table2.child WHERE Table2.glasses='yes' GROUP BY Table1.class but what I really want to do is: SELECT class, COUNT(child), COUNT(child wearing glasses) and frankly I have no idea how to do that in only one query. help?

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  • Using a nested group by statement or sub query to filter this result sets

    - by vivid-colours
    This question is a continuation of Changing this query to group rows and filter out all rows apart from the one with smallest value but with an extra bit at the end.... I have the following results set: 275 72.87368055555555555555555555555555555556 foo 70 275 72.87390046296296296296296296296296296296 foo 90 113 77.06431712962962962962962962962962962963 foo 80 113 77.07185185185185185185185185185185185185 foo 60 that I got from this query: SELECT id, (tbl2.date_modified - tbl1.date_submitted)/86400, some_value FROM tbl1, tbl2, tbl3 WHERE tbl1.id = tbl2.fid AND tbl1.id = tbl3.fid Notice there are 4 rows with 2 ids. I wanted to filter the rows to get only the minimum number in the second column. This fixed it: SELECT id, min((tbl2.date_modified - tbl1.date_submitted)/86400), max(some_value) FROM tbl1, tbl2, tbl3 WHERE tbl1.id = tbl2.fid AND tbl1.id = tbl3.fid GROUP BY tbl1.id so I got: 275 72.87368055555555555555555555555555555556 foo 70 113 77.06431712962962962962962962962962962963 foo 80 How can I change it to do the same but not include rows where the are other rows with some_value=90 ? I.e. 113 77.06431712962962962962962962962962962963 foo 80 I think I need some nested group or nested query ?! Many thanks :).

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  • Converting a certain SQL query into relational algebra

    - by Fumler
    Just doing an assignment for my database course and I just want to double check that I've correctly wrapped my head around relational algebra. The SQL query: SELECT dato, SUM(pris*antall) AS total FROM produkt, ordre WHERE ordre.varenr = produkt.varenr GROUP BY dato HAVING total >= 10000 The relational algebra: stotal >= 10000( ?R(dato, total)( sordre.varenr = produkt.varenr( datoISUM(pris*antall(produkt x ordre)))) Is this the correct way of doing it?

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  • to take values of checkbox in table attributes

    - by mwj
    i have a database patient with 3-4 tables n each table has about 8 attributes.... i have a table medical history which has attribute additional info ... under which i have 5 checkboxes.... all the values entered are taken up except the chekbox values..... plz help

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  • T_BOOLEAN_AND error?

    - by Ronnie Chester Lynwood
    whats wrong with this? anybody help me please.. if(stripos($nerde, $hf) !== false) && (stripos($nerde, $rs) !== false){ @mysql_query("update table set dltur = '3' where id = '".$ppl[id]."'"); } else { //dont do anything } i get T_BOOLEAN_AND error.

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  • Recalculate Counter Cache of 120k Records [Rails / ActiveRecord]

    - by Sebastian
    The following situation: I have a poi model, which has many pictures (1:n). I want to recalculate the counter_cache column, because the values are inconsistent. I've tried to iterate within ruby over each record, but this takes much too long and quits sometimes with some "segmentation fault" bugs. So i wonder, if its possible to do this with a raw sql query?

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  • 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?

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  • Php INNER JOING jqGrid help

    - by yanike
    I'm trying to get INNER JOIN to work with JQGRID, but I can't get it working. I want the code to get the first_name and last_name from members using the "efrom" from messages that matches the "id" from members. $col = array(); $col["title"] = "From"; $col["name"] = "messages.efrom"; $col["width"] = "70"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "First Name"; $col["name"] = "members.first_name"; $col["width"] = "80"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Last Name"; $col["name"] = "members.last_name"; $col["width"] = "80"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Subject"; $col["name"] = "messages.esubject"; $col["width"] = "300"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Date"; $col["name"] = "messages.edatetime"; $col["width"] = "150"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $g = new jqgrid(); $grid["sortname"] = 'messages.edatetime'; $g->select_command = "SELECT messages.efrom, messages.esubject, messages.edatetime, members.first_name, members.last_name FROM messages INNER JOIN members ON messages.efrom = members.id";

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  • Why is str_replace not replacing this string?

    - by Niall
    I have the following PHP code which should load the data from a CSS file into a variable, search for the old body background colour, replace it with the colour from a submitted form, resave the CSS file and finally update the colour in the database. The problem is, str_replace does not appear to be replacing anything. Here is my PHP code (stored in "processors/save_program_settings.php"): <?php require("../security.php"); $institution_name = mysql_real_escape_string($_POST['institution_name']); $staff_role_title = mysql_real_escape_string($_POST['staff_role_title']); $program_location = mysql_real_escape_string($_POST['program_location']); $background_colour = mysql_real_escape_string($_POST['background_colour']); $bar_border_colour = mysql_real_escape_string($_POST['bar_border_colour']); $title_colour = mysql_real_escape_string($_POST['title_colour']); $url = $global_variables['program_location']; $data_background = mysql_query("SELECT * FROM sents_global_variables WHERE name='background_colour'") or die(mysql_error()); $background_output = mysql_fetch_array($data_background); $css = file_get_contents($url.'/default.css'); $str = "body { background-color: #".$background_output['data']."; }"; $str2 = "body { background-color: #".$background_colour."; }"; $css2 = str_replace($str, $str2, $css); unlink('../default.css'); file_put_contents('../default.css', $css2); mysql_query("UPDATE sents_global_variables SET data='{$institution_name}' WHERE name='institution_name'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$staff_role_title}' WHERE name='role_title'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$program_location}' WHERE name='program_location'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$background_colour}' WHERE name='background_colour'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$bar_border_colour}' WHERE name='bar_border_colour'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$title_colour}' WHERE name='title_colour'") or die(mysql_error()); header('Location: '.$url.'/pages/start.php?message=program_settings_saved'); ?> Here is my CSS (stored in "default.css"): @charset "utf-8"; /* CSS Document */ body,td,th { font-family: Arial, Helvetica, sans-serif; font-size: 14px; color: #000; } body { background-color: #CCCCFF; } .main_table th { background:#003399; font-size:24px; color:#FFFFFF; } .main_table { background:#FFF; border:#003399 solid 1px; } .subtitle { font-size:20px; } input#login_username, input#login_password { height:30px; width:300px; font-size:24px; } input#login_submit { height:30px; width:150px; font-size:16px; } .timetable_cell_lesson { width:100px; font-size:10px; } .timetable_cell_tutorial_a, .timetable_cell_tutorial_b, .timetable_cell_break, .timetable_cell_lunch { width:100px; background:#999; font-size:10px; } I've run some checks using the following code in the PHP file: echo $css . "<br><br>" . $str . "<br><br>" . $str2 . "<br><br>" . $css2; exit; And it outputs (as you can see it's not changing anything in the CSS): @charset "utf-8"; /* CSS Document */ body,td,th { font-family: Arial, Helvetica, sans-serif; font-size: 14px; color: #000; } body { background-color: #CCCCFF; } .main_table th { background:#003399; font-size:24px; color:#FFFFFF; } .main_table { background:#FFF; border:#003399 solid 1px; } .subtitle { font-size:20px; } input#login_username, input#login_password { height:30px; width:300px; font-size:24px; } input#login_submit { height:30px; width:150px; font-size:16px; } .timetable_cell_lesson { width:100px; font-size:10px; } .timetable_cell_tutorial_a, .timetable_cell_tutorial_b, .timetable_cell_break, .timetable_cell_lunch { width:100px; background:#999; font-size:10px; } body { background-color: #CCCCFF; } body { background-color: #FF5719; } @charset "utf-8"; /* CSS Document */ body,td,th { font-family: Arial, Helvetica, sans-serif; font-size: 14px; color: #000; } body { background-color: #CCCCFF; } .main_table th { background:#003399; font-size:24px; color:#FFFFFF; } .main_table { background:#FFF; border:#003399 solid 1px; } .subtitle { font-size:20px; } input#login_username, input#login_password { height:30px; width:300px; font-size:24px; } input#login_submit { height:30px; width:150px; font-size:16px; } .timetable_cell_lesson { width:100px; font-size:10px; } .timetable_cell_tutorial_a, .timetable_cell_tutorial_b, .timetable_cell_break, .timetable_cell_lunch { width:100px; background:#999; font-size:10px; }

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  • 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 :)

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  • Need to map classes to different databases at runtime in Hibernate

    - by serg555
    I have MainDB database and unknown number (at compile time) of UserDB_1, ..., UserDB_N databases. MainDB contains names of those UserDB databases in some table (new UserDB can be created at runtime). All UserDB have exactly the same table names and fields. How to handle such situation in Hibernate? (database structure cannot be changed). Currently I am planning to create generic User classes not mapped to anything and just use native SQL for all queries: session.createSQLQuery("select * from " + db + ".user where id=1") .setResultTransformer(Transformers.aliasToBean(User.class)); Is there anything better I can do? Ideally I would want to have mappings for UserDB tables and relations and use HQL on required database.

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  • 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; } ?>

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  • User Getting Logged Out After Making First Comment

    - by John
    Hello, I am using a login system that works well. I am also using a comment system. The comment function does not show up unless the user is logged in (as shown in commentformonoff.php below). When a user makes a comment, the info is passed from the function "show_commentbox" to the file comments2a.php. Then, the info is passed to the file comments2.php. When the site is first pulled up on a browser, after logging in and making a comment, the user is logged out. After logging in a second time during the same browser session, the user is no longer logged out after making a comment. How can I keep the user logged in after making the first comment? Thanks in advance, John Commentformonoff.php: <?php if (!isLoggedIn()) { if (isset($_POST['cmdlogin'])) { if (checkLogin($_POST['username'], $_POST['password'])) { show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl); } else { echo "<div class='logintocomment'>Login to comment</div>"; } } else { echo "<div class='logintocomment'>Login to comment</div>"; } } else { show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl); } ?> Function "show_commentbox": function show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl) { echo '<form action="http://www...com/.../comments/comments2a.php" method="post"> <input type="hidden" value="'.$_SESSION['loginid'].'" name="uid"> <input type="hidden" value="'.$_SESSION['username'].'" name="u"> <input type="hidden" value="'.$submissionid.'" name="submissionid"> <input type="hidden" value="'.stripslashes($submission).'" name="submission"> <input type="hidden" value="'.$url.'" name="url"> <input type="hidden" value="'.$submittor.'" name="submittor"> <input type="hidden" value="'.$submissiondate.'" name="submissiondate"> <input type="hidden" value="'.$countcomments.'" name="countcomments"> <input type="hidden" value="'.$dispurl.'" name="dispurl"> <label class="addacomment" for="title">Add a comment:</label> <textarea class="checkMax" name="comment" type="comment" id="comment" maxlength="1000"></textarea> <div class="commentsubbutton"><input name="submit" type="submit" value="Submit"></div> </form> '; } Included in comments2a.php: $uid = mysql_real_escape_string($_POST['uid']); $u = mysql_real_escape_string($_POST['u']); $query = sprintf("INSERT INTO comment VALUES (NULL, %d, %d, '%s', NULL)", $uid, $subid, $comment); mysql_query($query) or die(mysql_error()); $lastcommentid = mysql_insert_id(); header("Location: comments2.php?submission=".$submission."&submissionid=".$submissionid."&url=".$url."&submissiondate=".$submissiondate."&comment=".$comment."&subid=".$subid."&uid=".$uid."&u=".$u."&submittor=".$submittor."&countcomments=".$countcomments."&dispurl=".$dispurl."#comment-$lastcommentid"); exit(); Included in comments2.php: if($_SERVER['REQUEST_METHOD'] == "POST"){header('Location: http://www...com/.../comments/comments2.php?submission='.$submission.'&submissionid='.$submissionid.'&url='.$url.'&submissiondate='.$submissiondate.'&submittor='.$submittor.'&countcomments='.$countcomments.'&dispurl='.$dispurl.'');} $uid = mysql_real_escape_string($_GET['uid']); $u = mysql_real_escape_string($_GET['u']);

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  • PHP While() Stop Looping

    - by Axel
    Hi, i have a php loop which displays only one record even if there is hundreds. here is the code: <?php $result1 = mysql_query("SELECT * FROM posts") or die(mysql_error()); $numexem = mysql_num_rows($result1); $s="0"; while($s<$numexem){ $postid=mysql_result($result1,$s,"id"); echo "Post id:".$postid; $result2 = mysql_query("SELECT * FROM pics WHERE postid='$postid'") or die(mysql_error()); $rows = mysql_fetch_array($result2) or die(mysql_error()); $pnum = mysql_num_rows($result2); echo " There is ".$pnum." Attached Pictures"; $s++; } ?> I'm wondering if the loop stop because there is other SQL query inside it or what? and i don't think so. Thanks

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  • 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);

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  • Compare structures of two databases?

    - by streetparade
    Hello, I wanted to ask whether it is possible to compare the complete database structure of two huge databases. We have two databases, the one is a development database, the other a production database. I've sometimes forgotten to make changes in to the production database, before we released some parts of our code, which results that the production database doesn't have the same structure, so if we release something we got some errors. Is there a way to compare the two, or synchronize?

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  • how to link table to table

    - by Niño Seymour L. Rodriguez
    I am a comsci student and I'm taking up database now. I got a problem in or should I say I dont know how to link table to table. It is not like you'll just use a foreign key and connect it to the primary key. The outcome should be like this: In the table Course there are three fields namely "course_id", "Description" and "subjects". When you click the name field Subject, a table named Subject should appear. Can you help me with this? hope you understnd my grammar, hehe..im not good in english......it will be a big help if you can answer it.........thank you po..............

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  • 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 ?

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  • Is it possible to LIMIT results from a JOIN query?

    - by Arms
    I've got a query that currently queries a Post table while LEFT JOINing a Comment table. It fetches all Posts and their respective Comments. However, I want to limit the number of Comments returned. I tried adding a sub-select, but ran into errors if I didn't LIMIT the results to 1. I'm really not sure how to go about this while still using only one query. Is this possible?

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  • current_date casting

    - by Armen Mkrtchyan
    Hi. string selectSql = "update " + table + " set state_" + mode + "_id=1 WHERE stoping_" + mode + " < current_date;"; when i call current_date, it return yyyy-MM-dd format, but i want to return dd.MM.yyyy format, how can i do that. please help. my program works fine when i am trying string selectSql = "update " + table + " set state_" + mode + "_id=1 WHERE stoping_" + mode + " < '16.04.2010';";

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  • Favouriting things in a database - most efficient method of keeping track?

    - by a2h
    I'm working on a forum-like webapp where I'd like to allow users to favourite an item so that they can keep track of it, and also so that others can see how many times an item's been favourited. The problem is, I'm unsure on the best practices for databases, which includes this situation. I have two ideas in my head on how to do this: Add an extra column to the user table and store things like so: "|2|5|73|" Add an extra table with at least two columns, one for referencing an item, the other for referencing a user. I feel uncomfortable about going for the second method as it involves an extra table, and potentially more queries would be required. Perhaps these beliefs aren't an issue, as I have little understanding of databases beyond simply working with table layouts and basic queries.

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  • PHP coding a price comparaison tool

    - by Tristan
    Hello, it's the first time I developp such tool you all know (the possibility to compare articles according to price and/or options) Since I never did that i want to tell me what do you think of the way i see that : On the database we would have : offer / price / option 1 / option 2 / option 3 / IDseller / IDoffer best buy / 15$ / full FTP / web hosting / php.ini / 10 / 1 .../..../.... And the request made by the client : "SELECT * FROM offers WHERE price <= 20 AND option1 = fullFTP"; I don't know if it seems OK to you. Plus i was wondering, how to avoid multiples entries for the same seller. Imagine you have multiple offers with a price <= 20 with the option FullFTP for the same seller, i don't want him to be shown 5 times on the comparator. If you have any advices ;) Thanks

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