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  • Best way to implement keywords for image upload gallery

    - by Dan Berlyoung
    I'm starting to spec out an image gallery type system similar to Facebook's. Members of the site will be able to create image galleries and upload images for others to view. Images will have keywords the the uploader can specify. Here's the question, what's the best way to model this? With image and keyword tables linked vi a HABTM relation? Or a single image table with the keywords saved as comma delimited values in a text field in the image record? Then search them using a LIKE or FULL TEXT index function? I want to be able to pull up all images containing a given keyword as well as generate a keyword cloud. I'm leaning toward the HABTM setup but I wanted to see what everyone else though. Thanks!!

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  • PHP / Zend Framework: Force prepend table name to column name in result array?

    - by Brian Lacy
    I am using Zend_Db_Select currently to retrieve hierarchical data from several joined tables. I need to be able to convert this easily into an array. Short of using a switch statement and listing out all the columns individually in order to sort the data, my thought was that if I could get the table names auto-prepended to the keys in the result array, that would solve my problem. So considering the following (assembled) SQL: SELECT user.*, contact.* FROM user INNER JOIN contact ON contact.user_id = user.user_id I would normally get a result array like this: [username] => 'bob', [contact_id] => 5, [user_id] => 2, [firstname] => 'bob', [lastname] => 'larsen' But instead I want this: [user.user_id] => 2, [user.username] => 'bob', [contact.contact_id] => 5, [contact.firstname] => 'bob', [contact.lastname] => 'larsen' Does anyone have an idea how to achieve this? Thanks!

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  • PHP error problem.

    - by TaG
    I get the following error on line 8: Undefined index: real_name which is $privacy_policy = mysqli_real_escape_string($mysqli, $_POST['privacy_policy']); I was wondering how can I fix this problem? Here is the PHP. if (isset($_POST['submitted'])) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT users.* FROM users WHERE user_id=3"); $privacy_policy = mysqli_real_escape_string($mysqli, $_POST['privacy_policy']); if (mysqli_num_rows($dbc) == 0) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"INSERT INTO users (user_id, privacy_policy) VALUES ('$user_id', '$privacy_policy')"); } if ($dbc == TRUE) { $dbc = mysqli_query($mysqli,"UPDATE users SET privacy_policy = '$privacy_policy' WHERE user_id = '$user_id'"); echo '<p class="changes-saved">Your changes have been saved!</p>'; } if (!$dbc) { print mysqli_error($mysqli); return; } } Here is the HTML. <form method="post" action="index.php"> <fieldset> <ul> <li><input type="checkbox" name="privacy_policy" id="privacy_policy" value="yes" <?php if (isset($_POST['privacy_policy'])) { echo 'checked="checked"'; } else if($privacy_policy == "yes") { echo 'checked="checked"'; } ?> /></li> <li><input type="submit" name="submit" value="Save Changes" class="save-button" /> <input type="hidden" name="submitted" value="true" /> <input type="submit" name="submit" value="Preview Changes" class="preview-changes-button" /></li> </ul> </fieldset> </form>

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  • Querying Last Entries group by DeviceId

    - by TTCG
    I have the database table logs as the following: I would like to extract the last entry of device, pollDate, status. For eg. deviceId, pollDate, status 1, 2010-95-06 10:53:28, 1 3, 2010-95-06 10:26:28, 1 I tried to run the following query but the distinct only selects the first records, not the latest SELECT DISTINCT deviceId, pollDate, status FROM logs GROUP By deviceId ORDER BY pollDate DESC So, could you please help me to extract the latest entries from the table? Thanks.

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  • How to add SQL elements to an array in PHP

    - by DanLeaningphp
    So this question is probably pretty basic. I am wanting to create an array from selected elements from a SQL table. I am currently using: $rcount = mysql_num_rows($result); for ($j = 0; $j <= $rcount; $j++) { $row = mysql_fetch_row($result); $patients = array($row[0] => $row[2]); } I would like this to return an array like this: $patients = (bob=>1, sam=>2, john=>3, etc...) Unfortunately, in its current form, this code is either copying nothing to the array or only copying the last element.

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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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  • 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

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  • Unique 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 updated version of the code: old php: // Generate Guid function NewGuid() { $s = strtoupper(md5(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(); echo $Guid; echo "<br><br><br>"; New PHP: // Generate Guid function NewGuid() { $s = strtoupper(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 the second (new php) code guarantee 100% uniqueness. Final 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(); echo $Guid; $alphabet = '123456789abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ'; function base_encode($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 $decoded; } echo base_encode($Guid, $alphabet); } So for more stronger uniqueness, i am using the $Guid as the key generator. That should be ok right?

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  • 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?

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  • PHP mysqli error return time

    - by Dori
    Hello. Can i ask a fundamental question. Why when I try to create a new mysqli object in php with invalid database infomation (say an incorrect database name) does it not return an error intstantly? I usually program server stuff in Java and something like this would throw back an error strait away, not after 20 seconds or so. For example $conn = new mysqli($host, $username, $password, $database); Thanks!

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  • Managing Foreign Keys

    - by jwzk
    So I have a database with a few tables. The first table contains the user ID, first name and last name. The second table contains the user ID, interest ID, and interest rating. There is another table that has all of the interest ID's. For every interest ID (even when new ones are added), I need to make sure that each user has an entry for that interest ID (even if its blank, or has defaults). Will foreign keys help with this scenario? or will I need to use PHP to update each and every record when I add a new key?

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  • Php mysqli and stored-procedure

    - by Mneva skoko
    I have a stored procedure: Create procedure news(in dt datetime,in title varchar(10),in desc varchar(200)) Begin Insert into news values (dt,title,desc); End Now my php: $db = new mysqli("","","",""); $dt = $_POST['date']; $ttl = $_POST['title']; $desc = $_POST['descrip']; $sql = $db-query("CALL news('$dt','$ttl','$desc')"); if($sql) { echo "data sent"; }else{ echo "data not sent"; } I'm new with php please help thank you

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  • cakephp pagination overide paginate() method

    - by islam
    iam using cakephp pagination and to paginate using custome query i override the paginate() method in my controller the problem is that every other action that use paginate() method not work any more how i can override this method without conflict with other that use this method from the same model

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  • Why does this query only select a single row?

    - by Joe
    SELECT * FROM tbl_houses WHERE (SELECT HousesList FROM tbl_lists WHERE tbl_lists.ID = '123') LIKE CONCAT('% ', tbl_houses.ID, '#') It only selects the row from tbl_houses of the last occuring tbl_houses.ID inside tbl_lists.HousesList I need it to select all the rows where any ID from tbl_houses exists within tbl_lists.HousesList

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  • Getting dynamic childs for a parent in SQL

    - by Islam
    I have a table called Categories which contains category_Id and parent_category_Id, so each category can has a child and the child can has a child and so on (it is dynamic). So if i have category A and category A has child B and child B has child C and child C has child D. I want to get all the child tree of A using SQL so when I give this query the id of A its result will be the ids of A's child which is B,C & D.....any ideas. Thanks in regards,

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  • Use Zip to Pre-Populate City/State Form with jQuery AJAX

    - by Paul
    I'm running into a problem that I can solve fine by just submitting a form and calling a db to retrieve/echo the information, but AJAX seems to be a bit different for doing this (and is what I need). Earlier in a form process I ask for the zip code like so: <input type="text" maxlength="5" size="5" id="zip" /> Then I have a button to continue, but this button just runs a javascript function that shows the rest of the form. When the rest of the form shows, I want to pre-populate the City input with their city, and pre-populate the State dropdown with their state. I figured I would have to find a way to set city/state to variables, and echo the variables into the form. But I can't figure out how to get/set those variables with AJAX as opposed to a form submit. Here's how I did it without ajax: $zip = mysql_real_escape_string($_POST['zip']); $q = " SELECT city FROM citystatezip WHERE zip = $zip"; $r = mysql_query($q); $row = mysql_fetch_assoc($r); $city = $row['city']; Can anybody help me out with using AJAX to set these variables? Thanks!

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  • 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?

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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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  • In SQL, in what situation do we want to Index a field in a table, or 2 fields in a table at the same

    - by Jian Lin
    In SQL, it is obvious that whenever we want to do a search on millions of record, say CustomerID in a Transactios table, then we want to add an index for CustomerID. Is another situation we want to add an index to a field when we need to do inner join or outer join using that field as a criteria? Such as Inner join on t1.custumerID = t2.customerID. Then if we don't have an index on customerID on both tables, we are looking at O(n^2) because we need to loop through the 2 tables sequentially. If we have index on customerID on both tables, then it becomes O( (log n) ^ 2 ) and it is much faster. Any other situation where we want to add an index to a field in a table? What about adding index for 2 fields combined in a table. That is, one index, for 2 fields together?

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  • Can you automatically create a mysqldump file that doesn't enforce foreign key constraints?

    - by Tai Squared
    When I run a mysqldump command on my database and then try to import it, it fails as it attempts to create the tables alphabetically, even though they may have a foreign key that references a table later in the file. There doesn't appear to be anything in the documentation and I've found answers like this that say to update the file after it's created to include: set FOREIGN_KEY_CHECKS = 0; ...original mysqldump file contents... set FOREIGN_KEY_CHECKS = 1; Is there no way to automatically set those lines or export the tables in the necessary order (without having to manually specify all table names as that can be tedious and error prone)? I could wrap those lines in a script, but was wondering if there is an easy way to ensure I can dump a file and then import it without manually updating it.

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  • checking if records exists in DB, in single step or 2 steps?

    - by Sinan
    Suppose you want to get a record from database which returns a large data and requires multiple joins. So my question would be is it better to use a single query to check if data exists and get the result if it exists. Or do a more simple query to check if data exists then id record exists, query once again to get the result knowing that it exists. Example: 3 tables a, b and ab(junction table) select * from from a, b, ab where condition and condition and condition and condition etc... or select id from a, b ab where condition then if exists do the query above. So I don't know if there is any reason to do the second. Any ideas how this affects DB performance or does it matter at all?

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  • PHP in Wordpress Posts - Is this okay?

    - by Thomas
    I've been working with some long lists of information and I've come up with a good way to post it in various formats on my wordpress blog posts. I installed the exec-PHP plugin, which allows you to run php in posts. I then created a new table (NEWTABLE) in my wordpress database and filled that table with names, scores, and other stuff. I was then able to use some pretty simple code to display the information in a wordpress post. Below is an example, but you could really do whatever you wanted. My question is - is there a problem with doing this? with security? or memory? I could just type out all the information in each post, but this is really much nicer. Any thoughts are appreciated. <?php $theResult = mysql_query("SELECT * FROM NEWTABLE WHERE Score < 100 ORDER BY LastName"); while($row = mysql_fetch_array($theResult)) { echo $row['FirstName']; echo " " . $row['LastName']; echo " " . $row['Score']; echo "<br />"; } ?>

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  • 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

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  • How to select DISTINCT rows without having the ORDER BY field selected

    - by JannieT
    So I have two tables students (PK sID) and mentors (PK pID). This query SELECT s.pID FROM students s JOIN mentors m ON s.pID = m.pID WHERE m.tags LIKE '%a%' ORDER BY s.sID DESC; delivers this result pID ------------- 9 9 3 9 3 9 9 9 10 9 3 10 etc... I am trying to get a list of distinct mentor ID's with this ordering so I am looking for the SQL to produce pID ------------- 9 3 10 If I simply insert a DISTINCT in the SELECT clause I get an unexpected result of 10, 9, 3 (wrong order). Any help much appreciated.

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