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  • how store date in myqsl database?

    - by Syom
    i have date in dd/mm/yyyy format. how can i store it in databse, if i fant to do some operations on them after? for example i must find out the rows, where date > something what type i must set to date field? thanks

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  • MULTIPLE CRITERIA TABLE JOIN

    - by user1447203
    I have a table listing clothing items (shirt, trousers, etc) named . Each item is identified with a unique CLOTHING.CLOTHING_ID. So a blue shirt is 01, a flowery shirt is 12 and jeans are 07 say. I have a second table identifying outfits with a column for shirts, for trousers, shoes etc. For example Outfit 1: shirt 01, trousers 07 (i.e. blue shirt with jeans) Outfit 2: shirt 12, trousers 07 (so flowery shirt with jeans). This table is named and each outfit is unique with OUTFIT_LIST.OUTFIT_ID. I want to produce a select statement that will list each outfit's contents, i.e. find the clothing specified in Outfit 1. Any help would be very much appreciated, and apologies in advance if I am missing a very simple solution. I have been playing with JOINS of all descriptions and CONCATS and so on with now luck - I am very new to this. Thanks.

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  • PHP: How to get the days of the week?

    - by fwaokda
    I'm wanting to store items in my database with a DATE value for a certain day. I don't know how to get the current Monday, or Tuesday, etc. from the current week. Here is my current database setup. menuentry id int(10) PK menu_item_id int(10) FK day_of_week date message varchar(255) So I have a class setup that holds all the info then I was going to do something like this... foreach ( $menuEntryArray as $item ) { if ( $item->getDate() == [DONT KNOW WHAT TO PUT HERE] ) { // code to display menu_item information } } So I'm just unsure what to put in "[DONT KNOW WHAT TO PUT HERE]" to compare to see if the date is specified for this week's Monday, or Tuesday, etc. The foreach above runs for each day of the week - so it'll look like this... Monday Item 1 Item 2 Item 3 Tuesday Item 1 Wednesday Item 1 Item 2 ... Thanks!

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  • Easy way to compute how close an auto_increment is to its maximum value?

    - by David M
    So yesterday we had a table that has an auto_increment PK for a smallint that reached its maximum. We had to alter the table on an emergency basis, which is definitely not how we like to roll. Is there an easy way to report on how close each auto_increment field that we use is to its maximum? The best way I can think of is to do a SHOW CREATE TABLE statement, parse out the size of the auto-incremented column, then compare that to the AUTO_INCREMENT value for the table. On the other hand, given that the schema doesn't change very often, should I store information about the columns' maximum values and get the current AUTO_INCREMENT with SHOW TABLE STATUS?

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  • How can I find all records for a model without doing a long list of "OR" conditions?

    - by gomezuk
    I'm having trouble composing a CakePHP find() which returns the records I'm looking for. My associations go like this: User -(has many)- Friends , User -(has many)- Posts I'm trying to display a list of all a user's friends recent posts, in other words, list every post that was created by a friend of the current user logged in. The only way I can think of doing this is by putting all the user's friends' user_ids in a big array, and then looping through each one, so that the find() call would look something like: $posts = $this->Post->find('all',array( 'conditions' => array( 'Post.user_id' => array( 'OR' => array( $user_id_array[0],$user_id_array[1],$user_id_array[2] # .. etc ) ) ) )); I get the impression this isn't the best way of doing things as if that user is popular that's a lot of OR conditions. Can anyone suggest a better alternative?

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  • login/logout problem in PHP

    - by user356170
    i have a question. I have a website in which i am giving security like login id and password( as usual). No what i want is that, 1) I don't want to allow a single user to login in different machine at the same time. 2) For this i am using a column in database which is keeping the current status of user(i.e. loging/logout). I am allowing user to login only when has session has not closed and status is login. 3) So my problem is that when i am logging out manually. it is closing the session as well as updating the database with status "logout". 4) but when i am closing the window from Cross buttonat top right corner. it is closing the ssion but table data is still "login". so later on i can't be able to login into the same user. 5) So how could i solve this problem. Please help me!

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  • TV Guide script - getting current date programmes to show

    - by whitstone86
    This is part of my TV guide script: //Connect to the database mysql_connect("localhost","root","PASSWORD"); //Select DB mysql_select_db("mytvguide"); //Select only results for today and future $result = mysql_query("SELECT programme, channel, episode, airdate, expiration, setreminder FROM mediumonair where airdate >= now()"); The episodes show up, so there are no issues there. However, it's getting the database to find data that's the issue. If I add a record for a programme that airs today this should show: Medium showing on TV4 8:30pm "Episode" Set Reminder Medium showing on TV4 May 18th - 6:25pm "Episode 2" Set Reminder Medium showing on TV4 May 18th - 10:25pm "Episode 3" Set Reminder Medium showing on TV4 May 19th - 7:30pm "Episode 3" Set Reminder Medium showing on TV4 May 20th - 1:25am "Episode 3" Set Reminder Medium showing on TV4 May 20th - 6:25pm "Episode 4" Set Reminder but this shows instead: Medium showing on TV4 May 18th - 6:25pm "Episode 2" Set Reminder Medium showing on TV4 May 18th - 10:25pm "Episode 3" Set Reminder Medium showing on TV4 May 19th - 7:30pm "Episode 3" Set Reminder Medium showing on TV4 May 20th - 1:25am "Episode 3" Set Reminder Medium showing on TV4 May 20th - 6:25pm "Episode 4" Set Reminder I almost have the SQL working; just not sure what the right code is here, to avoid the second mistake showing - as the record (which indicates a show currently airing) does not seem to work at present. Please can anyone help me with this? Thanks

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  • jQuery async ajax query and returning value problem (scope, closure)

    - by glebovgin
    Hi. Code not working because of async query and variable scope problem. I can't understand how to solve this. Change to $.ajax method with async:false - not an option. I know about closures, but how I can implement it here - don't know. I've seen all topics here about closures in js and jQuery async problems - but still nothing. Help, please. Here is the code: var map = null; var marker; var cluster = null; function refreshMap() { var markers = []; var markerImage = new google.maps.MarkerImage('/images/image-1_32_t.png', new google.maps.Size(32, 32)); $.get('/get_users.php',{},function(data){ if(data.status == 'error') return false; var users = data.users; // here users.length = 1 - this is ok; for(var i in users) { //here I have every values from users - ok var latLng = new google.maps.LatLng(users[i].lat, users[i].lng); var mark = new google.maps.Marker({ position: latLng, icon: markerImage }); markers.push(mark); alert(markers.length); // length 1 } },'json'); alert(markers.length); // length 0 //if I have alert() above - I get result cluster = new MarkerClusterer(map, markers, { maxZoom: null, gridSize: null }); } Thanks.

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  • where id = multiple artists

    - by pixel
    Any time there is an update within my music community (song comment, artist update, new song added, yadda yadda yadda), a new row is inserted in my "updates" table. The row houses the artist id involved along with other information (what type of change, time and date, etc). My users have a "favorite artists" section where they can do just that -- mark artists as their favorites. As such, I'd like to create a new feature that shows the user the changes made to their various favorite artists. How should I be doing this efficiently? SELECT * FROM table_updates WHERE artist_id = 1 and artist_id = 500 and artist_id = 60032 Keep in mind, a user could have 43,000 of our artists marked as a favorite. Thoughts?

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  • Find node level in a tree

    - by Álvaro G. Vicario
    I have a tree (nested categories) stored as follows: 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 I need to feed my client-side language (PHP) with node information (child+parent) so it can build the tree in memory. I can tweak my PHP code but I think the operation would be way simpler if I could just retrieve the rows in such an order that all parents come before their children. I could do that if I knew the level for each node: SELECT category_id, category_name, parent_id FROM category ORDER BY level -- No `level` column so far :( Can you think of a way (view, stored routine or whatever...) to calculate the node level? I guess it's okay if it's not real-time and I need to recalculate it on node modification.

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  • How can I fake sql data while preserving statements without commenting my server-side code?

    - by Fedor
    I have to use hardcoded values for certain fields because at this moment we don't have access to the real data. When we do get access, I don't want to go through a lot of work uncommenting. Is it possible to keep this statement the way it is, except use '25' as the alias for ratecode? IF(special.ratecode IS NULL, br.ratecode, special.ratecode) AS ratecode, I have about 8 or so IF statements similar to this and I'm just too lazy ( even with vim ) to re-append while commenting out each if statement line by line. I would have to do this: $sql = 'SELECT u.*,'; // IF ( special.ratecode IS NULL, br.ratecode, special.ratecode) AS ratecode $sql.= '25 AS ratecode';

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  • Is it possible to select data with max value for a column using Criteria in Hibernate?

    - by Palo
    Lets say I have the following mapping: <hibernate-mapping package="mypackage"> <class name="User" table="user"> <id name="id" column="id"> <generator class="native"></generator> </id> <property name="name" /> <property name="age" /> </class> </hibernate-mapping> Is it possible to select the oldest user (that is age is maximal) using Criteria in Hibernate? I can imagine that i could do this with 2 selects. (first select total number of users and then order the entries descending by age and select the first entry). But is it possible with a single select? Thank you Palo

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  • problem during data fetch

    - by nectar
    here is my code $sql="SELECT * FROM $tbl_name WHERE ownerId='$UserId'"; $result=mysql_query($sql,$link)or die(mysql_error()); $row = mysql_fetch_array($result, MYSQL_ASSOC); <?php while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo "<tr>"; echo "<td>".$row['pinId']."</td>"; echo "<td>".$row['usedby']."</td>"; echo "<td>".$row['status']."</td>"; echo "</tr>"; } ?> it is ignoring the first record means if 4 rows are in $row its ignoring the 1st one rest three are coming on page. ownerId is not primary key.

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  • How would I go about writing a conditional statement to check if visitor is coming from a particular

    - by Matthew
    Hello guys, What I have in mind is this... We are going to have people come from a particular site during a acquisition campaign and was wondering how I could conditionalize a certain section of my site to display a thank you message instead of the sign up form as they would have had the opportunity to fill this out before coming to my landing page. I have seen solutions like: $referal = mysql_real_escape_string($_SERVER['HTTP_REFERER']); I would like to know if this is the best way to get this to work??? - okay this is what i think might work. The third party website that is referring people to our landing page once the form on that site has been filled out can push into the record a hidden input value of "www.sample.com" or whatever... then I can have something check the for that particular value and fire off the conidtional. Does that even sound right?

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  • How to pass values from array into mysql with php

    - by moustafa
    my original code is this <tr> <th> <label for="user_level"> User Level: * <?php echo isset($valid_user_level) ? $valid_user_level : NULL; ?> </label> </th> </tr> <td> <select name="user_level" id="user_level" class="sel"> <option value="">Select one…</option> <option value="1">User</option> <option value="5">Admin</option> </select> </td> this give me the option to select one of choice from the drop down menu i.e. user and when user is selected and the submit button is pressed this will insert the value 1 into the database which will when the user logs in tell the system that they are are normal user. I want to change the code to the following <tr> <td> <select name="user_level" id="user_level" class="sel"> <option value="">Select one…</option> <?php if(!empty($level)) { foreach($level as $value) { echo "<option value='{$value}'"; echo getSticky(2,'user_level',$value); echo ">{$value}</option>"; } } ?> </select> </td> </tr> With this being my array query $level = array('User','Admin'); How can I pass the values of 1 for user level and 5 for admin in this code so when the user is selected it inouts 1 into the database?

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  • 2 IP are stored for a visitor : PROXY ?

    - by Tristan
    Hello, on my database i've decided to store IP of the visitors who answoers to polls. It's all working, but there is only 2 cases where not only 1 IP is stored, but there is 2 SAME ip for the same visitor MySQLL output (i replaced 2 numbers by XX) 10.188.XX.129, 10.188.XX.129 Here's the script to recieve the IP of the visitor : <?php function realip() { if (isset($_SERVER)) { if (isset($_SERVER["HTTP_X_FORWARDED_FOR"])) { $realip = $_SERVER["HTTP_X_FORWARDED_FOR"]; } elseif (isset($_SERVER["HTTP_CLIENT_IP"])) { $realip = $_SERVER["HTTP_CLIENT_IP"]; } else { $realip = $_SERVER["REMOTE_ADDR"]; } } else { if ( getenv( 'HTTP_X_FORWARDED_FOR' ) ) { $realip = getenv( 'HTTP_X_FORWARDED_FOR' ); } elseif ( getenv( 'HTTP_CLIENT_IP' ) ) { $realip = getenv( 'HTTP_CLIENT_IP' ); } else { $realip = getenv( 'REMOTE_ADDR' ); } } return $realip; } ? Thanks

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  • LinqtoSql Pre-compile Query problem with Count() on a group by

    - by Joe Pitz
    Have a LinqtoSql query that I now want to precompile. var unorderedc = from insp in sq.Inspections where insp.TestTimeStamp > dStartTime && insp.TestTimeStamp < dEndTime && insp.Model == "EP" && insp.TestResults != "P" group insp by new { insp.TestResults, insp.FailStep } into grp select new { FailedCount = (grp.Key.TestResults == "F" ? grp.Count() : 0), CancelCount = (grp.Key.TestResults == "C" ? grp.Count() : 0), grp.Key.TestResults, grp.Key.FailStep, PercentFailed = Convert.ToDecimal(1.0 * grp.Count() / tcount * 100) }; I have created this delegate: public static readonly Funct<SQLDataDataContext, int, string, string, DateTime, DateTime, IQueryable<CalcFailedTestResult>> GetInspData = CompiledQuery.Compile((SQLDataDataContext sq, int tcount, string strModel, string strTest, DateTime dStartTime, DateTime dEndTime, IQueryable<CalcFailedTestResult> CalcFailed) => from insp in sq.Inspections where insp.TestTimeStamp > dStartTime && insp.TestTimeStamp < dEndTime && insp.Model == strModel && insp.TestResults != strTest group insp by new { insp.TestResults, insp.FailStep } into grp select new { FailedCount = (grp.Key.TestResults == "F" ? grp.Count() : 0), CancelCount = (grp.Key.TestResults == "C" ? grp.Count() : 0), grp.Key.TestResults, grp.Key.FailStep, PercentFailed = Convert.ToDecimal(1.0 * grp.Count() / tcount * 100) }); The syntax error is on the CompileQuery.Compile() statement It appears to be related to the use of the select new {} syntax. In other pre-compiled queries I have written I have had to just use the select projection by it self. In this case I need to perform the grp.count() and the immediate if logic. I have searched SO and other references but cannot find the answer.

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  • PHP & HTML Purifier Error: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given

    - by TaG
    I'm trying to Integrate HTML Purifier http://htmlpurifier.org/ to filter my user submitted data but I get the following error below. And I was wondering how can I fix this problem? I get the following error. on line 22: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given line 22 is. if (mysqli_num_rows($dbc) == 0) { Here is the php code. if (isset($_POST['submitted'])) { // Handle the form. require_once '../../htmlpurifier/library/HTMLPurifier.auto.php'; $config = HTMLPurifier_Config::createDefault(); $config->set('Core.Encoding', 'UTF-8'); // replace with your encoding $config->set('HTML.Doctype', 'XHTML 1.0 Strict'); // replace with your doctype $purifier = new HTMLPurifier($config); $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT users.*, profile.* FROM users INNER JOIN contact_info ON contact_info.user_id = users.user_id WHERE users.user_id=3"); $about_me = mysqli_real_escape_string($mysqli, $purifier->purify($_POST['about_me'])); $interests = mysqli_real_escape_string($mysqli, $purifier->purify($_POST['interests'])); if (mysqli_num_rows($dbc) == 0) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"INSERT INTO profile (user_id, about_me, interests) VALUES ('$user_id', '$about_me', '$interests')"); } if ($dbc == TRUE) { $dbc = mysqli_query($mysqli,"UPDATE profile SET about_me = '$about_me', interests = '$interests' WHERE user_id = '$user_id'"); echo '<p class="changes-saved">Your changes have been saved!</p>'; } if (!$dbc) { // There was an error...do something about it here... print mysqli_error($mysqli); return; } }

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  • You can't specify target table 'b' for update in FROM clause

    - by elo
    I need to know where did i do wrong in this sql statement. I try to find solution in previous threads with similar problem but none can solve my problem. so i think maybe my statement is actually wrong. update table1 b left join table2 m on b.ICNO=m.ICNO set b.SalMoveMth = '01' where m.Status!='6' and (DATE_FORMAT(startDateSand,'%m')='10' or DATE_FORMAT(startDateSand,'%m')='11' or DATE_FORMAT(startDateSand,'%m')='12') and ((select SalMoveMth from table1 where ICNO=table2.ICNO order by SalMoveMthStDt desc limit 1)!='10'). Thank You.

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