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  • GdkEventKey. Key value question

    - by spajak
    Hi How to translate GdkEventKey.keyval with level 0 to key value with level == 0? For example: GDK_ISO_Left_Tab ==> GDK_Tab? Of course not using switch/case for every key. I need to get particular key, not a character.

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  • Using a user-defined type as a primary key

    - by Chris Kaminski
    Suppose I have a system where I have metadata such as: table: ====== key name address ... Then suppose I have a user-defined type described as so: datasource datasource-key A) are there systems where it's possible to have keys based on user-defined types? B) if so, how do you decompose the keys into a form suitable for querying? C) is this a case where I'm just better off with a composite primary key?

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  • Is foreign key reference from two different primary key from two different tables valid?

    - by arundex
    I have a foreign key that has to refer primary keys of two different tables. Table 1: animal animal_ id (primary key) Table 2: bird bird_ id (primary key) Table 3: Pet_info pet_id, type ENUM ('bird', 'animal') foreign key (pet_ id) references animal(animal_id), bird(bird_id) So, I need to check for pet_id either from animal or bird table depending on the need. Is this valid? Or should I go for some restructuring . . . NOTE: I referred this . . but I'm not sure whether I have to change my existing design

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  • MySQL ON DUPLICATE KEY UPDATE issue

    - by user644347
    Hi could some one look at this and tell me where I am going wrong. I have an SQL statement that when I echo using php I get this to screen INSERT INTO 'moviedb'.'genre' SET 'GenreID' = '18' , 'GenreName' = 'Drama' ON DUPLICATE KEY UPDATE 'GenreName' = 'Drama' WHERE 'GenreID' = '18' INSERT INTO 'moviedb'.'genre' SET 'GenreID' = '16' , 'GenreName' = 'Animation' ON DUPLICATE KEY UPDATE 'GenreName' = 'Animation' WHERE 'GenreID' = '16' And here is the statement $sql="INSERT INTO 'moviedb'.'genre' SET 'GenreID' = '{$genresID[$i]}' , 'GenreName' = '{$genreName[$i]}' ON DUPLICATE KEY UPDATE 'GenreName' = '{$genreName[$i]}' WHERE 'GenreID' = '{$genresID[$i]}'"; This is the error I recieve: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''moviedb'.'genre' SET 'GenreID' = '18' , 'GenreName' = 'Drama' ON DUPLICATE KEY ' at line 1 Any help would be greatly appreciated, thanks in advance.

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  • php Undefined index but i can access that key anyways

    - by Linas
    I just don't get it, i know what the error means i know how to fix it but it seems to be more of a php bug, i have an array returned from some function which looks like this when i use var_dump: array(1) { ["89512480-1884-11e2-aee6-618bca23e5b1"]=> array(5) { ["created"]=> string(10) "1350514509" ["expires"]=> string(10) "1351724109" ["token"]=> string(40) "81fa1895a6d5d4cb11efae8e71adf98fbdbfb5ca" ["user_agent"]=> string(40) "6ab18a2ea1d4b17d67713e416b07d46bf2f4d03b" ["username"]=> string(4) "test" } } array(0) { } but if i try for example: echo $token['89512480-1884-11e2-aee6-618bca23e5b1']['user_agent']; it shows me the value, but at the same time i get that annoying error, what the hell?

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  • Dealing with Android devices without MENU key

    - by hgpc
    I'm told by some users that my Android app is not usable because their device does not have a MENU key. What's the best way to deal with these devices? Is it possible to detect if the device lack a MENU key and show a menu button only in this case? And most importantly, how do you test this in the simulator? Thanks. Edit: Apparently the problem was that the users didn't know that the Menu key was called the Menu key.

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  • When to use a foreign key in MySQL

    - by Mel
    Is there official guidance or a threshold to indicate when it is best practice to use a foreign key in a MySQL database? Suppose you created a table for movies. One way to do it is to integrate the producer and director data into the same table. (movieID, movieName, directorName, producerName). However, suppose most directors and producers have worked on many movies. Would it be best to create two other tables for producers and directors, and use a foreign key in the movie table? When does it become best practice to do this? When many of the directors and producers are appearing several times in the column? Or is it best practice to employ a foreign key approach at the start? While it seems more efficient to use a foreign key, it also raises the complexity of the database. So when does the trade off between complexity and normalization become worth it? I'm not sure if there is a threshold or a certain number of cell repetitions that makes it more sensible to use a foreign key. I'm thinking about a database that will be used by hundreds of users, many concurrently. Many thanks!

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  • Merging two arrays in PHP

    - by Industrial
    Hi everyone, I am trying to create a new array from two current arrays. Tried array_merge, but it will not give me what I want. $array1 is a list of keys that I pass to a function. $array2 holds the results from that function, but doesn't contain any non-available resuls for keys. So, I want to make sure that all requested keys comes out with 'null':ed values, as according to the shown $result array. It goes a little something like this: $array1 = array('item1', 'item2', 'item3', 'item4'); $array2 = array( 'item1' => 'value1', 'item2' => 'value2', 'item3' => 'value3' ); Here's the result I want: $result = array( 'item1' => 'value1', 'item2' => 'value2', 'item3' => 'value3', 'item4' => '' ); It can be done this way, but I don't think that it's a good solution - I really don't like to take the easy way out and suppress PHP errors by adding @:s in the code. This sample would obviously throw errors since 'item4' is not in $array2, based on the example. foreach ($keys as $k => $v){ @$array[$v] = $items[$v]; } So, what's the fastest (performance-wise) way to accomplish the same result?

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  • How to catch key press on a form c# .net

    - by flavour404
    Hi, I have a parent form that contains a lot of controls. What I am trying to do is filter all of the key presses for that form. The trouble is that if the focus is on one of the controls on the form then the parent form is not getting the key press event, so how do I capture the key down event? Thanks, R.

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  • Map works, but throws a popup error at load complainig about wrong api key

    - by Filip
    Ok, I see a lot of people have this problem, but none answer I found either here or at stackoverflow. Problem: Map works super fine! But throws an error at load "This map needs a different api key... sign up at......" Already signed up, already got a key. On some forum post a guy told that in this case map loads twice, once with the wrong key, another with the right one. But Im sure my app loads it once and with the correct key. URL: http://ki.org.ua/ (it will redirect to /projects but I think that it aint a problem cuz i tested without redirection too) Thanks in advance for any suggestion or help.

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  • php weird bug where an array is not an array !

    - by iko
    I've been going mad trying to figure out why an array would not be an array in php. For a reason I can't understand I have a bug in a smarty class. The code is this : $compiled_tags = array(); for ($i = 0, $for_max = count($template_tags); $i < $for_max; $i++) { $this->_current_line_no += substr_count($text_blocks[$i], "\n"); // I tried array push instead to see // bug is here array_push($compiled_tags,$this->_compile_tag($template_tags[$i])); //$compiled_tags[] = $this->_compile_tag($template_tags[$i]); $this->_current_line_no += substr_count($template_tags[$i], "\n"); } the error message is Warning: array_push() expects parameter 1 to be array, integer given in .... OR before with [] Warning: Cannot use a scalar value as an array in .... I trying a var_debug on $compiled_tags and as soon I enter the for loop is not an array anymore but an integer. I tried renaming the variable, but same problem. I'm sure is something simple that I missed but I can't figure it out. Any help is (as always) welcomed !

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  • function.array-diff problems!

    - by SKY
    Hi, im currently getting these error on my site: Warning: array_keys() [function.array-keys]: The first argument should be an array on line 43 Warning: Invalid argument supplied for foreach() on line 44 Warning: array_diff() [function.array-diff]: Argument #1 is not an array on line 47 Warning: array_diff() [function.array-diff]: Argument #1 is not an array on line 48 And the source are: 42. $tmp = $this->network->get_user_follows($this->user->id); 43. $tmp = array_keys($tmp->followers); 44. foreach($tmp as &$v) { $v = intval($v); } 45. $tmp2 = array_keys($this->network->get_group_members($g->id)); 46. foreach($tmp2 as &$v) { $v = intval($v); } 47. $tmp = array_diff($tmp, $tmp2); 48. $tmp = array_diff($tmp, array(intval($this->user->id))); I want to know what is the problem and how i fix it. Thanks!

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  • Generate a valid array key from an URL string in PHP

    - by John Riche
    I have a PHP array with some predefined values: $aArray = array( 0 => 'value0', 1 => 'value1' ); I need to create a function where the string input will always return the same, valid, array key so that when I call: GiveMeAKey('http://www.google.com'); // May return 0 or 1 I receive always the same key (I don't care which one) from the array. Obvisously I can't store the relationship in a database and the string passed to the GiveMeAKey method can be any URL. I wonder if there is a way of doing that ?

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  • syntax for MySQL INSERT with an array of columns

    - by Mike_Laird
    I'm new to PHP and MySQL query construction. I have a processor for a large form. A few fields are required, most fields are user optional. In my case, the HTML ids and the MySQL column names are identical. I've found tutorials about using arrays to convert $_POST into the fields and values for INSERT INTO, but I can't get them working - after many hours. I've stepped back to make a very simple INSERT using arrays and variables, but I'm still stumped. The following line works and INSERTs 5 items into a database with over 100 columns. The first 4 items are strings, the 5th item, monthlyRental is an integer. $query = "INSERT INTO `$table` (country, stateProvince, city3, city3Geocode, monthlyRental) VALUES ( '$country', '$stateProvince', '$city3', '$city3Geocode', '$monthlyRental')"; When I make an array for the fields and use it, as follows: $colsx = array('country,', 'stateProvince,', 'city3,', 'city3Geocode,', 'monthlyRental'); $query = "INSERT INTO `$table` ('$colsx') VALUES ( '$country', '$stateProvince', '$city3', '$city3Geocode', '$monthlyRental')"; I get a MySQL error - check the manual that corresponds to your MySQL server version for the right syntax to use near ''Array') VALUES ( 'US', 'New York', 'Fairport, Monroe County, New York', '(43.09)' at line 1. I get this error whether the array items have commas inside the single quotes or not. I've done a lot of reading and tried many combinations and I can't get it. I want to see the proper syntax on a small scale before I go back to foreach expressions to process $_POST and both the fields and values are arrays. And yes, I know I should use mysql_real_escape_string, but that is an easy later step in the foreach. Lastly, some clues about the syntax for an array of values would be helpful, particularly if it is different from the fields array. I know I need to add a null as the first array item to trigger the MySQL autoincrement id. What else? I'm pretty new, so please be specific.

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  • inserts 'Array' into mysql table

    - by Noah Smith
    i want to insert an array into a mysql table. The array is produced by script scanning all the links, converting into absolute links and then displaying them in an array. i decided to mysql_query the array into the table but now i am stuck. it only posts 'Array', instead of every row from the array into a different row. Any ideas??! <?php require_once('simplehtmldom_1_5/simple_html_dom.php'); require_once('url_to_absolute/url_to_absolute.php'); $connect = mysql_connect("xxxx", "xxxx", "xxx") or die('Couldn\'t connect to MySQL Server: ' . mysql_error()); mysql_select_db("xxxx", $connect ) or die('Couldn\'t Select the database: ' . mysql_error( $connect )); $links = Array(); $URL = 'http://www.theqlick.com'; // change it for urls to grab // grabs the urls from URL $file = file_get_html($URL); foreach ($file->find('a') as $theelement) { $links[] = url_to_absolute($URL, $theelement->href); } print_r($links); mysql_query("INSERT INTO pages (url) VALUES ('$links[]')"); mysql_close($connect);

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