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  • Tracking Votes and only allowing 1 vote per member

    - by MikeAdams
    What I'm trying to do is count the votes when someone votes on a "page". I think I lost myself trying to figure out how to track when a member votes or not. I can't seem to get the code to tell when a member has voted. //Generate code ID $useXID = intval($_GET['id']); $useXrank = $_GET['rank']; //if($useXrank!=null && $useXID!=null) { $rankcheck = mysql_query('SELECT member_id,code_id FROM code_votes WHERE member_id="'.$_MEMBERINFO_ID.'" AND WHERE code_id="'.$useXID.'"'); if(!mysql_fetch_array($rankcheck) && $useXrank=="up"){ $rankset = mysql_query('SELECT * FROM code_votes WHERE member_id="'.$_MEMBERINFO_ID.'"'); $ranksetfetch = mysql_fetch_array($rankset); $rankit = htmlentities($ranksetfetch['ranking']); $rankit+="1"; mysql_query("INSERT INTO code_votes (member_id,code_id) VALUES ('$_MEMBERINFO_ID','$useXID')") or die(mysql_error()); mysql_query("UPDATE code SET ranking = '".$rankit."' WHERE ID = '".$useXID."'"); } elseif(!mysql_fetch_array($rankcheck) && $useXrank=="down"){ $rankset = mysql_query('SELECT * FROM code_votes WHERE member_id="'.$_MEMBERINFO_ID.'"'); $ranksetfetch = mysql_fetch_array($rankset); $rankit = htmlentities($ranksetfetch['ranking']); $rankit-="1"; mysql_query("INSERT INTO code_votes (member_id,code_id) VALUES ('$_MEMBERINFO_ID','$useXID')") or die(mysql_error()); mysql_query("UPDATE code SET ranking = '".$rankit."' WHERE ID = '".$useXID."'"); } // hide vote links since already voted elseif(mysql_fetch_array($rankcheck)){$voted="true";} //}

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  • [SOLVED]Django - Passing variables to template based on db

    - by George 'Griffin
    I am trying to add a feature to my app that would allow me to enable/disable the "Call Me" button based on whether or not I am at [home|the office]. I created a model in the database called setting, it looks like this: class setting(models.Model): key = models.CharField(max_length=200) value = models.CharField(max_length=200) Pretty simple. There is currently one row, available, the value of it is the string True. I want to be able to transparently pass variables to the templates like this: {% if available %} <!-- Display button --> {% else %} <!-- Display grayed out button --> {% endif %} Now, I could add logic to every view that would check the database, and pass the variable to the template, but I am trying to stay DRY. What is the best way to do this? UPDATE I created a context processor, and added it's path to the TEMPLATE_CONTEXT_PROCESSORS, but it is not being passed to the template def available(request): available = Setting.objects.get(key="available") if open.value == "True": return {"available":True} else: return {} UPDATE TWO If you are using the shortcut render_to_response, you need to pass an instance of RequestContext to the function. from the django documentation: If you're using Django's render_to_response() shortcut to populate a template with the contents of a dictionary, your template will be passed a Context instance by default (not a RequestContext). To use a RequestContext in your template rendering, pass an optional third argument to render_to_response(): a RequestContext instance. Your code might look like this: def some_view(request): # ... return render_to_response('my_template.html', my_data_dictionary, context_instance=RequestContext(request)) Many thanks for all the help!

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  • Looping Through Database Query

    - by DrakeNET
    I am creating a very simple script. The purpose of the script is to pull a question from the database and display all answers associated with that particular question. We are dealing with two tables here and there is a foreign key from the question database to the answer database so answers are associated with questions. Hope that is enough explanation. Here is my code. I was wondering if this is the most efficient way to complete this or is there an easier way? <html> <head> <title>Advise Me</title> <head> <body> <h1>Today's Question</h1> <?php //Establish connection to database require_once('config.php'); require_once('open_connection.php'); //Pull the "active" question from the database $todays_question = mysql_query("SELECT name, question FROM approvedQuestions WHERE status = active") or die(mysql_error()); //Variable to hold $todays_question aQID $questionID = mysql_query("SELECT commentID FROM approvedQuestions WHERE status = active") or die(mysql_error()); //Print today's question echo $todays_question; //Print comments associated with today's question $sql = "SELECT commentID FROM approvedQuestions WHERE status = active"; $result_set = mysql_query($sql); $result_num = mysql_numrows($result_set); for ($a = 0; $a < $result_num; $a++) { echo $sql; } ?> </body> </html>

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  • Java: Do something on event in SQL Database?

    - by wretrOvian
    Hello I'm building an application with distributed parts. Meaning, while one part (writer) maybe inserting, updating information to a database, the other part (reader) is reading off and acting on that information. Now, i wish to trigger an action event in the reader and reload information from the DB whenever i insert something from the writer. Is there a simple way about this? Would this be a good idea? : // READER while(true) { connect(); // reload info from DB executeQuery("select * from foo"); disconnect(); }

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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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  • Only first word of two strings gets added to db

    - by dkgeld
    When trying to add words to a database via php, only the first word of both strings gets added. I send the text via this code: public void sendTextToDB() { valcom = editText1.getText().toString(); valnm = editText2.getText().toString(); t = new Thread() { public void run() { try { url = new URL("http://10.0.2.2/HB/hikebuddy.php?function=setcomm&comment="+valcom+"&name="+valnm); h = (HttpURLConnection)url.openConnection(); if( h.getResponseCode() == HttpURLConnection.HTTP_OK){ is = h.getInputStream(); }else{ is = h.getErrorStream(); } h.disconnect(); } catch (Exception e) { // TODO Auto-generated catch block e.printStackTrace(); Log.d("Test", "CONNECTION FAILED 1"); } } }; t.start(); } When tested with spaces and commas etc. in a browser, the php function adds all text. The strings also return the full value when inserted into a dialog. How do I fix this? Thank you.

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  • mysql_connect() doesn't work when run by apache; works from command line.

    - by Skeeter
    Hi, I have a strange issue. I'm trying to write a simple php webpage on my server, but mysql_connect() doesn't connect to any server, either local or otherwise. Here's where it gets strange. If I take the same php script and run it from the commandline, the script works. phpinfo() indicates that both the file (being run by apache) and the commandline (run as root) are calling the same version of php, mysql is loaded, and the php.ini is the same. Furthermore, I'm running a MediaWiki installation on this same server, and it's using the mysqld installed locally and works just fine, so I'm completely at a loss as to why the code isn't working. The error I receive on runtime: Can't connect to MySQL server on 'xxx.xxx.xxx.xxx' (13) (The IP is x'd out for the privacy of the owner of the server I'm connecting to)

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  • Stop Submit With empty and error Input Values Using PHP

    - by user3615781
    I am using the following code for sending data to database, but it sends the data even the values of the fields are incorrect or empty. So can anyone help me solve this by using php? Here is my code: <?php //Connecting to sql db $connect = mysqli_connect("localhost","root","","form"); /* check connection */ if (!$connect) { die('Connect Error: ' . mysqli_connect_error()); } //Sending data to sql db $result = mysqli_query($connect,"INSERT INTO students(name,email,website,comment,gender) VALUES('$name','$email','$website','$comment','$gender')"); if (!$result) { die('Query Error:'. mysqli_error($connect)); } ?>

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  • Why is Zend Framework (Zend_Db_table) rejecting this SQL Query?

    - by Michael T. Smith
    I'm working on a simple JOIN of two tables (urls and companies). I am using this query call: print $this->_db->select()->from(array('u' => 'urls'), array('id', 'url', 'company_id')) ->join(array('c' => 'companies'), 'u.company_id = c.id'); which is out putting this query: SELECT `u`.`id`, `u`.`url`, `u`.`company_id`, `c`.* FROM `urls` AS `u` INNER JOIN `companies` AS `c` ON u.company_id = c.id Now, I'd prefer the c.* to not actually appear, but either way it doesn't matter. ZF dies with this error: SQLSTATE[42000]: Syntax error or access violation: 1064 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 '' at line 1" but I can run that query perfectly fine in my MySQL CLI. Any ideas how to fix up this query?

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  • How to find intersect rows when condition depend on some columns in one table

    - by user3695637
    Table subscribe subscriber | subscribeto (columns) 1 | 5 1 | 6 1 | 7 1 | 8 1 | 9 1 | 10 2 | 5 2 | 6 2 | 7 There are two users that have id 1 and 2. They subscribe to various user and I inserted these data to table subscribe. Column subscriber indicates who is subscriber and column subscribeto indicates who they've subscribe to. From the above table can conclude that; user id=1 subscribed to 6 users user id=2 subscribed to 3 users I want to find manual of subscription (like Facebook is manual friends) user 1 subscribe to user 5,6,7,8,9,10 user 2 subscribe to user 5,6,7 So, Manual subscription of user 1 and 2 are: 5,6,7 And I'm trying to create SQL statement.. I give you user table for my SQL statement and I think we can use only subscribe table but I can't figure out. Table user userid (columns) 1 2 3 ... ... SQL "select * from user where (select count( 1 ) from subscribe where subscriber = '1' and subscribeto = user.userid) and (select count( 1 ) from subscribe where subscriber = '2' and subscribeto = user.userid);" This SQL can work correctly, but it very slow for thousands of columns. Please provide better SQL for me, Thanks.

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  • Explanation of NSMutableURLRequest : setHTTPMethod

    - by kkmoslehpour
    I am new to objective c programming. I am currently trying to insert data into mysql database. I have read a couple links on this and could not find the exact answer I am looking for. Here is what I am trying to do: I have my app that user inputs a name in a text field and once the press the add button, it makes a connection to my php file and my php file does the rest of the work (using POST method) and adds the name to the mysql dataase. I have seen a lot of people use NSMutableURLRequest : setHTTPMethod :POST`` in their code (in my case my php file does that and I don't haveNSMutableURLRequest` included in my code and it works perfectly fine.) My questions are: Are there any benefits of including NSMutableURLRequest in your code when you can just call your php code to take care of the POST for you? What does NSMutableURLRequest exactly do and which way is more efficient?

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  • How can I handle this kind of exceptions (in Doctrine)

    - by ppavlovic
    Can you tell me how can I handle this kind of exceptions: Fatal error: Uncaught exception 'Doctrine_Connection_Exception' with message 'PDO Connection Error: SQLSTATE[HY000] [2013] Lost connection to MySQL server at 'reading initial communication packet', system error: 110' in ... It happens when connection with MySQL is lost during query. I need to handle this exception so I can show 500 error page so the crawlers do not cache page, and to redirect user to appropriate "Try again" page. P.S. I have a lot's of code, so I can not go trough all code to put try/catch block. I need something simple and yet effective.

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  • Dependency Injection: How to pass DB around?

    - by Stephane
    Edit: This is a conceptual question first and foremost. I can make applications work without knowing this, but I'm trying to learn the concept. I've seen lots of videos with related classes and that makes sense, but when it comes to classes wrapping around other classes, I can't seem to grasp where things should be instantiated/passed around. =-=-=-=-=-=-= Question: Let's say I have a simple page that loads data from a table, manipulates the result and displays it. Simple. I'm going to use '=' for instantiating a class and '-' for passing a class in using constructor injection. It seems to me that the database has to be passed from one end of the application to the other which doesn't seem right. Here's how I would do it if I wanted to separate concerns: index =>Controller =>Model Layer =>Database =>DAO->Database I have this rule in my head that says I'm not supposed to create objects inside other objects. So what do I do with the Database? Or even the Model for that matter? I'm obviously missing something so basic about this. I would love a simplified example so that I can move forward in my code. I feel really hamstrung by this.

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  • Codeigniter : database configuration

    - by nppCods
    I am beginner for CI so don’t have good knowledge. My problem is: I am not using any database. All the records are fetched from JSON. Therefore I don’t think I need to configure database… As said by CI, it doesn’t necessarily ask for database. So my database.php configuration is : $active_group = ‘default’; $active_record = FALSE; $db[‘default’][‘hostname’] = ‘’; $db[‘default’][‘username’] = ‘’; $db[‘default’][‘password’] = ‘’; $db[‘default’][‘database’] = ‘’; $db[‘default’][‘dbdriver’] = ‘mysql’; $db[‘default’][‘dbprefix’] = ‘’; $db[‘default’][‘pconnect’] = TRUE; $db[‘default’][‘db_debug’] = TRUE; $db[‘default’][‘cache_on’] = FALSE; $db[‘default’][‘cachedir’] = ‘’; $db[‘default’][‘char_set’] = ‘utf8’; $db[‘default’][‘dbcollat’] = ‘utf8_general_ci’; $db[‘default’][‘swap_pre’] = ‘’; $db[‘default’][‘autoinit’] = TRUE; $db[‘default’][‘stricton’] = FALSE; It works in my local system but doesn’t work in live server. I am so surprised. Then I provided hostname, username, password and dbdriver, it works. My question is that, is that necessary to provide all the details if I am not using database? Thank you for your suggestion.

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  • Problem with user logins after db Restore

    - by JJgates
    I have two SQL 2005 instances that reside on different networks. I need to backup a database from instance A and restore it to a database in instance B on a weekly basis so that both databases hold the same data. After the restore, logins SIDS on database B are changed and therefore users can't log into database B and connection strings for the web application it supports are broken. Is there a work around for this? Thanks.

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  • SQL most popular

    - by Brae
    I have a mysql table with items in relation to their order. CREATE DATABASE IF NOT EXISTS `sqltest`; USE `sqltest`; DROP TABLE IF EXISTS `testdata`; CREATE TABLE `testdata` ( `orderID` varchar(10) DEFAULT NULL, `itemID` varchar(10) DEFAULT NULL, `qtyOrdered` int(10) DEFAULT NULL, `sellingPrice` decimal(10,2) DEFAULT NULL ) INSERT INTO `testdata`(`orderID`,`itemID`,`qtyOrdered`,`sellingPrice`) values ('1','a',1,'7.00'),('1','b',2,'8.00'),('1','c',3,'3.00'),('2','a',1,'7.00'),('2','c',4,'3.00'); Intended Result: A = (1+1)2 B = 2 C = (2+4)6 <- most popular How do I add up all the qty's for each item and result the highest one? It should be fairly strait forward but I'm new to SQL and I can't work this one out :S Solution needs to be mysql and or php. I guess there needs to be some sort of temporary tally variable for each item ID, but that seems like it could get messy with too many items.

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  • Same Data Appear only once.

    - by friendishan
    I have the following code which produces following output:- <? $tablaes = mysql_query("SELECT * FROM members where id='$order[user_id]'"); $user = mysql_fetch_array($tablaes); $idsd=$user['id']; $rPaid=mysql_query("SELECT SUM(`price`) AS total FROM order_history WHERE type!='rent_referral' AND date>'" . strtotime($time1) . "' AND date<'" . strtotime($time2) . "'"); $hdPaid = mysql_fetch_array($rPaid); $sPaid=mysql_query("SELECT SUM(`price`) AS total FROM order_history WHERE user_id='$idsd' AND type!='rent_referral' AND date>'" . strtotime($time1) . "' AND date<'" . strtotime($time2) . "'"); while ($hPaid = mysql_fetch_array($sPaid)) { ?> <td><?=$user['username']?></td> <td><?=$hPaid['total']?></td> <? } ?> </tr> It appears like this http://dl.dropbox.com/u/14384295/darrenan.jpg I want same data to appear only once.. Like Username: Vegas and price with him only once.

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  • PHP Localhost Application connect to server database

    - by Cross Vander
    I want to ask about how to connect my localhost application (C:xampp/htdocs/myproject) to database at my server host (www.someweb.somedomain)? Am I possible to do that? If it is, how to connect it at my php config? Right now my config (server.php) is: <?php $host = "my web IP public:3306"; $user = "root"; $pass = ""; $db = "dispatcherDB"; $conn = mysql_connect($host, $user, $pass) or die ("Cant connect to mySQL"); mysql_select_db($db); ?> what I got: Warning: mysql_connect(): No connection could be made because the target machine actively refused it. in C:\xampp\htdocs\XMS\server.php on line 7 So, what must I filled for $host? I'm trying using website IP, it still can't connect. Maybe there's someone here have experience at this problem? Sorry for my bad English

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  • Won't connect to the database

    - by user1657958
    I'm confused...I'm using the same code in a different document and in there it's not a problem to get a connection to the database. But in the new document it's just not working...(password, username, database name...all is checked and correct) :-/ <?php define ("DB_HOST", "db1234567.db.hello.com"); // set database host define ("DB_USER", "db1234567"); // set database user define ("DB_PASS","password123"); // set database password define ("DB_NAME","db1234567"); // set database name $link = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die("Couldn't make connection."); $db = mysql_select_db(DB_NAME, $link) or die("Couldn't select database"); ?> In the browser I get this: "Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'db1234567'@'123.123.12.12 (using password: YES) in /homepages/12/1234567/test/test.php on line 8 Couldn't make connection." Would be cool if someone could help me :) I'm not seeing any error... Thx!

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