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  • PHP Sockets Errors (connection refused and No such file or directory)

    - by Purefan
    Hello all, I am writing a server app (broadcaster) and a client (relayer). Several relayers can connect to the broadcaster at the same time, send information and the broadcaster will redirect the message to a matching relayer (for example relayer1 sends to broadcaster who sends to relayer43, relayer2 - broadcaster - relayer73...) The server part is working as I have tested it with a telnet client and although its at this point only an echo server it works. Both relayer and broadcaster sit on the same server so I am using AF_UNIX sockets, both files are in different folders though. I have tried two approaches for the relayer and both have failed, the first one is using socket_create: public function __construct() { // where is the socket server? $this->_sHost = 'tcp://127.0.0.1'; $this->_iPort = 11225; // open a client connection $this->_hSocket = socket_create(AF_UNIX, SOCK_STREAM, 0); echo 'Attempting to connect to '.$this->_sHost.' on port '.$this->_iPort .'...'; $result = socket_connect($this->_hSocket, $this->_sHost, $this->_iPort); if ($result === false) { echo "socket_connect() failed.\nReason: ($result) " . socket_strerror(socket_last_error($this->_hSocket)) . "\n"; } else { echo "OK.\n"; } This returns "Warning: socket_connect(): unable to connect [2]: No such file or directory in relayer.class.php on line 27" and (its running from command line) it often also returns a segmentation fault. The second approach is using pfsockopen: public function __construct() { // where is the socket server? $this->_sHost = 'tcp://127.0.0.1'; $this->_iPort = 11225; // open a client connection $fp = pfsockopen ($this->_sHost, $this->_iPort, $errno, $errstr); if (!$fp) { $result = "Error: could not open socket connection"; } else { // get the welcome message fgets ($fp, 1024); // write the user string to the socket fputs ($fp, 'Message ' . __LINE__); // get the result $result .= fgets ($fp, 1024); // close the connection fputs ($fp, "END"); fclose ($fp); // trim the result and remove the starting ? $result = trim($result); $result = substr($result, 2); // now print it to the browser } which only returns the error "Warning: pfsockopen(): unable to connect to tcp://127.0.0.1:11225 (Connection refused) in relayer.class.php on line 33 " In all tests I have tried with different host names, 127.0.0.1, localhost, tcp://127.0.0.1, 192.168.0.199, tcp://192.168.0.199, none of it has worked. Any ideas on this?

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  • Error in connecting Eclipse to SQL Server

    - by user3721900
    This is the syntax error Jun 10, 2014 5:15:51 PM org.apache.catalina.core.AprLifecycleListener init INFO: The APR based Apache Tomcat Native library which allows optimal performance in production environments was not found on the java.library.path: C:\Program Files (x86)\Java\jre7\bin;C:\Windows\Sun\Java\bin;C:\Windows\system32;C:\Windows;C:/Program Files (x86)/Java/jre7/bin/client;C:/Program Files (x86)/Java/jre7/bin;C:/Program Files (x86)/Java/jre7/lib/i386;C:\Program Files (x86)\NVIDIA Corporation\PhysX\Common;C:\Program Files (x86)\Intel\iCLS Client\;C:\Program Files\Intel\iCLS Client\;C:\Program Files (x86)\AMD APP\bin\x86_64;C:\Program Files (x86)\AMD APP\bin\x86;C:\Program Files\Common Files\Microsoft Shared\Windows Live;C:\Program Files (x86)\Common Files\Microsoft Shared\Windows Live;C:\Windows\system32;C:\Windows;C:\Windows\System32\Wbem;C:\Windows\System32\WindowsPowerShell\v1.0\;C:\Program Files (x86)\Windows Live\Shared;C:\Program Files (x86)\ATI Technologies\ATI.ACE\Core-Static;C:\Program Files\Intel\Intel(R) Management Engine Components\DAL;C:\Program Files\Intel\Intel(R) Management Engine Components\IPT;C:\Program Files (x86)\Intel\Intel(R) Management Engine Components\DAL;C:\Program Files (x86)\Intel\Intel(R) Management Engine Components\IPT;C:\Program Files\Microsoft\Web Platform Installer\;C:\Program Files (x86)\Microsoft ASP.NET\ASP.NET Web Pages\v1.0\;C:\Program Files (x86)\Windows Kits\8.0\Windows Performance Toolkit\;C:\Program Files\Microsoft SQL Server\110\Tools\Binn\;C:\Program Files (x86)\Microchip\MPLAB C32 Suite\bin;C:\Program Files\Java\jdk1.7.0_25\bin;C:\Program Files (x86)\Java\jdk1.7.0_03\bin;c:\Program Files (x86)\Microsoft SQL Server\100\Tools\Binn\VSShell\Common7\IDE\;c:\Program Files (x86)\Microsoft SQL Server\100\Tools\Binn\;c:\Program Files\Microsoft SQL Server\100\Tools\Binn\;c:\Program Files (x86)\Microsoft SQL Server\100\DTS\Binn\;c:\Program Files\Microsoft SQL Server\100\DTS\Binn\;C:\Program Files (x86)\Google\google_appengine\;C:\Users\Patrick\Desktop\2013-2014 2nd Sem Files\Eclipsee\eclipse;;. Jun 10, 2014 5:15:51 PM org.apache.tomcat.util.digester.SetPropertiesRule begin WARNING: [SetPropertiesRule]{Server/Service/Engine/Host/Context} Setting property 'source' to 'org.eclipse.jst.jee.server:B2B' did not find a matching property. Jun 10, 2014 5:15:51 PM org.apache.coyote.AbstractProtocol init INFO: Initializing ProtocolHandler ["http-bio-8080"] Jun 10, 2014 5:15:51 PM org.apache.coyote.AbstractProtocol init INFO: Initializing ProtocolHandler ["ajp-bio-8009"] Jun 10, 2014 5:15:51 PM org.apache.catalina.startup.Catalina load INFO: Initialization processed in 544 ms Jun 10, 2014 5:15:51 PM org.apache.catalina.core.StandardService startInternal INFO: Starting service Catalina Jun 10, 2014 5:15:51 PM org.apache.catalina.core.StandardEngine startInternal INFO: Starting Servlet Engine: Apache Tomcat/7.0.42 Jun 10, 2014 5:15:52 PM org.apache.coyote.AbstractProtocol start INFO: Starting ProtocolHandler ["http-bio-8080"] Jun 10, 2014 5:15:52 PM org.apache.coyote.AbstractProtocol start INFO: Starting ProtocolHandler ["ajp-bio-8009"] Jun 10, 2014 5:15:52 PM org.apache.catalina.startup.Catalina start INFO: Server startup in 374 ms com.microsoft.sqlserver.jdbc.SQLServerException: Incorrect syntax near '`'. This is my code package b2b.fishermall; public class ConnectionString extends SqlStringCommands { public String getDriver(){ return "com.microsoft.sqlserver.jdbc.SQLServerDriver"; } public String getURL() { return "jdbc:sqlserver://localhost:1433;databaseName=B2B;integratedSecurity=true;"; } public String getUsername() { return ""; } public String getDbPassword() { return ""; } }

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  • Getting the responseText from XMLHttpRequest-Object

    - by Sammy46
    I wrote a cgi-script with c++ to return the query-string back to the requesting ajax object. I also write the query-string in a file in order to see if the cgi script works correctly. But when I ask in the html document for the response Text to be shown in a messagebox i get a blank message. here is my code: js: <script type = "text/javascript"> var XMLHttp; if(navigator.appName == "Microsoft Internet Explorer") { XMLHttp = new ActiveXObject("Microsoft.XMLHTTP"); } else { XMLHttp = new XMLHttpRequest(); } function getresponse () { XMLHttp.open ("GET", "http://localhost/cgi-bin/AJAXTest?" + "fname=" + document.getElementById('fname').value + "&sname=" + document.getElementById('sname').value,true); XMLHttp.send(null); } XMLHttp.onreadystatechange=function(){ if(XMLHttp.readyState == 4) { document.getElementById('response_area').innerHTML += XMLHttp.readyState; var x= XMLHttp.responseText alert(x) } } </script> First Names(s)<input onkeydown = "javascript: getresponse ()" id="fname" name="name"> <br> Surname<input onkeydown = "javascript: getresponse();" id="sname"> <div id = "response_area"> </div> C++: int main() { QFile log("log.txt"); if(!log.open(QIODevice::WriteOnly | QIODevice::Text)) { return 1; } QTextStream outLog(&log); QString QUERY_STRING= getenv("QUERY_STRING"); //if(QUERY_STRING!=NULL) //{ cout<<"Content-type: text/plain\n\n" <<"The Query String is: " << QUERY_STRING.toStdString()<< "\n"; outLog<<"Content-type: text/plain\n\n" <<"The Query String is: " <<QUERY_STRING<<endl; //} return 0; } I'm happy about every advice what to do! EDIT: the output to my logfile works just fine: Content-type: text/plain The Query String is: fname=hello&sname=world I just noticed that if i open it with IE8 i get the query-string. But only on the first "keydown" after that IE does nothing.

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  • PHP & MySQL - saving and looping problems.

    - by R.I.P.coalMINERS
    I'm new to PHP and MySQL I want a user to be able to store multiple names and there meanings in a MySQL database tables named names using PHP I will dynamically create form fields with JQuery every time a user clicks on a link so a user can enter 1 to 1,000,000 different names and there meanings which will be stored in a table called names. Since I asked my last question I figured out how to store my values from my form using the for loop but every time I loop my values when I add one or more dynamic fields the second form field named meaning will not save the value entered also my dynamic form fields keep looping doubling, tripling and so on the entered values into the database it all depends on how many form fields are added dynamically. I was wondering how can I fix these problems? On a side note I replaced the query with echo's to see the values that are being entered. Here is the PHP code. <?php if(isset($_POST['submit'])) { $mysqli = mysqli_connect("localhost", "root", "", "site"); $dbc = mysqli_query($mysqli,"SELECT * FROM names WHERE userID='$userID'"); $name = $_POST['name']; $meaning = $_POST['meaning']; if(isset($name['0']) && mysqli_num_rows($dbc) == 0 && trim($name['0'])!=='' && trim($meaning['0'])!=='') { for($n = 0; $n < count($name); $n++) { for($m = 0; $m < count($meaning); $m++) { echo $name[$n] . '<br />'; echo $meaning[$m] . '<br /><br />'; break; } } } } ?> And here is the HTML code. <form method="post" action="index.php"> <ul> <li><label for="name">Name: </label><input type="text" name="name[]" id="name" /></li> <li><label for="meaning">Meaning: </label><input type="text" name="meaning[]" id="meaning" /></li> <li><input type="submit" name="submit" value="Save" /></li> </ul> </form> If needed I will place the JQuery code.

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  • mysql_fetch_array() expects parameter 1 to be resource problem

    - by user225269
    I don't get it, I see no mistakes in this code but there is this error, please help: mysql_fetch_array() expects parameter 1 to be resource problem <?php $con = mysql_connect("localhost","root","nitoryolai123$%^"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("school", $con); $result = mysql_query("SELECT * FROM student WHERE IDNO=".$_GET['id']); ?> <?php while ($row = mysql_fetch_array($result)) { ?> <table class="a" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#D3D3D3"> <tr> <form name="formcheck" method="get" action="updateact.php" onsubmit="return formCheck(this);"> <td> <table border="0" cellpadding="3" cellspacing="1" bgcolor=""> <tr> <td colspan="16" height="25" style="background:#5C915C; color:white; border:white 1px solid; text-align: left"><strong><font size="2">Update Students</td> <tr> <td width="30" height="35"><font size="2">*I D Number:</td> <td width="30"><input name="idnum" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $_GET['id']; ?>"></td> </tr> <tr> <td width="30" height="35"><font size="2">*Year:</td> <td width="30"><input name="yr" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $row["YEAR"]; ?>"></td> <?php } ?> I'm just trying to load the data in the forms but I don't know why that error appears. What could possibly be the mistake in here?

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  • Parse error in PHP login form

    - by user225269
    I'm trying to have a login form in php. But my current code doesnt work. Here is the form: <form name="form1" method="post" action="loginverify.php"> <td><font size="3">Username:</td> <td></td> <td><input type="text" name="uname" value="" maxlength="15"/><br/></td> <td><font size="3">Password:</td> <td></td> <td><input type="text" name="pword" value="" maxlength="15"/><br/></td> <tr> <td>&nbsp;</td> <td>&nbsp;</td> <td><input type="submit" name="Submit" value="Login" /></td> </form> And the verify.php <?php session_start(); ?> <?php $host="localhost"; $username="root"; $password="nitoryolai123$%^"; $db_name="login"; $tbl="users"; $connection=mysql_connect($host, $username, $password) or die("cannot connect"); mysql_select_db($db_name, $connection) or die("cannot select db"); $user=$_POST['uname'] $pass=$_POST['pword'] $sql="SELECT Username, Password from users where Username='$user' and Password='$pass'"; $result=mysql_query[$sql]; $count=mysql_num_rows($result); if($count==1){ $SESSION['Username']=$user; echo"<a href='searchmain.php'> CONTINUE</a>"; } else{ echo"wrong username or password"; echo"<a href='loginform.php'>Back</a>"; } ?> Is there something wrong with my code. I get this parse error at line 15, which is this: $pass=$_POST['pword'] But when I try to remove it.It goes to line 16 or line 17 again. What do I do

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  • Page not redirecting properly(php)

    - by user225269
    I want to do the login page this way so that I won't be having trouble posting the username in the userpage. But everytime I try to access login.php. I get an error in firefox, that the page is not redirecting properly. What do I do? This works when I separate them into two. Into something like, login.php and verifylogin.php as the form action. But if I do it like this, I get redirection errors: <?php $host="localhost"; $username="root"; $password="nitoryolai123$%^"; $db_name="school"; $tbl_name="users"; mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $uname = mysql_real_escape_string($_POST['username']); $pword = mysql_real_escape_string($_POST['password']); $SQL = "SELECT * FROM users WHERE username = '$uname' AND password = '$pword'"; $result = mysql_query($SQL); $num_rows = mysql_num_rows($result); if ($result) { if ($num_rows > 0) { session_start(); $_SESSION['login'] = "1"; header ("Location: userpage.php"); } else { session_start(); $_SESSION['login'] = ""; header ("Location: login.php"); } } else { $errorMessage = "Error logging on"; } ?> <tr> <form name="form1" method="post" action="login.php"> <td> <table> <tr> <td><strong><font size="2">Login User</strong></td> </tr> <tr> <td width="30" height="35"><font size="2">Username:</td> <td width="30"><input name="username" type="text" id="username" maxlength="17"></td> </tr> <tr> <td width="30" height="35" ><font size="2">Password:</td> <td width="30"><input name="password" type="password" id="password" maxlength="17"></td> </tr> <td><td align="right" width="30"><input type="submit" name="Submit" value="Submit" /></td> <td><input type="reset" name="Reset" value="Reset"></td></td> </tr> </form> please help, thanks.

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  • C# ApplicationContext usage

    - by rd42
    Apologies if my terminology is off, I'm new to C#. I'm trying to use an ApplicationContext file to store mysql conn values, like dbname, username, password. The class with mysql conn string is "using" the namespace for the ApplicationContext, but when I print out the connection string, the values are making it. A friend said, "I'm not initializing it" but couldn't stay to expand on what "it" was. and the "Console.WriteLine("1");" in ApplicationContext.cs never shows up. Do I need to create an ApplicationContext object and the call Initialize() on that object? Thanks for any help. ApplicationContext.cs: namespace NewApplication.Context { class ApplicationContext { public static string serverName; public static string username; public static string password; public static void Initialize() { //need to read through config here try { Console.WriteLine("1"); XmlDocument xDoc = new XmlDocument(); xDoc.Load(".\\Settings.xml"); XmlNodeList serverNodeList = xDoc.GetElementsByTagName("DatabaseServer"); XmlNodeList usernameNodeList = xDoc.GetElementsByTagName("UserName"); XmlNodeList passwordNodeList = xDoc.GetElementsByTagName("Password"); } catch (Exception ex) { // MessageBox.Show(ex.ToString()); //TODO: Future write to log file username = "user"; password = "password"; serverName = "localhost"; } } } } MySQLManager.cs: note: dbname is the same as the username as you'll see in the code, I copied this from a friend who does that. using System; using System.Collections.Generic; using System.Linq; using System.Text; using MySql.Data; using MySql.Data.MySqlClient; using NewApplication.Context; namespace NewApplication.DAO { class MySQLManager { private static MySqlConnection conn; public static MySqlConnection getConnection() { if (conn == null || conn.State == System.Data.ConnectionState.Closed) { string connStr = "server=" + ApplicationContext.serverName + ";user=" + ApplicationContext.username + ";database=" + ApplicationContext.username + ";port=3306;password=" + ApplicationContext.password + ";"; conn = new MySqlConnection(connStr); try { Console.WriteLine("Connecting to MySQL... "); Console.WriteLine("Connection string: " + connStr + "\n"); conn.Open(); // Perform databse operations // conn.Close(); } catch (Exception ex) { Console.WriteLine(ex.ToString()); } } return conn; } } } and, thanks for still reading, this is the code that uses the two previous files: class LogDAO { MySqlConnection conn; public LogDAO() { conn = MySQLManager.getConnection(); } Thank you, rd42

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  • Criticise/Recommendations for my code

    - by aLk
    Before i go any further it would be nice to know if there is any major design flaws in my program so far. Is there anything worth changing before i continue? Model package model; import java.sql.*; import java.util.*; public class MovieDatabase { @SuppressWarnings({ "rawtypes", "unchecked" }) public List queryMovies() throws SQLException { Connection connection = null; java.sql.Statement statement = null; ResultSet rs = null; List results = new ArrayList(); try { DriverManager.registerDriver(new com.mysql.jdbc.Driver()); connection = DriverManager.getConnection("jdbc:mysql://localhost:3306/test", "root", "password"); statement = connection.createStatement(); String query = "SELECT * FROM movie"; rs = statement.executeQuery(query); while(rs.next()) { MovieBean bean = new MovieBean(); bean.setMovieId(rs.getInt(1)); bean.setTitle(rs.getString(2)); bean.setYear(rs.getInt(3)); bean.setRating(rs.getInt(4)); results.add(bean); } } catch(SQLException e) { } return results; } } Servlet public class Service extends HttpServlet { @SuppressWarnings("rawtypes") protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { response.setContentType("text/html"); PrintWriter out = response.getWriter(); out.println("Movies!"); MovieDatabase movies = new MovieDatabase(); try { List results = movies.queryMovies(); Iterator it = results.iterator(); while(it.hasNext()) { MovieBean movie = new MovieBean(); movie = (MovieBean)it.next(); out.println(movie.getYear()); } } catch(SQLException e) { } } } Bean package model; @SuppressWarnings("serial") public class MovieBean implements java.io.Serializable { protected int movieid; protected int rating; protected int year; protected String title; public MovieBean() { } public void setMovieId(int movieidVal) { movieid = movieidVal; } public void setRating(int ratingVal) { rating = ratingVal; } public void setYear(int yearVal) { year = yearVal; } public void setTitle(String titleVal) { title = titleVal; } public int getMovieId() { return movieid; } public int getRating() { return rating; } public int getYear() { return year; } public String getTitle() { return title; } }

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  • php web services not getting data from iphone application

    - by user317192
    Hi, I am connecting with a php web service from my iphone application, I am doing a simple thing i.e. 1. Getting user inputs for: username password in a text field from the iphone form and sending the same to the PHP Post request web service. At the web service end I receive nothing other than blank fields that are inserted into the MySQL Database....... The code for sample web service is: ***********SAMPLE CODE FOR WEB SERVICES***** mysql_select_db("eventsfast",$con); $username = $_REQUEST['username']; $password = $_REQUEST['password']; echo $username; echo $password; $data = $_REQUEST; $fp = fopen("log.txt", "w"); fwrite($fp, $data['username']); fwrite($fp, $data['password']); $sql="INSERT INTO users(username,password) VALUES('{$username}','{$password}')"; if(!mysql_query($sql,$con)) { die('Error:'.mysql_error()); } echo json_encode("1 record added to users table"); mysql_close($con); echo "test"; ? ***************PHP******** ****** **************IPHONE EVENT CODE******* import "postdatawithphpViewController.h" @implementation postdatawithphpViewController @synthesize userName,password; -(IBAction) postdataid) sender { NSLog(userName.text); NSLog(password.text); NSString * dataTOB=[userName.text stringByAppendingString:password.text]; NSLog(dataTOB); NSData * postData=[dataTOB dataUsingEncoding:NSUTF8StringEncoding allowLossyConversion:YES]; NSString *postLength = [NSString stringWithFormat:@"%d", [postData length]]; NSLog(postLength); NSMutableURLRequest *request = [[[NSMutableURLRequest alloc] init] autorelease]; NSURL *url = [NSURL URLWithString:[NSString stringWithFormat:@"http://localhost:8888/write.php"]]; [request setURL:url]; [request setHTTPMethod:@"POST"]; [request setValue:postLength forHTTPHeaderField:@"Content-Length"]; [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"]; [request setHTTPBody:postData]; NSURLResponse *response; NSError *error; [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error]; if(error==nil) NSLog(@"Error is nil"); else NSLog(@"Error is not nil"); NSLog(@"success!"); } Please help.............

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  • How to upload a file into database by using Servlet?

    - by user1765496
    Hi all iam working on servlets, so i need to upload a file by using servlet as follows my code. package com.limrasoft.image.servlets; import java.io.*; import javax.servlet.*; import javax.servlet.http.*; import javax.servlet.annotation.*; import java.sql.*; @WebServlet(name="serv1",value="/s1") public class Account extends HttpServlet{ public void doPost(HttpServletRequest req,HttpServletResponse res)throws ServletException,IOException{ try{ Class.forName("oracle.jdbc.driver.OracleDriver"); Connecection con=null; try{ con=DriverManager.getConnection("jdbc:oracle:thin:@localhost:1521:xe","system","sajid"); PrintWriter pw=res.getWriter(); res.setContentType("text/html"); String s1=req.getParameter("un"); string s2=req.getParameter("pwd"); String s3=req.getParameter("g"); String s4=req.getParameter("uf"); PreparedStatement ps=con.prepareStatement("insert into account(?,?,?,?)"); ps.setString(1,s1); ps.setString(2,s2); ps.setString(3,s3); File file=new File("+s4+"); FileInputStream fis=new FileInputStream(fis); int len=(int)file.length(); ps.setBinaryStream(4,fis,len); int c=ps.executeUpdate(); if(c==0){pw.println("<h1>Registratin fail");} else{pw.println("<h1>Registration fail");} } finally{if(con!=null)con.close();} } catch(ClassNotFoundException ce){pw.println("<h1>Registration Fail");} catch(SQLException se){pw.println("<h1>Registration Fail");} pw.flush(); pw.close(); } } I have written the above code for file upload into database, but it giving error as "HTTP Status 500 - Servlet3.java (The system cannot find the file specified)" Could you plz help me to do this code,thanks in advanse.

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  • pdo connection scope

    - by Scarface
    Hey guys I have a connection class I found for pdo. I am calling the connection method on the page that the file is included on. The problem is that within functions the $conn variable is not defined even though I stated the method was public (bare with me I am very new to OOP), and I was wondering if anyone had an elegant solution other then using global in every function. Any suggestions are greatly appreciated. CONNECTION class PDOConnectionFactory{ // receives the connection public $con = null; // swich database? public $dbType = "mysql"; // connection parameters // when it will not be necessary leaves blank only with the double quotations marks "" public $host = "localhost"; public $user = "user"; public $senha = "password"; public $db = "database"; // arrow the persistence of the connection public $persistent = false; // new PDOConnectionFactory( true ) <--- persistent connection // new PDOConnectionFactory() <--- no persistent connection public function PDOConnectionFactory( $persistent=false ){ // it verifies the persistence of the connection if( $persistent != false){ $this->persistent = true; } } public function getConnection(){ try{ // it carries through the connection $this->con = new PDO($this->dbType.":host=".$this->host.";dbname=".$this->db, $this->user, $this->senha, array( PDO::ATTR_PERSISTENT => $this->persistent ) ); // carried through successfully, it returns connected return $this->con; // in case that an error occurs, it returns the error; }catch ( PDOException $ex ){ echo "We are currently experiencing technical difficulties. We have a bunch of monkies working really hard to fix the problem. Check back soon: ".$ex->getMessage(); } } // close connection public function Close(){ if( $this->con != null ) $this->con = null; } } PAGE USED ON include("includes/connection.php"); $db = new PDOConnectionFactory(); $conn = $db->getConnection(); function test(){ try{ $sql = 'SELECT * FROM topic'; $stmt = $conn->prepare($sql); $result=$stmt->execute(); } catch(PDOException $e){ echo $e->getMessage(); } } test();

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  • php code is not fetching data from mysql database using wamp server

    - by john
    I want to display a table from database in phpMyAdmin by putting the following conditions that in every different options in drop down menu it displays different table from database by pressing the button of search. But it is not doing so. <p class="h2">Quick Search</p> <div class="sb2_opts"> <p></p> <form method="post" action="" > <p>Enter your source and destination.</p> <p>From:</p> <select name="from"> <option value="Islamabad">Islamabad</option> <option value="Lahore">Lahore</option> <option value="murree">Murree</option> <option value="Muzaffarabad">Muzaffarabad</option> </select> <p>To:</p> <select name="To"> <option value="Islamabad">Islamabad</option> <option value="Lahore">Lahore</option> <option value="murree">Murree</option> <option value="Muzaffarabad">Muzaffarabad</option> </select> <input type="submit" value="search" /> </form> </form> </table> <?php $con=mysqli_connect("localhost","root","","test"); if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } if(isset($_POST['from']) and isset($_POST['To'])) { $from = $_POST['from'] ; $to = $_POST['To'] ; $table = array($from, $to); switch ($table) { case array ("Islamabad", "Lahore") : $result = mysqli_query($con,"SELECT * FROM flights"); echo "</flights>"; //table name is flights break; case array ("Islamabad", "Murree") : $result = mysqli_query($con,"SELECT * FROM isb to murree"); echo "</isb to murree>"; //table name isb to murree ; break; case array ("Islamabad", "Muzaffarabad") : $result = mysqli_query($con,"SELECT * FROM isb to muzz"); echo "</isb to muzz>"; break; //..... //...... default: echo "Your choice is nor valid !!"; } } mysqli_close($con); ?>

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  • How do I create links in the cells of a PHP generated table?

    - by typoknig
    I have a table generated from some PHP code that lists a SMALL amount of important information for employees. I want to make it so each row, or at least one element in each row can be clicked on so the user will be redirected to ALL of the information (pulled from MySQL database) related to the employee who was clicked on. I am not sure how would be the best way to go about this, but I am open to suggestions. I would like to stick to PHP and/or JavaScript. Below is the code for my table: <table> <tr> <td id="content_heading" width="25px">ID</td> <td id="content_heading" width="150px">Last Name</td> <td id="content_heading" width="150px">First Name</td> <td id="content_heading" width="75px">SSN</td> </tr> <?php $user = 'user'; $pass = 'pass'; $server = 'localhost'; $link = mysql_connect($server, $user, $pass); if (!$link){ die('Could not connect to database!' . mysql_error()); } mysql_select_db('mydb', $link); $query = "SELECT * FROM employees"; $result = mysql_query($query); mysql_close($link); $num = mysql_num_rows($result); for ($i = 0; $i < $num; $i++){ $row = mysql_fetch_array($result); $class = (($i % 2) == 0) ? "table_odd_row" : "table_even_row"; echo "<tr class=".$class.">"; echo "<td>".$row[id]."</td>"; echo "<td>".$row[l_name]."</td>"; echo "<td>".$row[f_name]."</td>"; echo "<td>".$row[ssn]."</td>"; echo "</tr>"; } ?> </table>

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  • How to check if a checkbox/ radio button is checked in php

    - by user225269
    I have this html code: <tr> <td><label><input type="text" name="id" class="DEPENDS ON info BEING student" id="example">ID</label></td> </tr> <tr> <td> <label> <input type="checkbox" name="yr" class="DEPENDS ON info BEING student"> Year</label> </td> </tr> But I don't have any idea on how do I check this checkboxes if they are checked using php, and then output the corresponding data based on the values that are checked. Please help, I'm thinking of something like this. But of course it won't work, because I don't know how to equate checkboxes in php if they are checked: <?php $con = mysql_connect("localhost","root","nitoryolai123$%^"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("school", $con); $id = mysql_real_escape_string($_POST['idnum']); if($_POST['id'] == checked & $_POST['yr'] ==checked ){ $result2 = mysql_query("SELECT * FROM student WHERE IDNO='$id'"); echo "<table border='1'> <tr> <th>IDNO</th> <th>YEAR</th> </tr>"; while($row = mysql_fetch_array($result2)) { echo "<tr>"; echo "<td>" . $row['IDNO'] . "</td>"; echo "<td>" . $row['YEAR'] . "</td>"; echo "</tr>"; } echo "</table>"; } mysql_close($con); ?>

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  • Converting my lightweight MySQL DB wrapper into MySQLi. Pesky Problems

    - by Chaplin
    Here is the original code: http://pastebin.com/DNxtmApY. I'm not that interested in prepared statements at the moment, I just want this wrapper updating to MySQLi so once MySQL becomes depreciated I haven't got to update a billion websites. Here is my attempt at converting to MySQLi. <? $database_host = "127.0.0.1"; $database_user = "user"; $database_pass = "pass"; $database_name = "name"; $db = new database($database_host, $database_user, $database_pass, $database_name); class database { var $link, $result; function database($host, $user, $pass, $db) { $this->link = mysqli_connect($host, $user, $pass, $db) or $this->error(); mysqli_select_db($db, $this->link) or $this->error(); } function query($query) { $this->result = mysqli_query($query, $this->link) or $this->error(); $this->_query_count++; return $this->result; } function countRows($result = "") { if ( empty( $result ) ) $result = $this->result; return mysqli_num_rows($result); } function fetch($result = "") { if ( empty( $result ) ) $result = $this->result; return mysqli_fetch_array($result); } function fetch_num($result = "") { if ( empty( $result ) ) $result = $this->result; return mysqli_fetch_array($result, mysqli_NUM); } function fetch_assoc($result = "") { if ( empty( $result ) ) $result = $this->result; return mysqli_fetch_array($result, mysqli_ASSOC); } function escape($str) { return mysqli_real_escape_string($str); } function error() { if ( $_GET["debug"] == 1 ){ die(mysqi_error()); } else { echo "Error in db code"; } } } function sanitize($data) { //apply stripslashes if magic_quotes_gpc is enabled if(get_magic_quotes_gpc()) $data = stripslashes($data); // a mysqli connection is required before using this function $data = trim(mysqli_real_escape_string($data)); return $data; } However it chucks all sorts of errors: Warning: mysql_query(): Access denied for user 'www-data'@'localhost' (using password: NO) in /home/count/Workspace/lib/classes/user.php on line 7 Warning: mysql_query(): A link to the server could not be established in /home/count/Workspace/lib/classes/user.php on line 7 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/count/Workspace/lib/classes/user.php on line 8 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given in /home/count/Workspace/lib/classes/database.php on line 31

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  • Create unique links for indvidual elements of a PHP generated table

    - by typoknig
    I have a table generated from some PHP code that lists a SMALL amount of important information for employees. I want to make it so each row, or at least one element in each row can be clicked on so the user will be redirected to ALL of the information (pulled from MySQL database) related to the employee who was clicked on. I am not sure how would be the best way to go about this, but I am open to suggestions. I would like to stick to PHP and/or JavaScript. Below is the code for my table: <table> <tr> <td id="content_heading" width="25px">ID</td> <td id="content_heading" width="150px">Last Name</td> <td id="content_heading" width="150px">First Name</td> <td id="content_heading" width="75px">SSN</td> </tr> <?php $user = 'user'; $pass = 'pass'; $server = 'localhost'; $link = mysql_connect($server, $user, $pass); if (!$link){ die('Could not connect to database!' . mysql_error()); } mysql_select_db('mydb', $link); $query = "SELECT * FROM employees"; $result = mysql_query($query); mysql_close($link); $num = mysql_num_rows($result); for ($i = 0; $i < $num; $i++){ $row = mysql_fetch_array($result); $class = (($i % 2) == 0) ? "table_odd_row" : "table_even_row"; echo "<tr class=".$class.">"; echo "<td>".$row[id]."</td>"; echo "<td>".$row[l_name]."</td>"; echo "<td>".$row[f_name]."</td>"; echo "<td>".$row[ssn]."</td>"; echo "</tr>"; } ?> </table>

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  • Socket connection to a telnet-based server hangs on read

    - by mixwhit
    I'm trying to write a simple socket-based client in Python that will connect to a telnet server. I can test the server by telnetting to its port (5007), and entering text. It responds with a NAK (error) or an AK (success), sometimes accompanied by other text. Seems very simple. I wrote a client to connect and communicate with the server, but it hangs on the first attempt to read the response. The connection is successful. Queries like getsockname and getpeername are successful. The send command returns a value that equals the number of characters I'm sending, so it seems to be sending correctly. But in the end, it always hangs when I try to read the response. I've tried using both file-based objects like readline and write (via socket.makefile), as well as using send and recv. With the file object I tried making it with "rw" and reading and writing via that object, and later tried one object for "r" and another for "w" to separate them. None of these worked. I used a packet sniffer to watch what's going on. I'm not versed in all that I'm seeing, but during a telnet session I can see my typed text and the server's text coming back. During my Python socket connection, I can see my text going to the server, but packets back don't seem to have any text in them. Any ideas on what I'm doing wrong, or any strategies to try? Here's the code I'm using (in this case, it's with send and recv): #!/usr/bin/python host = "localhost" port = 5007 msg = "HELLO EMC 1 1" msg2 = "HELLO" import socket import sys try: skt = socket.socket(socket.AF_INET, socket.SOCK_STREAM) except socket.error, e: print("Error creating socket: %s" % e) sys.exit(1) try: skt.connect((host,port)) except socket.gaierror, e: print("Address-related error connecting to server: %s" % e) sys.exit(1) except socket.error, e: print("Error connecting to socket: %s" % e) sys.exit(1) try: print(skt.send(msg)) print("SEND: %s" % msg) except socket.error, e: print("Error sending data: %s" % e) sys.exit(1) while 1: try: buf = skt.recv(1024) print("RECV: %s" % buf) except socket.error, e: print("Error receiving data: %s" % e) sys.exit(1) if not len(buf): break sys.stdout.write(buf)

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  • MySQLi Insert prepared statement via PHP

    - by Jimmy
    Howdie do, This is my first time dealing with MySQLi inserts. I had always used mysql and directly ran the queries. Apparently, not as secure as MySQLi. Anywho, I'm attempting to pass two variables into the database. For some reason my prepared statement keeps erroring out. I'm not sure if it's a syntax error, but the insert just won't work. I've updated the code to make the variables easier to read Also, the error is specifically, Error preparing statement. I've updated the code, but it's not a pHP error. It's a MySQL error as the script runs but fails at the execution of: if($stmt = $mysqli - prepare("INSERT INTO subnets (id, subnet, mask, sectionId, description, vrfId, masterSubnetId, allowRequests, vlanId, showName, permissions, pingSubnet, isFolder, editDate) VALUES ('', ?, ?, '1', '', '0', '0', '0', '0', '0', '{"3":"1","2":"2"}', '0', '0', 'NULL')")) { I'm going to enable MySQL error checking. I honestly didn't know about that. <?php error_reporting(E_ALL); function InsertIPs($decimnal,$cidr) { error_reporting(E_ALL); $mysqli = new mysqli("localhost","jeremysd_ips","","jeremysd_ips"); if(mysqli_connect_errno()) { echo "Connection Failed: " . mysqli_connect_errno(); exit(); } if($stmt = $mysqli -> prepare("INSERT INTO subnets (id, subnet, mask,sectionId,description,vrfld,masterSubnetId,allowRequests,vlanId,showName,permissions,pingSubnet,isFolder,editDate) VALUES ('',?,?,'1','','0','0','0','0','0', '{'3':'1','2':'2'}', '0', '0', NULL)")) { $stmt-> bind_param('ii',$decimnal,$cidr); if($stmt-> execute()) { echo "Row added successfully\n"; } else { $stmt->error; } $stmt -> close; } } else { echo 'Error preparing statement'; } $mysqli -> close(); } $SUBNETS = array ("2915483648 | 18"); foreach($SUBNETS as $Ip) { list($TempIP,$TempMask) = explode(' | ',$Ip); echo InsertIPs($Tempip,$Tempmask); } ?> This function isn't supposed to return anything. It's just supposed to perform the Insert. Any help would be GREATLY appreciated. I'm just not sure what I'm missing here

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  • Allow users to pull temporary data then delete table?

    - by JM4
    I don't know the best way to title this question but am trying to accomplish the following goal: When a client logs into their profile, they are presented with a link to download data from an existing database in CSV format. The process works, however, I would like for this data to be 'fresh' each time they click the link so my plan was - once a user has clicked the link and downloaded the CSV file, the database table would 'erase' all of its data and start fresh (be empty) until the next set of data populated it. My EXISTING CSV creation code: <?php $host = 'localhost'; $user = 'username'; $pass = 'password'; $db = 'database'; $table = 'tablename'; $file = 'export'; $link = mysql_connect($host, $user, $pass) or die("Can not connect." . mysql_error()); mysql_select_db($db) or die("Can not connect."); $result = mysql_query("SHOW COLUMNS FROM ".$table.""); $i = 0; if (mysql_num_rows($result) > 0) { while ($row = mysql_fetch_assoc($result)) { $csv_output .= $row['Field'].", "; $i++; } } $csv_output .= "\n"; $values = mysql_query("SELECT * FROM ".$table.""); while ($rowr = mysql_fetch_row($values)) { for ($j=0;$j<$i;$j++) { $csv_output .= '"'.$rowr[$j].'",'; } $csv_output .= "\n"; } $filename = $file."_".date("Y-m-d",time()); header("Content-type: application/vnd.ms-excel"); header("Content-disposition: csv" . date("Y-m-d") . ".csv"); header( "Content-disposition: filename=".$filename.".csv"); print $csv_output; exit; ?> any ideas?

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  • What should I learn & use to become a pro in PHP & Python Web development?

    - by pecker
    Hello, I'll just show some code to show how I do web development in PHP. <html> <head> <title>Example #3 TDavid's Very First PHP Script ever!</title> </head> <? print(Date("m/j/y")); require_once("somefile.php"); $mysql_db = "DATABASE NAME"; $mysql_user = "YOUR MYSQL USERNAME"; $mysql_pass = "YOUR MYSQL PASSWORD"; $mysql_link = mysql_connect("localhost", $mysql_user, $mysql_pass); mysql_select_db($mysql_db, $mysql_link); $result = mysql_query("SELECT impressions from tds_counter where COUNT_ID='$cid'", $mysql_link); if(mysql_num_rows($result)) { mysql_query("UPDATE tds_counter set impressions=impressions+1 where COUNT_ID='$cid'", $mysql_link); $row = mysql_fetch_row($result); if(!$inv) { print("$row[0]"); } } ?> <body> </body> </html> Thats it. I write every file like this. Recently, I learnt OOP and started using classes & objects in PHP. I hear that there are many frameworks there for PHP. They say that one must use these libraries. But I feel they are just making things complicated. Anyway, this is how I've been doing my web development. Now, I want to improve this. and make it professional. Also I want to move to Python. I searched SO archives and found everyone suggesting Django. But, can any one give me some idea about how web development in Python works? user (client) request for page --- webserver(-embedded PHP interpreter) ---- Server side(PHP) Script --- MySQL Server. Now, is it that instead of PHP interpreter there is python interpreter & instead of php script there is python script, which contains both HTML & python (embedded in some kind of python tags). Python script connects to database server and fetches some data which will be printed as HTML. or is it different in python world? Is this Django thing like frameworks for PHP? Can't one code in python without using Django. Because, I never encountered any post without django Please give me some kick start.

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  • login not working when changing from mysql to mysqli

    - by user1438647
    I have a code below where it logs a teacher in by matching it's username and password in the database, if correct, then log in, if incorrect, then display a message. <?php session_start(); $username="xxx"; $password="xxx"; $database="mobile_app"; $link = mysqli_connect('localhost',$username,$password); mysqli_select_db($link, $database) or die( "Unable to select database"); foreach (array('teacherusername','teacherpassword') as $varname) { $$varname = (isset($_POST[$varname])) ? $_POST[$varname] : ''; } ?> <form action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post" id="teachLoginForm"> <p>Username</p><p><input type="text" name="teacherusername" /></p> <!-- Enter Teacher Username--> <p>Password</p><p><input type="password" name="teacherpassword" /></p> <!-- Enter Teacher Password--> <p><input id="loginSubmit" type="submit" value="Login" name="submit" /></p> </form> <?php if (isset($_POST['submit'])) { $query = " SELECT * FROM Teacher t WHERE (t.TeacherUsername = '".mysqli_real_escape_string($teacherusername)."') AND (t.TeacherPassword = '".mysqli_real_escape_string($teacherpassword)."') "; $result = mysqli_query($link, $query); $num = mysqli_num_rows($result); $loged = false; while($row = mysqli_fetch_array($result)) { if ($_POST['teacherusername'] == ($row['TeacherUsername']) && $_POST['teacherpassword'] == ($row['TeacherPassword'])) { $loged = true; } $_SESSION['teacherforename'] = $row['TeacherForename']; $_SESSION['teachersurname'] = $row['TeacherSurname']; $_SESSION['teacherusername'] = $row['TeacherUsername']; } if ($loged == true){ header( 'Location: menu.php' ) ; }else{ echo "The Username or Password that you Entered is not Valid. Try Entering it Again."; } mysqli_close($link); } ?> Now the problem is that even if the teacher has entered in the correct username and password, it still doesn't let the teacher log in. When the code above was the old mysql() code, it worked fine as teacher was able to login when username and password match, but when trying to change the code into mysqli then it causes login to not work even though username and password match. What am I doing wrong?

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  • gauge chart is not displaying any thing

    - by Sandy
    i am trying to display the latest speed in mysql database on guage chart. i have tried so many things but gauge is not display plz any can help me...my code is attached and php part shows the correct value but dont know why guage is not display <?php $host="localhost"; // Host name $username="root"; // Mysql username $password=""; // Mysql password $db_name="mysql"; // Database name $tbl_name="gpsdb"; // Table name // Connect to server and select database. $con=mysql_connect("$host", "$username")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $data = mysql_query("SELECT speed FROM gpsdb WHERE DeviceId=1234 ORDER BY TIME DESC LIMIT 1") or die(mysql_error()); while ($nt = mysql_fetch_assoc($data)) { $speed = $nt['speed']; $jsonTable = json_encode($speed); echo $jsonTable; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="content-type" content="text/html; charset=utf-8"/> <title> Google Visualization API Sample </title> <script type="text/javascript" src="//www.google.com/jsapi"></script> <script type="text/javascript"> google.load('visualization', '1', {packages: ['gauge']}); </script> <script type="text/javascript"> function drawVisualization() { // Create and populate the data table. var data = new google.visualization.DataTable(<?=$speed?>); // Create and draw the visualization. new google.visualization.Gauge(document.getElementById('visualization')). draw(data); } google.setOnLoadCallback(drawVisualization); </script> </head> <body style="font-family: Arial;border: 0 none;"> <div id="visualization" style="width: 600px; height: 300px;"></div> </body> </html>

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  • IE is doing strange things with JQuery

    - by Syncopated
    So ... The thing is, the code works in FireFox, no problems. But when I open the same page, it gives me the following error: "Undefined is null or not an object." But when I copy the code to a localhost page, it works fine. Also when I clear my cache in IE it works, but only once, if I refresh after that one load, it gives me the same error. Here is the code: <script type="text/javascript" src="datepicker/js/jquery-1.4.2.min.js"></script> <script type="text/javascript"> var count3 = 0; var count5 = 0; var count2 = 0; var count4 = 0; $(document).ready(function(){ $('#switch3').click(function(){ $('#switchDiv3').slideToggle(350); if(count3 == 0){ count3 = 1; document.getElementById('switchImage3').src = "images/ArrowDown.png"; return; } else { count3 = 0; document.getElementById('switchImage3').src = "images/ArrowRight.png"; return; } }); ... (this is the code for each item that is generated) </script> And the code that determines the div that should hide: <table width="100%" border="0" cellspacing="0" cellpadding="0" align="center"> <tr> <td width="20" align="center" valign="top" style="padding-right: 3px"> <a style="cursor: pointer;" id="switch3"><img width="20" height="20" src="images/ArrowRight.png" id="switchImage3" style="border-style: solid; border-width: 1px; border-color: black;"/></a> </td> <td> <div id="switchDiv3"> <div align="left"> (Contents of the div here) </div> </div> </td> </tr> </table> Any help is appreciated! Thanks in advance

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  • PHP/MySQL Interview - How would you have answered?

    - by martincarlin87
    I was asked this interview question so thought I would post it here to see how other users would answer: Please write some code which connects to a MySQL database (any host/user/pass), retrieves the current date & time from the database, compares it to the current date & time on the local server (i.e. where the application is running), and reports on the difference. The reporting aspect should be a simple HTML page, so that in theory this script can be put on a web server, set to point to a particular database server, and it would tell us whether the two servers’ times are in sync (or close to being in sync). This is what I put: // Connect to database server $dbhost = 'localhost'; $dbuser = 'xxx'; $dbpass = 'xxx'; $dbname = 'xxx'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die (mysql_error()); // Select database mysql_select_db($dbname) or die(mysql_error()); // Retrieve the current time from the database server $sql = 'SELECT NOW() AS db_server_time'; // Execute the query $result = mysql_query($sql) or die(mysql_error()); // Since query has now completed, get the time of the web server $php_server_time = date("Y-m-d h:m:s"); // Store query results in an array $row = mysql_fetch_array($result); // Retrieve time result from the array $db_server_time = $row['db_server_time']; echo $db_server_time . '<br />'; echo $php_server_time; if ($php_server_time != $db_server_time) { // Server times are not identical echo '<p>Database server and web server are not in sync!</p>'; // Convert the time stamps into seconds since 01/01/1970 $php_seconds = strtotime($php_server_time); $sql_seconds = strtotime($db_server_time); // Subtract smaller number from biggest number to avoid getting a negative result if ($php_seconds > $sql_seconds) { $time_difference = $php_seconds - $sql_seconds; } else { $time_difference = $sql_seconds - $php_seconds; } // convert the time difference in seconds to a formatted string displaying hours, minutes and seconds $nice_time_difference = gmdate("H:i:s", $time_difference); echo '<p>Time difference between the servers is ' . $nice_time_difference; } else { // Timestamps are exactly the same echo '<p>Database server and web server are in sync with each other!</p>'; } Yes, I know that I have used the deprecated mysql_* functions but that aside, how would you have answered, i.e. what changes would you make and why? Are there any factors I have omitted which I should take into consideration? The interesting thing is that my results always seem to be an exact number of minutes apart when executed on my hosting account: 2012-12-06 11:47:07 2012-12-06 11:12:07

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