Search Results

Search found 1909 results on 77 pages for 'die'.

Page 24/77 | < Previous Page | 20 21 22 23 24 25 26 27 28 29 30 31  | Next Page >

  • PHP export to text file - Only saving first line.

    - by wertz8090
    I'm trying to export some extracted $_POST information into a text file, however my code is only capturing the first variable and ignoring the rest. I'm saving the information to the text file in this manner: $values = "First Name: $fName\r\n"; $values .= "Last Name: $lName\r\n"; $values .= "Address: $address\r\n"; etc. This is the code I use to write to the text file: $fp = @fopen("person.data", "w") or die("Couldn't open person.data for writing!"); $numBytes = @fwrite($fp, $values) or die("Couldn't write values to file!"); @fclose($fp); Any ideas on why it would only save the first $values ($fName) variable but not the rest of them? It actually saves the first part of the $values string for all of them (so I see Last Name:, Address:, etc. on separate lines in the text file) but the called variables $lName and $address do not appear.

    Read the article

  • PHP New Line will not work

    - by user364333
    Hi I am trying to create some code that first reads the existing contents of the file in and then adds the new line of code on a new line but the code i am using just adds it on the new text on to the already existing line instead of the new line... Here is the code i am using: <?php $id = $_GET['id']; $userfile = "user1.txt"; $fo = fopen($userfile, 'r') or die("can't open favourites file"); $currentdata = fread($fo, filesize($userfile)); fclose($fo); $fw = fopen($userfile, 'w') or die("can't open favourites file"); $currentprocessed = "$currentdata\n"; fwrite($fw, $currentprocessed); fwrite($fw, $id); fclose($fw); ?> I have tried a whole range of different ideas but nothing has worked, any help would be appreciated.

    Read the article

  • PHP MySQL Syntax Error 'You have an error in your SQL syntax'

    - by Alec
    I cannot figure out the issue with my code here. I am trying to take info from the table, then subtract 1 second from Current_Time which looks like '2:00'. The problem is, I get: "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Current_Time) VALUES('22')' at line 1" I don't even understand where it gets 22 from. Thanks, I really appreciate it. if (isset($_GET['id']) && isset($_GET['time'])) { mysql_select_db("aleckaza_pennyauction", $connection); $query = "SELECT Current_Time FROM Live_Auctions WHERE ID='1'"; $results = mysql_query($query) or die(mysql_error()); while ($row = mysql_fetch_array($results)) { $newTime = $row['Current_Time'] - 1; $query = "INSERT INTO Live_Auctions(Current_Time) VALUES('".$newTime."')"; $results = mysql_query($query) or die(mysql_error()); } }

    Read the article

  • php class crash course

    - by rabidmachine9
    Hello people, I'm sorry for this question but I'm getting crazy trying to write my first php clash, it is supposed to connect me to server and to a database but I'm always getting this error: Parse error: syntax error, unexpected ',', expecting T_PAAMAYIM_NEKUDOTAYIM in /Applications/XAMPP/xamppfiles/htdocs/classTest/test.php on line 9 thanks in advance, here is the code: <?php // include 'DBConnect.php'; class DBConnect { function connection($hostname = "localhost", $myname = "root", $pass = ''){ mysql_connect(&hostname, &myname, &pass) or die("Could not connect!"); //Connect to mysql } function choose($dbnam){ mysql_select_db(&dbnam) or die("Couldn't find database!"); //fnd specific db } } $a = new DBConnect(); $connect = $a->connection("localhost", "root"); $a->choose('mydb'); ?>

    Read the article

  • Using Java, can I have one JVM spawn another, and then have the original one exit?

    - by CarlG
    I have a need to have a running JVM start another JVM and then exit. I'm currently trying to do this via Runtime.getRuntime().exec(). The other JVM starts, but my original JVM won't exit until the "child" JVM process stops. It appears that using Runtime.getRuntime().exec() creates a parent-child relationship between the processes. Is there some way to de-couple the spawned process so that the parent can die, or some other mechanism to spawn a process without any relationship to the creating process? Note that this seems exactly like this question: http://stackoverflow.com/questions/2566502/using-java-to-spawn-a-process-and-keep-it-running-after-parent-quits but the accepted answer there doesn't actually work, at least not on my system (Windows 7, Java 5 and 6). It seems that maybe this is a platform-dependent behavior. I'm looking for a platform independent way to reliably invoke the other process and let my original process die.

    Read the article

  • PHP loop position

    - by Jordan Pagaduan
    Can someone help me on this. I'm made an image uploader and i want the image to make another tr if it reach to 5 pics so it will not overflow. Here is my code: Can someone help me on this. I'm made an image uploader and i want the image to make another tr if it reach to 5 pics so it will not overflow. Here is my code: $dbc = mysql_connect("localhost" , "root" , "") or die (mysql_error()); mysql_select_db('blog_data') or die (mysql_error()); $sql = "SELECT * FROM img_uploaded"; $result = mysql_query($sql); while($rows=mysql_fetch_array($result)) { if ($rows) { echo "<tr><td><img src='user_images/".$rows['img_name'] . "' width='100' height='100'></td></tr>"; } else { echo "<td><img src='user_images/".$rows['img_name'] . "' width='100' height='100'></td>"; } }

    Read the article

  • PHP MySQL Insert Help

    - by user364333
    Hey I am trying to make a page that inserts some strings into a MySQL table but it just dosn't seem to be working for me. Here is the code I am using at the moment. <?php mysql_connect($address, $username, $password); @mysql_select_db($database) or die("Unable to select database"); $query = "insert INTO user (movieid, moviename)('" . $id . "','" . $name . "') or die(mysql_error())"; mysql_query($query); mysql_close(); ?> Where am i going wrong?

    Read the article

  • How can I insert a line at the beginning of a file with Perl's Tie::File?

    - by thebourneid
    I'm trying to insert/add a line 'COMMENT DUMMY' at the beginnig of a file as a first row if /PATTERN/ not found. I know how to do this with OPEN CLOSE function. Probably after reading the file it should look something like this: open F, ">", $fn or die "could not open file: $!"; ; print F "COMMENT DUMMY\n", @array; close F; But I have a need to implement this with the use of the Tie::File function and don't know how. use strict; use warnings; use Tie::File; my $fn = 'test.txt'; tie my @lines, 'Tie::File', $fn or die "could not tie file: $!"; untie @lines;

    Read the article

  • ajax how to read out $_post variable

    - by alex
    Hi, I am trying to filter/search a database with ajax $.ajax({ type: "POST", url: "filterSearch.php", queryString: qry, success: function(data){ alert( "Data Saved: " + data ); $('#searchResult').html(data); // Fill the search results box } }); Now in filterSearch.php i have the following test codes if(isset($_POST['queryString'])) { echo "TEST"; } if($_POST['runquery']==1) { $sql = "SELECT * FROM fs_vacatures WHERE here-the-like-query?"; $msg = $sql; echo $msg; die(); } die(); But nor TEST or the $sql is return in the alert??

    Read the article

  • Strange behaviour of mb_detect_order() in PHP

    - by termopro
    I would like to detect encoding of some text (using PHP). For that purpose i use mb_detect_encoding() function. The problem is that the function returns different results if i change the order of possible encodings with mb_detect_order() function. Consider the following example $html = <<< STR ????????????????????????????????????????????????????????????????????????????????????????????????????????? STR; mb_detect_order(array('UTF-8','EUC-JP', 'SJIS', 'eucJP-win', 'SJIS-win', 'JIS', 'ISO-2022-JP','ISO-8859-1','ISO-8859-2')); $originalEncoding = mb_detect_encoding($str); die($originalEncoding); // $originalEncoding = 'UTF-8' However if you change the order of encodings in mb_detect_order() the results will be different: mb_detect_order(array('EUC-JP','UTF-8', 'SJIS', 'eucJP-win', 'SJIS-win', 'JIS', 'ISO-2022-JP','ISO-8859-1','ISO-8859-2')); die($originalEncoding); // $originalEncoding = 'EUC-JP' So my questions are: Why is that happening ? Is there a way in PHP to correctly and unambiguously detect encoding of text ?

    Read the article

  • SFTP not supported error with PHP & cURL

    - by Bad Programmer
    I followed the advice from this Stack Overflow question thread, but I keep hitting a snag. I am receiving the following error message: Unsupported protocol: sftp Here is my code: $ch = curl_init(); if(!$ch) { $error = curl_error($ch); die("cURL session could not be initiated. ERROR: $error.""); } $fp = fopen($docname, 'r'); if(!$fp) { $error = curl_error($ch); die("$docname could not be read."); } curl_setopt($ch, CURLOPT_URL, "sftp://$user_name:$user_pass@$server:22/$docname"); curl_setopt($ch, CURLOPT_UPLOAD, 1); curl_setopt($ch, CURLOPT_PROTOCOLS, CURLPROTO_SFTP); curl_setopt($ch, CURLOPT_INFILE, $fp); curl_setopt($ch, CURLOPT_INFILESIZE, filesize($docname)); //this is where I get the failure $exec = curl_exec ($ch); if(!$exec) { $error = curl_error($ch); die("File $docname could not be uploaded. ERROR: $error."); } curl_close ($ch); I used the curl_version() function to see my curl information, and found that sftp doesn't seem to be in the array of supported protocols: [version_number] => 462597 [age] => 2 [features] => 1597 [ssl_version_number] => 0 [version] => 7.15.5 [host] => x86_64-redhat-linux-gnu [ssl_version] => OpenSSL/0.9.8b [libz_version] => 1.2.3 [protocols] => Array ( [0] => tftp [1] => ftp [2] => telnet [3] => dict [4] => ldap [5] => http [6] => file [7] => https [8] => ftps ) Is this a matter of my version of cURL being outdated, or is the SFTP protocol simply not supported at all? Any advice is greatly appreciated.

    Read the article

  • 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>

    Read the article

  • Stop Submit With empty and error Input Values Using PHP

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

    Read the article

  • Where I get wrong?

    - by Ole Jak
    I have PHP code like this require_once( "includes" . DIRECTORY_SEPARATOR . "constants.php"); $userID = validKey( $key ); $query = "select a.username, b.streamID from user a, streams b where a.id = b.userID;)"; $connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if (!$connection) { die("Database connection failed: " . mysql_error()); } $db_select = mysql_select_db(DB_NAME,$connection); if (!$db_select) {die( "Database selection failed: " . mysql_error()); } $generated_table = mysql_fetch_array(mysql_query($query, $connection)); // line (*) print_r($generated_table); But PHP gives me error/worning and does not print anything out Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given on line (*) 1) Why It does not printed me anything out 2) Where am I wrong?

    Read the article

  • cant made mySQL connection

    - by Andika Evangelion Thirteenth S
    I've use following code $con = mysql_connect("localhost","root","");if (!$con) {die('Could not connect: ' . mysql_error());} It's works. but when $db_host='localhost';$db_id='root';$db_pass=''; $con = mysql_connect($db_host,$db_id,$db_pass);if (!$con) {die('Could not connect: ' . mysql_error());} it didn't works, trying to swap ("),('),and empty in mysql_connect() and in $var and vice versa. Any help would be appreciated. Thanks.

    Read the article

  • Guice Problem with Tests

    - by D3orn
    Hey I wrote a LudoGame and now I like to test it with a little GuiceInjection^^ I have an Interface IDie for my die. Now for the game I only need an IDie instead of a realdie = in tests I simply give the LudoGame a MokeDie to set up the Numbers I like to roll. The IDie has only one method: roll() which returns a int. BUT the mokeDie now has another public method: sendNextNumber() (should be clear what this does^^) Now I like to @Inject a Die and if @UseMokeDie is before a Test I'll like to pass the MokeDie but I'm very new to Guice... Need some advices please! Thx for Answers

    Read the article

  • ftp php file uploadin

    - by Issak
    <?php ini_set('max_execution_time', '0'); $host = '234.546.155.485'; $usr = 'fgfgfgdf'; $pwd = 'fghghh'; // file to move: $file = 'http://vsomesite.com/file.flv'; $ftp_path = '/public_html/video57242/test.flv'; // connect to FTP server (port 21) $conn_id = ftp_connect($host, 21) or die ("Cannot connect to host"); // send access parameters ftp_login($conn_id, $usr, $pwd) or die("Cannot login"); // turn on passive mode transfers (some servers need this) // ftp_pasv ($conn_id, true); // perform file upload $upload = ftp_put($conn_id, $ftp_path, $local_file, FTP_ASCII); // check upload status: print (!$upload) ? 'Cannot upload' : 'Upload complete'; ?> upload fails when the file is remote not local. what problem?

    Read the article

  • MySQL Access in Php to a database created in C-Panel

    - by Rhys Drury
    Basically, i'm having trouble connecting to a mysql database using a php web page. I created the database in C-panel using the wizard i'm connecting like this $db_host = "localhost"; //your host Database address $db_username = "xxxx"; //your account username $db_pass = "xxxxx"; //your account password $db_name = "xxxxx"; //your database name @mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql"); @mysql_select_db("$db_name") or die ("no database"); But all my page seems to do is trigger the " could not connect to mysql" my page does have wordpress installed at the minute but I'm planning to get rid of it as I'm creating my site myself. I'm just baffled why it can't connect, because in Phpmyadmin ( a feature on C-panel) it says the database is in localhost.

    Read the article

  • 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

    Read the article

  • undefined GET id?

    - by Azzyh
    <?php $s = $_GET["s"]; if($s) { $hent_b = mysql_query("SELECT * FROM member_battles WHERE state = '1' ORDER BY id DESC LIMIT 0,200") or die(mysql_error()); }else{ $hent_b = mysql_query("SELECT * FROM member_battles WHERE state = '0' ORDER BY id DESC LIMIT 0,200") or die(mysql_error()); } while($vis = mysql_Fetch_array($hent_b)) { ?> I have this now i want when i enter my site (index.php) it should not come up undefined $_GET["s"]; how do i do this? but i want when you do index.php?s then it should change the query

    Read the article

  • Won't connect to the database

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

    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

  • MYSQL accentuated characters é display as %E9

    - by Jk_
    Hi guys, I'm pushing data from as3 to MSQL through a little php script! My problem is that all my accentuated characters are displayed as weird iso characters. Example : é is displayed %E9 Obvisously the collation of my field is set to utf8_general_ci Even when I try to INSERT the data from a simple php script without as3, I get the same mistake. <?php mysql_connect("***", "***", "***") or die("Error :" .mysql_error()); mysql_select_db("***"); $query ="INSERT INTO test (message) values ('éèàïû')"; mysql_query($query) or die("Error updating DB"); ?> Any idea on what am I doing wrong and how I could fix that? Thanks in advance. Jk_

    Read the article

  • Do email forms need to be santized before sending?

    - by levi
    I have a client that keeps getting reports from godaddy's "websiteprotection.com" stating how the website is insecure. Your website contains pages that do not properly sanitize visitor-provided input to make sure it contains no malicious content or scripts. Cross-site scripting vulnerabilities let malicious users execute arbitrary HTML or script code in another visitor's browser. Output: The request string used to detect this flaw was : /cross_site_scripting.?nasl.asp The output was : HTTP/1.1 404 Not Found\r Date: Wed, 21 Mar 2012 08:12:02 GMT\r Server: Apache\r X-Pingback:http://?CLIENTSWEBSITE.com/?xmlrpc.php\r Expires: Wed, 11 Jan 1984 05:00:00 GMT\r Cache-Control: no-cache, must-revalidate, max-age=0\r Pragma: no-cache\r Set-Cookie: PHPSESSID=?1jsnhuflvd59nb4trtquston50; path=/\r Last-Modified: Wed, 21 Mar 2012 08:12:02 GMT\r Keep-Alive: timeout=15, max=100\r Connection: Keep-Alive\r Transfer-Encoding: chunked\r Content-Type: text/html; charset=UTF-8\r \r <div id="contact-form" class="widget"><form action="http://?CLIENTSWEBSITE.c om/<script>cross_site_?scripting.nasl</script>.asp" id="contactForm" meth od="post"> It looks like it has an issue with the contact form. All the contact form does is posts an ajax request to the same page, and than a PHP script mails the data (no database stuff). Is there any a security issues here? Any ideas on how I can satisfy the security scanner? Here is the form and script: <form action="<?php echo $this->getCurrentUrl(); ?>" id="contactForm" method="post"> <input type="text" name="Name" id="Name" value="" class="txt requiredField name" /> //Some more text inputs <input type="hidden" name="sendadd" id="sendadd" value="<?php echo $emailadd ; ?>" /> <input type="hidden" name="submitted" id="submitted" value="true" /><input class="submit" type="submit" value="Send" /> </form> // Some initial JS validation, if that passes an ajax post is made to the script below //If the form is submitted if(isset($_POST['submitted'])) { //Check captcha if (isset($_POST["captchaPrefix"])) { $capt = new ReallySimpleCaptcha(); $correct = $capt->check( $_POST["captchaPrefix"], $_POST["Captcha"] ); if( ! $correct ) { echo false; die(); } else { $capt->remove( $_POST["captchaPrefix"] ); } } $dateon = $_POST["dateon"]; $ToEmail = $_POST["sendadd"]; $EmailSubject = 'Contact Form Submission from ' . get_bloginfo('title'); $mailheader = "From: ".$_POST["Email"]."\r\n"; $mailheader .= "Reply-To: ".$_POST["Email"]."\r\n"; $mailheader .= "Content-type: text/html; charset=iso-8859-1\r\n"; $MESSAGE_BODY = "Name: ".$_POST["Name"]."<br>"; $MESSAGE_BODY .= "Email Address: ".$_POST["Email"]."<br>"; $MESSAGE_BODY .= "Phone: ".$_POST["Phone"]."<br>"; if ($dateon == "on") {$MESSAGE_BODY .= "Date: ".$_POST["Date"]."<br>";} $MESSAGE_BODY .= "Message: ".$_POST["Comments"]."<br>"; mail($ToEmail, $EmailSubject, $MESSAGE_BODY, $mailheader) or die ("Failure"); echo true; die(); }

    Read the article

  • Using PHP to connect to RADIUS works on one server but not another

    - by JDS
    I have a fleet of webservers that server a LAMP webapp broken into multiple customer apps by virtualhost/domain. The platform is Ubuntu 10.04 VM + PHP 5.3 + Apache 2.2.14, on top of VMware ESX (v4 I think). This stuff's not too important, though -- I'm just setting up the background. I have one customer that connects to a RADIUS server for authentication. We've found that the app responds as if some number of web servers are configured correctly and some are not. i.e. Apparently random authentication failures or successes, with no rhyme or reason. I did a lot of analysis of our fleet, and resolved it down to the differences between two specific web servers. I'll call them "A" and "B". "A" works. "B" does not. "Works" means "connects to and gets authentication data successfully from the RADIUS server". Ultimately, I'm looking for one thing that is different, and I've exhausted everything that I can come up with, so, looking for something else. Here are things I've looked at PHP package versions (all from Ubuntu repos). These are exactly the same across servers. PECL package. There are no PECL packages that aren't installed by apt. Other libraries or packages. Nothing that was network-related or RADIUS-related was different among servers. (There were some minor package differences, though.) Network or hosting environment. I found that some of the working servers were on the same physical environment as some not-working ones (i.e. same ESX containers). So, probably, the physical network layer is not the problem. Test case. I created a test case as follows. It works on the working servers, and fails on the not-working servers, very consistently. <?php $radius = radius_auth_open(); $username = 'theusername'; $password = 'thepassword'; $hostname = '12.34.56.78'; $radius_secret = '39wmmvxghg'; if (! radius_add_server($radius,$hostname,0,$radius_secret,5,3)) { die('Radius Error 1: ' . radius_strerror($radius) . "\n"); } if (! radius_create_request($radius,RADIUS_ACCESS_REQUEST)) { die('Radius Error 2: ' . radius_strerror($radius) . "\n"); } radius_put_attr($radius,RADIUS_USER_NAME,$username); radius_put_attr($radius,RADIUS_USER_PASSWORD,$password); switch (radius_send_request($radius)) { case RADIUS_ACCESS_ACCEPT: echo 'GOOD LOGIN'; break; case RADIUS_ACCESS_REJECT: echo 'BAD LOGIN'; break; case RADIUS_ACCESS_CHALLENGE: echo 'CHALLENGE REQUESTED'; break; default: die('Radius Error 3: ' . radius_strerror($radius) . "\n"); } ?>

    Read the article

< Previous Page | 20 21 22 23 24 25 26 27 28 29 30 31  | Next Page >