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• How do I get long command lines to wrap to the next line?

  - by BrianH

  Edit It was my .bashrc file. I've copied the same profile from machine to machine, and I used special characters in my $PS1 that are somehow throwing it off. I'm now sticking with the standard bash variables for my $PS1. Thanks to @ændrük for the tip on the .bashrc! ...End Edit... Something I have noticed in Ubuntu for a long time that has been frustrating to me is when I am typing a command at the command line that gets longer (wider) than the terminal width, instead of wrapping to a new line, it goes back to column 1 on the same line and starts over-writing the beginning of my command line. (It doesn't actually overwrite the actual command, but visually, it is overwriting the text that was displayed). It's hard to explain without seeing it, but let's say my terminal was 20 characters wide (Mine is more like 120 characters - but for the sake of an example), and I want to echo the English alphabet. What I type is this: echo abcdefghijklmnopqrstuvwxyz But what my terminal looks like before I hit the key is: pqrstuvwxyzghijklmno When I hit enter, it echos abcdefghijklmnopqrstuvwxyz so I know the command was received properly. It just wrapped my typing after the "o" and started over on the same line. What I would expect to happen, if I typed this command in on a terminal that was only 20 characters wide would be this: echo abcdefghijklmno pqrstuvwxyz Background: I am using bash as my shell, and I have this line in my ~/.bashrc: set -o vi to be able to navigate the command line with VI commands. I am currently using Ubuntu 10.10 server, and connecting to the server with Putty. In any other environment I have worked in, if I type a long command line, it will add a new line underneath the line I am working on when my command gets longer than the terminal width and when I keep typing I can see my command on 2 different lines. But for as long as I can remember using Ubuntu, my long commands only occupy 1 line. This also happens when I am going back to previous commands in the history (I hit Esc, then 'K' to go back to previous commands) - when I get to a previous command that was longer than the terminal width, the command line gets mangled and I cannot tell where I am at in the command. The only work-around I have found to see the entire long command is to hit "Esc-V", which opens up the current command in a VI editor. I don't think I have anything odd in my .bashrc file. I commented out the "set -o vi" line, and I still had the problem. I downloaded a fresh copy of Putty and didn't make any changes to the configuration - I just typed in my host name to connect, and I still have the problem, so I don't think it's anything with Putty (unless I need to make some config changes) Has anyone else had this problem, and can anyone think of how to fix it? Thanks in advance! Brian

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• How to get the selected option value of a drop down box in PHP code

  - by Angeline Aarthi

  I have a dropdown box which lists a set of logos,like flower,butterfly etc. <p class="title1">Logo</p> <select name="logoMenu" class="select" size="7"> <?php foreach($logos as $logo):?> <option id="<?php echo $logo['Subproperty']['id'];?>" value="<?php echo $logo['Subproperty']['values'];?>"><?php echo $logo['Subproperty']['values'];?> </option> <?php endforeach;?> </select> Suppose If I select the logo 'Flower' from the drop down box, I want the flower pic to be displayed in a div.This is the div that I have to display the pictures. <div id="theme_logos" class="float_left spaceleft" style="display:none;"> <?php foreach($defaultLogos as $logo): //if($logo['Subproperty']['values']==clicked option value){?> <img height="50" width="50" src="/FormBuilder/app/webroot/img/themes/<?php echo $logo['Subproperty']['images'];?>" class="float_left user_profile_image user_profile_image" alt="Default50"/> <?php endforeach;?> </div> The problem with this code is that it displaya all the pictures found in the table. Because im My controller code, I give only the property id as that of 'Logo',but do not give which logo. $this->set('defaultLogos',$this->Subproperty->find('all',array('conditions'=>array('Subproperty.property_id'=>1,'Subproperty.values'=>"Flower")))); Here I have hard coded as 'flower' so that I get the flower picture alone.. If I select the logo from the drop down box, how to pass that selected value to the controller code? Or if I get the selected logo name thro' jquery,how to use that value in the if condition inside the for each loop? someone help me out with this.. I'm using CakePHP framework. $("#logoMenu option").click(function(){ selectedLogo=$(this).attr("value"); $('#subproperty_id').val($(this).attr("id")); if(selectedLogo=="Your logo"){ $("#themes_upload").show(); } else{ alert(selectedLogo); $("#themes_upload").hide(); $("#theme_logos").show(); } }); EDIT Now I have tried an ajax post where I pass the selected logo to the same action of the controller. I get the value when I alert the passed value in the success function of the ajax function. I the picture doesn't appear. $("#logoMenu option").click(function(){ selectedLogo=$(this).attr("value"); $('#subproperty_id').val($(this).attr("id")); if(selectedLogo=="Your logo"){ $("#themes_upload").show(); } else{ alert(selectedLogo); $.ajax({ type: "POST", url: "http://localhost/FormBuilder/index.php/themes/themes/", async: false, data: "selectedLogo="+selectedLogo, success: function(msg){ alert( "Data Saved: " + msg); } }); $("#themes_upload").hide(); $("#theme_logos").show(); } }); function themes(){ $this->set('themes',$this->Theme->find('all')); $logo=$this->params['form']['selectedLogo']; echo "logo:".$logo; $this->set('defaultLogos',$this->Subproperty->find('all',array('conditions'=>array('Subproperty.property_id'=>1,'Subproperty.values'=>$logo)))); } But when I trry to display the img in the page,it doesn't appear. Is it because the div show command is after the ajax request?

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• Not recognize JDK after installation in ubuntu 12.10

  - by HFDev

  I did these steps without error: 1-JDK path : Downloads/jdk-6u37-linux-x64.bin 2-Commands in Terminal : mkdir Programs cd Programs bash ../Downloads/jdk-6u37-linux-x64.bin ln -s jdk-6u37-linux-x64 jdk 3-Set JAVA_HOME and Path: in HomeView MenuShow Hidden Files Then open .bashrc in text editor. I added the following lines to end of file. export JAVA_HOME=$Home/Programs/jdk export PATH=:$JAVA_HOME/bin:$PATH --------------------------------------------------------------------------------------- This is the result of executing the command echo $JAVA_HOME : /Programs/jdk This is the result of executing the command echo $PATH : :/Programs/jdk/bin:/usr/lib/lightdm/lightdm:/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games:/usr/local/games1 ---------------------------------------------------------------------------------------- And problem is : This is the result of executing the command java -version : The program 'java' can be found in the following packages: * default-jre * gcj-4.6-jre-headless * gcj-4.7-jre-headless * openjdk-7-jre-headless * openjdk-6-jre-headless Try: sudo apt-get install

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• disable shutdown/suspend if there is other user logged in via ssh

  - by Denwerko

  I remember that in versions of ubuntu around 9.04 was possible to disable user to shutdown ( and maybe suspend too ) system if there was other user logged in.Something like policykit or similar. Is it possible to do in 11.04 ? Thanks edit: if someone needs ( for own risk ), little change in /usr/lib/pm-utils/bin/pm-action will allow user to suspend machine if he is only user logged in or when user will run sudo pm-suspend. Probably not best piece of code, but for now works. diff -r 805887c5c0f6 pm-action --- a/pm-action Wed Jun 29 23:32:01 2011 +0200 +++ b/pm-action Wed Jun 29 23:37:23 2011 +0200 @@ -47,6 +47,14 @@ exit 1 fi +if [ "$(id -u )" == 0 -o `w -h | cut -f 1 -d " " | sort | uniq | wc -l` -eq 1 ]; then + echo "either youre root or root isnt here and youre only user, continuing" 1&2 + else + echo "Not suspending, root is here or there is more users" 1&2 + exit 2 + fi + + remove_suspend_lock() { release_lock "${STASHNAME}.lock" Question still stands, is it possible to forbid shutdown or suspend when there is more than one user logged in ( without rewriting system file )?

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• How to select all options from a drop list in php / mysql

  - by Mirage81

  Thanks to stackoverflow.com's frienly experts I've managed to create my first php + mysql application. The code searches a mysql database for last names and cities. The choices are made through two drop lists like these: Choose city: All cities Liverpool Manchester Choose last name: All last names Lennon Gallagher The code would return eg. all the Lennons living in Liverpool. However, I haven't been able to make the options "All cities" and "All last names" to work so that the code would return eg. all the Lennons living in any city or all the people living in Liverpool. So, how can that be done? The code so far: index.php <?php $conn = mysql_connect('localhost', 'user', 'password') or die("Connection failed"); mysql_select_db("database", $conn) or die("Switch database failed"); //this gets the cities from the database to the drop list $query = "SELECT DISTINCT city FROM user".mysql_real_escape_string($city); $result = mysql_query($query, $conn); $options=""; while ($row=mysql_fetch_array($result)) { $city=$row["city"]; $options.="<OPTION VALUE=\"$city\">".$city; } //this gets the last names from the database to the drop list $query2 = "SELECT DISTINCT lastname FROM user".mysql_real_escape_string($lastname); $result2 = mysql_query($query2, $conn); $options2=""; while ($row2=mysql_fetch_array($result2)) { $lastname=$row2["lastname"]; $options2.="<OPTION VALUE=\"$lastname\">".$lastname; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"> <html> <head> <meta content="text/html; charset=ISO-8859-1" http-equiv="content-type"> <title>test</title> </head> <body> <form action="get.php" method="post"> <p> <select name="city"> <option value=0>Choose <option value=1>All cities <?=$options?> </select> </p> <p> <select name="lastname"> <option value=0>Choose <option value=1>All last names <?=$options2?> </select> </p> <p> <input value="Search" type="submit"> </p> </form> <br> </body> </html> get.php <?php $conn = mysql_connect('localhost', 'user', 'password') or die("Connection failed"); mysql_select_db("database", $conn) or die("Switch database failed"); $query = "SELECT * FROM user WHERE city = '".mysql_real_escape_string($_POST['city'])."' AND lastname = '".mysql_real_escape_string($_POST['lastname'])."'"; $result = mysql_query($query, $conn); echo $rowcount; $zerorows=true; while ($row = mysql_fetch_assoc($result)) { $zerorows=false; echo '<b>City: </b>'.htmlspecialchars($row[city]).'<br />'; echo '<b>Last name: </b>'.htmlspecialchars($row[lastname]).'<br />'; echo '<b>Information: </b>'.htmlspecialchars($row[information]).'<br />'.'<br />'; } if($zerorows) echo "No results"; mysql_close($conn); ?>

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• What's the equivalent name of "procedure" in OOP?

  - by AeroCross

  In several of my programming courses in the University, my teachers always told me the following: A function and a procedure are basically the same thing: the only difference is that a function returns a value, and the procedure doesn't. That means that this: function sum($a, $b) { return $a + $b; } ... is a function, and this: function sum($a, $b) { echo $a + $b; } ... is a procedure. In the same train of thought, I've seen that a method is the equivalent of a function in the OOP world. That means that this: class Example { function sum($a, $b) { return $a + $b; } } Is a method — but how do you call this? class Example { function sum($a, $b) { echo $a + $b; } } What's the equivalent name, or how do you call a method that doesn't returns anything?

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• Form Not Submitting

  - by John

  Hello, When I try to click on the "submit" button for the form below, nothing happens. Any ideas why not? Thanks in advance, John submit.php: <?php require_once "header.php"; $u = $_SESSION['username']; if (!isLoggedIn()) { // user is not logged in. if (isset($_POST['cmdlogin'])) { // retrieve the username and password sent from login form & check the login. if (checkLogin($_POST['username'], $_POST['password'])) { show_userbox2(); } else { echo "Incorrect Login information !"; show_loginform(); } } else { show_loginform(); } } else { . show_userbox2(); } echo '<div class="submittitle">Submit an item.</div>'; echo '<form action="http://www...com/.../submit2.php" method="post"> <input type="hidden" value="'.$_SESSION['loginid'].'" name="uid"> <div class="submissiontitle"><label for="title">Story Title:</label></div> <div class="submissionfield"><input name="title" type="title" id="title" maxlength="1000"></div> <div class="urltitle"><label for="url">Link:</label></div> <div class="urlfield"><input name="url" type="url" id="url" maxlength="500"></div> <div class="submissionbutton"><input name="submit" type="submit" value="Submit"></div> </form> '; ?> submit2.php: <?php //if($_SERVER['REQUEST_METHOD'] == "POST"){header('Location: http://www...com/.../submit2.php');} require_once "header.php"; if (isLoggedIn() == true) { $remove_array = array('http://www.', 'http://', 'https://', 'https://www.', 'www.'); $cleanurl = str_replace($remove_array, "", $_POST['url']); $cleanurl = strtolower($cleanurl); $cleanurl = preg_replace('/\/$/','',$cleanurl); $title = $_POST['title']; //$url = $_POST['url']; $uid = $_POST['uid']; $title = mysql_real_escape_string($title); $cleanurl = mysql_real_escape_string($cleanurl); $site1 = 'http://' . $cleanurl; $displayurl = parse_url($site1, PHP_URL_HOST); function isURL($url1 = NULL) { if($url1==NULL) return false; $protocol = '(http://|https://)'; $allowed = '[-a-z0-9]{1,63}'; $regex = "^". $protocol . // must include the protocol '(' . $allowed . '\.)'. // 1 or several sub domains with a max of 63 chars '[a-z]' . '{2,6}'; // followed by a TLD if(eregi($regex, $url1)==true) return true; else return false; } if(isURL($site1)==true) mysql_query("INSERT INTO submission VALUES (NULL, '$uid', '$title', '$cleanurl', '$displayurl', NULL)"); else echo "<p class=\"topicu\">Not a valid URL.</p>\n"; } else { // user is not loggedin show_loginform(); } if (!isLoggedIn()) { // user is not logged in. if (isset($_POST['cmdlogin'])) { // retrieve the username and password sent from login form & check the login. if (checkLogin($_POST['username'], $_POST['password'])) { show_userbox(); } else { echo "Incorrect Login information !"; show_loginform(); } } else { // User is not logged in and has not pressed the login button // so we show him the loginform show_loginform(); } } else { // The user is already loggedin, so we show the userbox. show_userbox(); } require_once "footer.php"; ?>

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• Internet Connectivity Indicicator on Unity

  - by Sathish

  How can I check whether my internet connection is active on Ubuntu. If I am connected to a wired or wi-fi network, the indicator applet shows that I'm connected. But there is now way to find out the internet is working or not. I have some problem in my internet connectivity and I frequently lose my connection. I found this link is useful Internet connectivity indicator applet But I don't know that where should I use this code! #!/bin/bash if ping -c 1 -W 2 google.com > /dev/null; then echo "Up" else echo "Down" fi

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• How to pass results of bc to a variable

  - by shaolin

  I'm writing a script and I would like to pass the results from bc into a variable. I've declared 2 variables (var1 and var2) and have given them values. In my script I want to pass the results from bc into another variable say var3 so that I can work with var3 for other calculations. So far I have been able write the result to a file which is not what I'm looking for and also I've been able to echo the result in the terminal but I just want to pass the result to a variable at moment so that I can work with that variable. echo "scale=2;$var1/var2" | bc

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• XMLHttpRequest not working, trying to test database connection [closed]

  - by Frederick Marcoux

  I'm currently creating my own CMS for personnal use but I'm blocked at a code. I'm trying to make a installation script but the AJAX request to test if database works, doesn't work... There's my JS code: function testDB() { "use strict"; var host = document.getElementById('host').value; var username = document.getElementById('username').value; var password = document.getElementById('password').value; var db = document.getElementById('db_name').value; var xmlhttp = new XMLHttpRequest(); var url = "test_db.php"; var params = "host="+host+"&username="+username+"&password="+password+"&db="+db; xmlhttp.open("POST", url, true); xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); xmlhttp.setRequestHeader("Content-length", params.length); xmlhttp.setRequestHeader("Connection", "close"); xmlhttp.send(params); $('#loader').removeAttr('style'); if (xmlhttp.responseText !== '') { if (xmlhttp.readyState===4 && xmlhttp.status===200) { $('#next').removeAttr('disabled'); $('#test').attr('disabled', 'disabled'); $('#test').text('Connection Successful!'); $('#test').addClass('btn-success'); $('#login').addClass('success'); $('#login1').addClass('success'); $('#db').addClass('success'); $('#loader').attr('style', 'display: none;'); } else { $('#next').attr('disabled', 'disabled'); $('#test').removeClass('btn-success'); $('#test').removeAttr('disabled'); $('#test').text('Test Connection'); $('#login').removeClass('success'); $('#login1').removeClass('success'); $('#db').removeClass('success'); $('#loader').attr('style', 'display: none;'); } } else { $('#next').attr('disabled', 'disabled'); $('#next').attr('disabled', 'disabled'); $('#test').removeClass('btn-success'); $('#test').removeAttr('disabled'); $('#test').text('Test Connection'); $('#login').removeClass('success'); $('#login1').removeClass('success'); $('#db').removeClass('success'); $('#loader').attr('style', 'display: none;'); } } And there's my PHP code: <?php $link = mysql_connect($_POST['host'], $_POST['username'], $_POST['password']); if (!$link) { echo ''; } else { if (mysql_select_db($_POST['db'])) { echo 'Connection Successful!'; } else { echo ''; } } mysql_close($link); ?> I don't know why it doesn't work but I tried with JQuery $.ajax, $.get, $.post but nothing work...

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• Joomla 2.5 -- Adding a custom field to menu-item-edit-form

  - by philipp

  I would like to add a new Field (Select list of all menu-items) to the menu item-edit form. To do so I was setting up an system plugin with the following directory structure: languageroot/languageroot.php languageroot/form/form.xml As you can see in the posted code, that is all very basic to try out. Only after adding the following lines: <li <?php echo $this-form-getLabel( 'langroot-text', 'main' )? <?php echo $this-form-getInput('langroot-text', 'main' ); ? </li to: /admininstrator/components/com_menus/views/item/tmpl/edit.php a textfield shows up. Is it possible to inject the field without touching the edit.php? Is there anywhere a good tutorial about the JForm api? Is a system-plugin the right kind, or could it be a content plugin, or should it even be a component?

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• Sudo Non-Password access to /sys/power/state

  - by John

  On my computer, pm-hibernate appears to be broken, however using the command echo disk > /sys/power/state appears to work perfectly. Now I just need regular user access to it, using sudo. How do I do this? The command sudo echo disk > /sys/power/state simply returns bash: /sys/power/state: Permission denied. Also, I need this in a regularly used script, how can I make it so that I don't have to type in my password for it to work???

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• ERROR CHECKING !!

  - by moata_u

  am trying catch any error when run command in order to write an log file / report i was trying write this code : FUNCTION FOR VALIDATION function valid (){ if [ $? -eq 0 ]; then echo "$var1" ": status : OK" else echo "$var1" ": status : ERROR" fi COMMAND FUNCTION function save(){ sed -i "/:@/c connection.url=jdbc:oracle:thin:@$ip:1521:$dataBase" $search var1="adding database ip" valid $var1 sed -i "/connection.username/c connection.username=$name" #$search retval=$? var1="addning database SID" valid $var1 $retval } save OUTPUT adding database ip : status : OK sed: no input file i want out put in this way: adding database ip : status : OK sed: no input file : status : ERROR" (OR) adding database ip : status : OK addning database SID : status : ERROR" I was tried toooo much but not working with me :(((

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• Creating basic ACPI event makes the system unusably slow

  - by skerit

  I want to change a few settings on my laptop when I switch to battery power. I created a new event in /etc/acpi/events/cust-battery and it looks like this: event=battery action=/home/skerit/power.sh I put a simple command in the power.sh file: echo This is a test >> /home/skerit/powertest Now, when I tail this file it shows "This is a test" 4-5 times upon switching to battery power. However, the system becomes totally unstable. It slows down significantly. I can't change anything in the terminal. The terminal and certain parts of the screen (like the gnome system monitor applet) go blank from time to time. What can be the cause of that? It's a simple echo that gets executed a few times!

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• Macbook Pro late 2011 screen brightness issue

  - by buchzumlesen

  The keys to set the screen brightness work properly but everytime I reboot, the screen brightness is reverted to 100%, which is very annoying. I've already tried to add the following lines to /etc/rc.local but with no success (only the keyboard backlight stays off): #!/bin/sh -e # # rc.local # # This script is executed at the end of each multiuser runlevel. # Make sure that the script will "exit 0" on success or any other # value on error. # # In order to enable or disable this script just change the execution # bits. # # By default this script does nothing. echo '1' > /sys/devices/platform/applesmc.768/leds/smc::kbd_backlight/brightness echo '6' > /sys/class/backlight/acpi_video0/brightness rfkill block bluetooth exit 0` This worked for me when I was using Ubuntu 12.04 and also did after the upgrade to 12.10 but then after rebooting the screen brightness always reverted to 100%. Would be nice if anyone knows how to fix this. My device: Macbook Pro 13" Late 2011 Thanks in advance!

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• Why isn't garbage collection being activated in my code? [migrated]

  - by Netmoon

  I have a foreach statement in my code, where each iteration calculates huge amounts of data and goes to the next iteration. I run this code, but when I read the log, I see there's a memory leak error. PHP.net says when this happens, using gc_enabled() is a good way to handle this. I've added these lines to last line of the foreach block: echo "Check GC enabled : " . gc_enabled(); echo "Number of affected cycles : " . gc_collect_cycles(); And this is the output: Check GC enabled : 1 Number of affected cycles : 0 Why do cycles exist, but the affected cycles is 0?

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• how can I exit from a php script and continue right after the script?

  - by Samir Ghobril

  Hey guys, I have this piece of code, and when I add return after echo(if there is an error and I need to continue right after the script) I can't see the footer, do you know what the problem is? <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd"> <html lang="en" > <head> <title>Login | JM Today </title> <link href="Mainstyles.css" type="text/css" rel="stylesheet" /> </head> <body> <div class="container"> <?php include("header.php"); ?> <?php include("navbar.php"); ?> <?php include("cleanquery.php") ?> <div id="wrap"> <?php ini_set('display_errors', 'On'); error_reporting(E_ALL | E_STRICT); $conn=mysql_connect("localhost", "***", "***") or die(mysql_error()); mysql_select_db('jmtdy', $conn) or die(mysql_error()); if(isset($_POST['sublogin'])){ if(( strlen($_POST['user']) >0) && (strlen($_POST['pass']) >0)) { checklogin($_POST['user'], $_POST['pass']); } elseif((isset($_POST['user']) && empty($_POST['user'])) || (isset($_POST['pass']) && empty($_POST['pass']))){ echo '<p class="statusmsg">You didn\'t fill in the required fields.</p><br/><input type="button" value="Retry" onClick="location.href='."'login.php'\">"; return; } } else{ echo '<p class="statusmsg">You came here by mistake, didn\'t you?</p><br/><input type="button" value="Retry" onClick="location.href='."'login.php'\">"; return; } function checklogin($username, $password){ $username=mysql_real_escape_string($username); $password=mysql_real_escape_string($password); $result=mysql_query("select * from users where username = '$username'"); if($result != false){ $dbArray=mysql_fetch_array($result); $dbArray['password']=mysql_real_escape_string($dbArray['password']); $dbArray['username']=mysql_real_escape_string($dbArray['username']); if(($dbArray['password'] != $password ) || ($dbArray['username'] != $username)){ echo '<p class="statusmsg">The username or password you entered is incorrect. Please try again.</p><br/><input type="button" value="Retry" onClick="location.href='."'login.php'\">"; return; } $_SESSION['username']=$username; $_SESSION['password']=$password; if(isset($_POST['remember'])){ setcookie("jmuser",$_SESSION['username'],time()+60*60*24*356); setcookie("jmpass",$_SESSION['username'],time()+60*60*24*356); } } else{ echo'<p class="statusmsg"> The username or password you entered is incorrect. Please try again.</p><br/>input type="button" value="Retry" onClick="location.href='."'login.php'\">"; return; } } ?> </div> <br/> <br/> <?php include("footer.php") ?> </div> </body> </html>

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• change password code error.....

  - by shimaTun

  I've created a code to change a password. Now it seem contain an error.before i fill the form. the page display the error message: Parse error: parse error, unexpected $end in C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php on line 222 this the code: <?php # userChangePass.php //this page allows logged in user to change their password. $page_title='Change Your Password'; //if no first_name variable exists, redirect the user if(!isset($_SESSION['nameuser'])){ header("Location: http://" .$_SERVER['HTTP_HOST']. dirname($_SERVER['PHP_SELF'])."/index.php"); ob_end_clean(); exit(); }else{ if(isset($_POST['submit'])) {//handle form. require_once('connectioncomplaint.php'); //connec to the database //check for a new password and match againts the confirmed password. if(eregi ("^[[:alnum:]]{4,20}$", stripslashes(trim($_POST['password1'])))){ if($_POST['password1'] == $_POST['password2']){ $p =escape_data($_POST['password1']); }else{ $p=FALSE; echo'<p><font color="red" size="+1"> Your password did not match the confirmed password!</font></p>'; } }else{ $p=FALSE; echo'<p><font color="red" size="+1"> Please Enter a valid password!</font></p>'; } if($p){ //if everything OK. //make the query $query="UPDATE access SET password=PASSWORD('$p') WHERE userid={$_SESSION['userid']}"; $result=@mysql_query($query);//run the query. if(mysql_affected_rows() == 1) {//if it run ok. //send an email,if desired. echo '<p><b>your password has been changed.</b></p>'; //include('templates/footer.inc');//include the HTML footer. exit(); }else{//if it did not run ok $message= '<p>Your password could not be change due to a system error.We apolpgize for any inconvenience.</p><p>' .mysql_error() .'</p>'; } mysql_close();//close the database connection. }else{//failed the validation test. echo '<p><font color="red" size="+1"> Please try again.</font></p>'; } }//end of the main Submit conditional. ?> And code for form: <h1>Change Your Password</h1> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <fieldset> <p><b>New Password:</b><input type="password" name="password1" size="20" maxlength="20" /> <small>Use only letters and numbers.Must be between 4 and 20 characters long.</small></p> <p><b>Confirm New Password:</b><input type="password" name="password2" size="20" maxlength="20" /></p> </fieldset> <div align="center"> <input type="submit" name="submit" value="Change My Password" /></div> </form><!--End Form-->

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• PHP database selection issue

  - by Citroenfris

  I'm in a bit of a pickle with freshening up my PHP a bit, it's been about 3 years since I last coded in PHP. Any insights are welcomed! I'll give you as much information as I possibly can to resolve this error so here goes! Files config.php database.php news.php BLnews.php index.php Includes config.php - news.php database.php - news.php news.php - BLnews.php BLnews.php - index.php Now the problem with my current code is that the database connection is being made but my database refuses to be selected. The query I have should work but due to my database not getting selected it's kind of annoying to get any data exchange going! database.php <?php class Database { //------------------------------------------- // Connects to the database //------------------------------------------- function connect() { if (isset($dbhost) && isset($dbuser) && isset($dbpass)) { $con = mysql_connect($dbhost, $dbuser, $dbpass) or die("Could not connect: " . mysql_error()); } }// end function connect function selectDB() { if (isset($dbname) && isset($con)) { $selected_db = mysql_select_db($dbname, $con) or die("Could not select test DB"); } } } // end class Database ?> News.php <?php // include the config file and database class include 'config.php'; include 'database.php'; ... ?> BLnews.php <?php // include the news class include 'news.php'; // create an instance of the Database class and call it $db $db = new Database; $db -> connect(); $db->selectDB(); class BLnews { function getNews() { $sql = "SELECT * FROM news"; if (isset($sql)) { $result = mysql_query($sql) or die("Could not execute query. Reason: " .mysql_error()); } return $result; } ?> index.php <?php ... include 'includes/BLnews.php'; $blNews = new BLnews(); $news = $blNews->getNews(); ?> ... <?php while($row = mysql_fetch_array($news)) { echo '<div class="post">'; echo '<h2><a href="#"> ' . $row["title"] .'</a></h2>'; echo '<p class="post-info">Posted by <a href="#"> </a> | <span class="date"> Posted on <a href="#">' . $row["date"] . '</a></span></p>'; echo $row["content"]; echo '</div>'; } ?> Well this is pretty much everything that should get the information going however due to the mysql_error in $result = mysql_query($sql) or die("Could not execute query. Reason: " .mysql_error()); I can see the error and it says: Could not execute query. Reason: No database selected I honestly have no idea why it would not work and I've been fiddling with it for quite some time now. Help is most welcomed and I thank you in advance! Greets Lemon

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• How do you change brightness, color and sharpness from command line?

  - by YumYumYum

  I am controlling my PC with SSH and scripting. How can i change the brightness, color and sharpness from command line? Try 1: failed $ sudo redshift -t 5000:5000 -g .5 Cannot list GNOME panel applets. Initialization of gnome-clock failed. Trying next provider... Latitude and longitude must be set. Try 2: failed $ cat brightness 20 $ cat max_brightness 20 $ echo 1 | sudo tee /sys/class/backlight/acpi_video0/brightness 1 $ echo 20 | sudo tee /sys/class/backlight/acpi_video0/brightness Any alternative way to do? Follow up: http://jonls.dk/redshift/ [command] [1000K to 10000K] [effects 0.1 to 10.0] | | / / / ^ ^ ^ ^ ^ redshift -t 1000:1000 -l 0:0 -g .1; Dark redshift -t 1000:1000 -l 0.0 -g 5; Bright

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• Output a php multi-dimensional array to a html table

  - by Fireflight

  I have been banging my head against the wall with this one for nearly a week now, and am no closer than I was the first day. I have a form that has 8 columns and a variable number of rows which I need to email to the client in a nicely formatted email. The form submits the needed fields as a multidimensional array. Rough example is below: <input name="order[0][topdiameter]" type="text" id="topdiameter0" value="1" size="5" /> <input name="order[0][bottomdiameter]" type="text" id="bottomdiameter0" value="1" size="5" /> <input name="order[0][slantheight]" type="text" id="slantheight0" value="1" size="5" /> <select name="order[0][fittertype]" id="fittertype0"> <option value="harp">Harp</option> <option value="euro">Euro</option> <option value="bulbclip">Regular</option> </select> <input name="order[0][washerdrop]" type="text" id="washerdrop0" value="1" size="5" /> <select name="order[0][fabrictype]" id="fabrictype"> <option value="linen">Linen</option> <option value="pleated">Pleated</option> </select> <select name="order[0][colours]" id="colours0"> <option value="beige">Beige</option> <option value="white">White</option> <option value="eggshell">Eggshell</option> <option value="parchment">Parchment</option> </select> <input name="order[0][quantity]" type="text" id="quantity0" value="1" size="5" /> This form is formatted in a table, and rows can be added to it dynamically. What I've been unable to do is get a properly formatted table out of the array. This is what I'm using now (grabbed from the net). <?php if (isset($_POST["submit"])) { $arr= $_POST['order'] echo '<table>'; foreach($arr as $arrs) { echo '<tr>'; foreach($arrs as $item) { echo "<td>$item</td>"; } echo '</tr>'; } echo '</table>; }; ?> This works perfectly for a single row of data. If I try submitting 2 or more rows from the form then one of the columns disappears. I'd like the table to be formatted as: | top | Bottom | Slant | Fitter | Washer | Fabric | Colours | Quantity | ------------------------------------------------------------------------ |value| value | value | value | value | value | value | value | with additional rows as needed. But, I can't find any examples that will generate that type of table! It seems like this should be something fairly straightforward, but I just can't locate an example that works the way I need it too.

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• Grub won't boot windows after update from 11.10 to 12.04

  - by Holger

  thanks for your time and reading this, here's the deal: i upgraded from 11.10 to 12.04 and everything worked out until i rebooted, i had 11.10 sucessfully running as a dual boot with windows vista. when i rebooted, my GRUB was shot to hell, what ever option i selected it said partion not found or something similar... booting into a live version on a thumb drive and running bootrepair from there fixed the issue... but only for ubuntu, when i try to boot into windows it only goes back to GRUB. i'm not at home, and heres a list of what i have here with me... 1 4gb thumb drive, empty 1 8gb thumb drive, windows vista installer bootable 1 old laptop, the one i try to save, optical drive is not existent 2 Mbps internet connection can you help me get back into my windows without having to reinstall windows? or at least show me a way how to use my illustrator through a virtual machine or something? here's my grub cfg # # DO NOT EDIT THIS FILE # # It is automatically generated by grub-mkconfig using templates # from /etc/grub.d and settings from /etc/default/grub # ### BEGIN /etc/grub.d/00_header ### if [ -s $prefix/grubenv ]; then set have_grubenv=true load_env fi set default="0" if [ "${prev_saved_entry}" ]; then set saved_entry="${prev_saved_entry}" save_env saved_entry set prev_saved_entry= save_env prev_saved_entry set boot_once=true fi function savedefault { if [ -z "${boot_once}" ]; then saved_entry="${chosen}" save_env saved_entry fi } function recordfail { set recordfail=1 if [ -n "${have_grubenv}" ]; then if [ -z "${boot_once}" ]; then save_env recordfail; fi; fi } function load_video { insmod vbe insmod vga insmod video_bochs insmod video_cirrus } insmod part_msdos insmod ext2 set root='(hd0,msdos2)' search --no-floppy --fs-uuid --set=root 1063e402-b14f-45e5-92b6-d20a2e3a717e if loadfont /usr/share/grub/unicode.pf2 ; then set gfxmode=auto load_video insmod gfxterm insmod part_msdos insmod ext2 set root='(hd0,msdos2)' search --no-floppy --fs-uuid --set=root 1063e402-b14f-45e5-92b6-d20a2e3a717e set locale_dir=($root)/boot/grub/locale set lang=de_DE insmod gettext fi terminal_output gfxterm if [ "${recordfail}" = 1 ]; then set timeout=-1 else set timeout=10 fi ### END /etc/grub.d/00_header ### ### BEGIN /etc/grub.d/05_debian_theme ### set menu_color_normal=white/black set menu_color_highlight=black/light-gray if background_color 44,0,30; then clear fi ### END /etc/grub.d/05_debian_theme ### ### BEGIN /etc/grub.d/10_linux ### function gfxmode { set gfxpayload="${1}" if [ "${1}" = "keep" ]; then set vt_handoff=vt.handoff=7 else set vt_handoff= fi } if [ "${recordfail}" != 1 ]; then if [ -e ${prefix}/gfxblacklist.txt ]; then if hwmatch ${prefix}/gfxblacklist.txt 3; then if [ ${match} = 0 ]; then set linux_gfx_mode=keep else set linux_gfx_mode=text fi else set linux_gfx_mode=text fi else set linux_gfx_mode=keep fi else set linux_gfx_mode=text fi export linux_gfx_mode if [ "${linux_gfx_mode}" != "text" ]; then load_video; fi menuentry 'Ubuntu, mit Linux 3.2.0-24-generic' --class ubuntu --class gnu-linux --class gnu --class os { recordfail gfxmode $linux_gfx_mode insmod gzio insmod part_msdos insmod ext2 set root='(hd0,msdos2)' search --no-floppy --fs-uuid --set=root 1063e402-b14f-45e5-92b6-d20a2e3a717e linux /boot/vmlinuz-3.2.0-24-generic root=UUID=1063e402-b14f-45e5-92b6-d20a2e3a717e ro quiet splash $vt_handoff initrd /boot/initrd.img-3.2.0-24-generic } menuentry 'Ubuntu, mit Linux 3.2.0-24-generic (Wiederherstellungsmodus)' --class ubuntu --class gnu-linux --class gnu --class os { recordfail insmod gzio insmod part_msdos insmod ext2 set root='(hd0,msdos2)' search --no-floppy --fs-uuid --set=root 1063e402-b14f-45e5-92b6-d20a2e3a717e echo 'Linux 3.2.0-24-generic wird geladen …' linux /boot/vmlinuz-3.2.0-24-generic root=UUID=1063e402-b14f-45e5-92b6-d20a2e3a717e ro recovery nomodeset echo 'Initiale Ramdisk wird geladen …' initrd /boot/initrd.img-3.2.0-24-generic } submenu "Previous Linux versions" { menuentry 'Ubuntu, mit Linux 3.0.0-19-generic' --class ubuntu --class gnu-linux --class gnu --class os { recordfail gfxmode $linux_gfx_mode insmod gzio insmod part_msdos insmod ext2 set root='(hd0,msdos2)' search --no-floppy --fs-uuid --set=root 1063e402-b14f-45e5-92b6-d20a2e3a717e linux /boot/vmlinuz-3.0.0-19-generic root=UUID=1063e402-b14f-45e5-92b6-d20a2e3a717e ro quiet splash $vt_handoff initrd /boot/initrd.img-3.0.0-19-generic } menuentry 'Ubuntu, mit Linux 3.0.0-19-generic (Wiederherstellungsmodus)' --class ubuntu --class gnu-linux --class gnu --class os { recordfail insmod gzio insmod part_msdos insmod ext2 set root='(hd0,msdos2)' search --no-floppy --fs-uuid --set=root 1063e402-b14f-45e5-92b6-d20a2e3a717e echo 'Linux 3.0.0-19-generic wird geladen …' linux /boot/vmlinuz-3.0.0-19-generic root=UUID=1063e402-b14f-45e5-92b6-d20a2e3a717e ro recovery nomodeset echo 'Initiale Ramdisk wird geladen …' initrd /boot/initrd.img-3.0.0-19-generic } } ### END /etc/grub.d/10_linux ### ### BEGIN /etc/grub.d/20_linux_xen ### ### END /etc/grub.d/20_linux_xen ### ### BEGIN /etc/grub.d/20_memtest86+ ### menuentry "Memory test (memtest86+)" { insmod part_msdos insmod ext2 set root='(hd0,msdos2)' search --no-floppy --fs-uuid --set=root 1063e402-b14f-45e5-92b6-d20a2e3a717e linux16 /boot/memtest86+.bin } menuentry "Memory test (memtest86+, serial console 115200)" { insmod part_msdos insmod ext2 set root='(hd0,msdos2)' search --no-floppy --fs-uuid --set=root 1063e402-b14f-45e5-92b6-d20a2e3a717e linux16 /boot/memtest86+.bin console=ttyS0,115200n8 } ### END /etc/grub.d/20_memtest86+ ### ### BEGIN /etc/grub.d/30_os-prober ### menuentry "Windows Vista (loader) (on /dev/sda1)" --class windows --class os { insmod part_msdos insmod ntfs set root='(hd0,msdos1)' search --no-floppy --fs-uuid --set=root 2C9E66B39E6674EC chainloader +1 } ### END /etc/grub.d/30_os-prober ### ### BEGIN /etc/grub.d/40_custom ### # This file provides an easy way to add custom menu entries. Simply type the # menu entries you want to add after this comment. Be careful not to change # the 'exec tail' line above. ### END /etc/grub.d/40_custom ### ### BEGIN /etc/grub.d/41_custom ### if [ -f $prefix/custom.cfg ]; then source $prefix/custom.cfg; fi ### END /etc/grub.d/41_custom ###

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• Compare a variable that can have numeric or string as value

  - by Tarun

  I have a variable named Seconds_Behind_Master from one of my scripts. The problem is that this variable can either have a numeric value or can also take a string NULL as its value. Now, when I try to execute this script in shell it gets executed but gives a warning like this: [: Illegal number: NULL I believe it is due to the fact that in this case the value is NULL but when it compares it with numeral value 60 it gives this warning. How can I rectify it? Here is the piece of code: Seconds_Behind_Master=$Show_Slave_Status | grep "Seconds_Behind_Master" | awk -F": " {' print $2 '} if [ "$Seconds_Behind_Master" -ge "60" ]; then echo "replication delayed greater than or equal to 60." else if [ "$Seconds_Behind_Master" = "NULL" ]; then echo "Delay is Null." fi fi

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• OpenVPN IPv6 over IPv4 tunnel

  - by user66779

  Today I installed OpenVPN 2.3rc2 on both my windows 7 client machine and centos 6 server. This new version of OpenVPN provides full compatibility for IPv6. The Problem: I am currently able to connect to the server (through the IPv4 tunnel) and ping the IPv6 address which is assigned to my client and I can also ping the tun0 interface on the server. However, I cannot browse to any IPv6 websites. My vps provider has given me this: 2607:f840:0044:0022:0000:0000:0000:0000/64 is routed to this server (2607:f840:0:3f:0:0:0:eda). This is ifconfig after setup with OpenVPN running: eth0 Link encap:Ethernet HWaddr 00:16:3E:12:77:54 inet addr:208.111.39.160 Bcast:208.111.39.255 Mask:255.255.255.0 inet6 addr: 2607:f740:0:3f::eda/64 Scope:Global inet6 addr: fe80::216:3eff:fe12:7754/64 Scope:Link UP BROADCAST RUNNING MULTICAST MTU:1500 Metric:1 RX packets:2317253 errors:0 dropped:7263 overruns:0 frame:0 TX packets:1977414 errors:0 dropped:0 overruns:0 carrier:0 collisions:0 txqueuelen:1000 RX bytes:1696120096 (1.5 GiB) TX bytes:1735352992 (1.6 GiB) Interrupt:29 lo Link encap:Local Loopback inet addr:127.0.0.1 Mask:255.0.0.0 inet6 addr: ::1/128 Scope:Host UP LOOPBACK RUNNING MTU:16436 Metric:1 RX packets:0 errors:0 dropped:0 overruns:0 frame:0 TX packets:0 errors:0 dropped:0 overruns:0 carrier:0 collisions:0 txqueuelen:0 RX bytes:0 (0.0 b) TX bytes:0 (0.0 b) tun0 Link encap:UNSPEC HWaddr 00-00-00-00-00-00-00-00-00-00-00-00-00-00-00-00 inet addr:10.8.0.1 P-t-P:10.8.0.2 Mask:255.255.255.255 inet6 addr: 2607:f740:44:22::1/64 Scope:Global UP POINTOPOINT RUNNING NOARP MULTICAST MTU:1500 Metric:1 RX packets:739567 errors:0 dropped:0 overruns:0 frame:0 TX packets:1218240 errors:0 dropped:1542 overruns:0 carrier:0 collisions:0 txqueuelen:100 RX bytes:46512557 (44.3 MiB) TX bytes:1559930874 (1.4 GiB) So OpenVPN is sucessfully creating a tun0 interface and assigning clients IPv6 addresses using 2607:f840:44:22::/64. The first client to connect is getting 2607:f840:44:22::1000 and the second 2607:f840:44:22::1001, and so on... plus 1 each time. After connecting as the first client, I can ping from my windows client machine 2607:f740:44:22::1 and 2607:f740:44:22::1000. However, I have no access to IPv6 websites. I believe the problem is that the tun0 IPv6 addressees are not being forwarded to the eth0 interface. This is the firewall running on the server: #!/bin/sh # # iptables configuration script # # Flush all current rules from iptables # iptables -F iptables -t nat -F # # Allow SSH connections on tcp port 22 # iptables -A INPUT -i eth0 -p tcp --dport 22 -j ACCEPT iptables -A OUTPUT -o eth0 -p tcp --sport 22 -j ACCEPT # # Set access for localhost # iptables -A INPUT -i lo -j ACCEPT # # Accept connections on 1195 for vpn access from client # iptables -A INPUT -i eth0 -p udp --dport 1195 -m state --state NEW,ESTABLISHED -j ACCEPT iptables -A OUTPUT -o eth0 -p udp --sport 1195 -m state --state ESTABLISHED -j ACCEPT # # Apply forwarding for OpenVPN Tunneling # iptables -A FORWARD -m state --state RELATED,ESTABLISHED -j ACCEPT iptables -A FORWARD -s 10.8.0.0/24 -j ACCEPT iptables -t nat -A POSTROUTING -o eth0 -j SNAT --to 209.111.39.160 iptables -A FORWARD -j REJECT # # Enable forwarding # echo 1 > /proc/sys/net/ipv4/ip_forward # # Set default policies for INPUT, FORWARD and OUTPUT chains # iptables -P INPUT ACCEPT iptables -P FORWARD ACCEPT iptables -P OUTPUT ACCEPT # # IPv6 # IP6TABLES=/sbin/ip6tables $IP6TABLES -F INPUT $IP6TABLES -F FORWARD $IP6TABLES -F OUTPUT echo -n "1" >/proc/sys/net/ipv6/conf/all/forwarding echo -n "1" >/proc/sys/net/ipv6/conf/all/proxy_ndp echo -n "0" >/proc/sys/net/ipv6/conf/all/autoconf echo -n "0" >/proc/sys/net/ipv6/conf/all/accept_ra $IP6TABLES -A INPUT -i eth0 -m state --state ESTABLISHED,RELATED -j ACCEPT $IP6TABLES -A INPUT -i eth0 -p tcp --dport 22 -j ACCEPT $IP6TABLES -A INPUT -i eth0 -p icmpv6 -j ACCEPT $IP6TABLES -P INPUT ACCEPT $IP6TABLES -P FORWARD ACCEPT $IP6TABLES -P OUTPUT ACCEPT Server.conf: server-ipv6 2607:f840:44:22::/64 server 10.8.0.0 255.255.255.0 port 1195 proto udp dev tun ca ca.crt cert server.crt key server.key dh dh2048.pem ifconfig-pool-persist ipp.txt push "redirect-gateway def1 bypass-dhcp" push "dhcp-option DNS 208.67.222.222" push "dhcp-option DNS 208.67.220.220" keepalive 10 60 tls-auth ta.key 0 cipher AES-256-CBC comp-lzo user nobody group nobody persist-key persist-tun status openvpn-status.log log-append openvpn.log verb 5 Client.conf: client dev tun nobind keepalive 10 60 hand-window 15 remote 209.111.39.160 1195 udp persist-key persist-tun ca ca.crt key client1.key cert client1.crt remote-cert-tls server tls-auth ta.key 1 comp-lzo verb 3 cipher AES-256-CBC I'm not sure where I am going wrong, it could be the firewall, or something missing from server or client.conf. This version of OpenVPN was only released yesterday, and there's little info on the internet about how to setup an IPv6 over IPv4 vpn tunnel. I've read the manual for this new version of OpenVPN (parts pertaining to IPv6) and it provides very little info too. Thanks for any help.

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• Why ~/.bash_profile is not getting sourced when opening a terminal in Ubuntu 11.04?

  - by Viriato

  Problem I have an Ubuntu 11.04 Virtual Machine and I wanted to set up my Java development environment. I did as follows sudo apt-get install openjdk-6-jdk Added the following entries to ~/.bash_profile export JAVA_HOME=/usr/lib/jvm/java-6-openjdk export PATH=$PATH:$JAVA_HOME/bin Save the changes and exit Open up a terminal again and typed the following echo $JAVA_HOME (blank) echo $PATH (displayed, but not the JAVA_HOME value) Nothing happened, like if the export of JAVA_HOME and it's addition to the PATH were never done. Solution I had to go to ~/.bashrc and add the following entry towards the end of file #Source bash_profile to set JAVA_HOME and add it to the PATH because for some reason is not being picked up . ~/.bash_profile Questions Why did I have to do that? I thought bash_profile, bash_login or profile in absence of those two get executed first before bashrc. Was in this case my terminal a non-login shell? If so, why when doing su after the terminal and putting the password it did not execute profile where I had also set the exports mentioned above?

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