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  • Fancybox The requested content cannot be loaded. Please try again later

    - by user964836
    <link rel="stylesheet" href="<?php echo $this->getSkinUrl(''); ?>js/fancybox/source/jquery.fancybox.css?v=2.0.6" type="text/css" media="screen" /> <script type="text/javascript" src="<?php echo $this->getSkinUrl(''); ?>js/fancybox/source/jquery.fancybox.pack.js?v=2.0.6"></script> <script type="text/javascript"> $(document).ready(function() { $("a#image").fancybox(); $("a#image").trigger('click'); $("a#image").hide(); }); <a id="image" href="banner-about-cart.png"><img src="<?php echo $this->getSkinUrl() ?>images/banner-about-cart.png" alt=""/></a> any idea? the image loding i test whitout fancy plugin i can see image in site, i usind magento as well

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  • uploading different types of files mostly pdfs

    - by Anders Kitson
    I would like to upload different types of files pressumably pdfs to a certain directory I am currently trying to get this one script working that I found on snipplr but it is not working as I assumed it would, here is my code. <?php if( isset($_POST['submit']) ) { $target_path = "../downloads/"; $target_path = $target_path . basename( $_FILES['uploadedfile']['name']); if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) { echo "The file ". basename( $_FILES['uploadedfile']['name'])." has been uploaded"; } else{ echo "There was an error uploading the file, please try again!"; } } ?> <form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post" enctype="multipart/form-data"> <input type="file"> <input type="submit" name="submit" value="submit" /> </form>

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  • How to check whether a mysql select statement result has a value or not in php?

    - by udaya
    I have a textbox UserName and a Check Availability button next to it..... I have checked the availability by passing UserName textbox value..... But it doesn't seem to work.... Here is what i am doing? echo $UserName = $_GET['CheckUsername']; $_SESSION['state'] = $State; $queryres = "SELECT dUser_name FROM tbl_login WHERE dUser_name='$UserName'"; $result = mysql_query($queryres,$cn) or die("Selection Query Failed !!!"); if($result==true) // this condition doesn't seem to work { echo "User Name Available"; } else { echo "Sorry user name taken"; }

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  • Solve code issue - string identity testing

    - by kovibb
    Please for help with shell script. I really dont understand where is problem. Where I write to the Input variable for example /home/Kovi/Deskop/netlist.exe I still get true answer but the extension "exe" is not "scs". I also try "==" but it didnt working. Please some advice. Thank you very much. echo -n "Insert Entire Path (with file name and it's extension) of Input Netlist > " read Input if [ "${Input##*.}"="scs" ]; then echo "Patch verification is succesful.; else echo "Patch verification is failed, script was aborted. File dont exist or is empty or unreadable or is not spectre netlist." error_exit; fi

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  • if then endif , other smart solution in place of if then ....

    - by yael
    I have the following VB script How to write this VB script with case syntax? In order to perform professional writing in place if then…. yael Set fso = CreateObject("Scripting.FileSystemObject") If (fso.FileExists("C:\file1 ")) Then Verification=ok Else WScript.Echo("file1") Wscript.Quit(100) End If If (fso.FileExists("C:\file2 ")) Then Verification=ok Else WScript.Echo("file2") Wscript.Quit(100) End If If (fso.FileExists("C:\file3 ")) Then Verification=ok Else WScript.Echo("file3") Wscript.Quit(100) End If . . . .

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  • Function argument treated as undeclared

    - by Mikulas Dite
    I've prepared this simple example which is not working for me #include <stdio.h> #include <stdlib.h> FILE *fp; char filename[] = "damy.txt"; void echo (char[] text) { fp = fopen(filename, "a"); fwrite(text, 1, strlen(text), fp); fclose(fp); printf(text); } int main () { echo("foo bar"); return 0; } It's supposed to write both to command window and to file. However, this gives compilation error - the text used in echo() is not declared. Does c need another declaration of the variable?

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  • Getting Undefined index errors when trying to use $_POST

    - by Apoorv Agrawal
    I'm getting the following error messages: Notice: Undefined index: name in C:\xampp\htdocs\ApoorvProject\PostMethod.php on line 7 Notice: Undefined index: email in C:\xampp\htdocs\ApoorvProject\PostMethod.php on line 9 This is what line 7 and line 9 contains: echo $_POST["name"]; echo "<br>"; echo $_POST["email"]; This is the body of my HTML documnent: <form action="PostMethod.php" method="post"> Name: <input type="text" name="name" /> Email: <input type="text" name="email" /> <input type="submit" /> </form> PostMethod.php is the name of my PHP file.

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  • Creating a session user login php

    - by user2419393
    I'm stuck on how to create a session for a user who logs in. I got the part of checking to make sure the log in information corresponds with the database information, but is stuck on how to take the email address and store into a session. Here is my php code below. <?php include '../View/header.php'; session_start(); require('../model/database.php'); $email = $_POST['username']; $password = $_POST['password']; $sql = "SELECT emailAddress FROM customers WHERE emailAddress ='$email' AND password = '$password'"; $result = mysql_query($sql, $db); if (!$result) { echo "DB Error, could not query the database\n"; echo 'MySQL Error: ' . mysql_error(); exit; } while ($row = mysql_fetch_assoc($result)) { echo $row['emailAddress']; } mysql_free_result($result); ?>

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  • PHP - How to retrieve session in php

    - by Klaus Jasper
    I created a table that contains id - names - jobs and page that shows the names only and beside each name there is button Job and session that contains the id. this is my code $query = mysql_query("SELECT * FROM table"); while($fetch = mysql_fetch_array("$query")){ $name = $fetch['names']; $id = $fetch['id']; echo '</br>'; echo $name; $_SESSION['name'] = $id; echo "<button>Job</button>"; } I want when the user click on button Job redirect to a page that contains the job of that session. so how can I do it?

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  • IE display problem

    - by bah
    Hi, I have a page, and I just added pagination to it, but there's a prob in IE 7 & 8. When I add pagination code echo "<div class='pagination'>"; echo $pagination->GetPageLinks(); echo "</div>"; it breaks somehow layout in IE, I have looked all over it and I couldn't find what's wrong there. oh, and its css div.pagination { text-align : center; } div.pagination a, div.pagination span { padding : 5px; display : inline-block; } page url - adelija.puslapiai.lt, it's on index page, if you could take a look, that'd be great.

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  • Passing CMD Results to Variable in a Batch File

    - by TripleNad
    I am trying to install an application and a group of services using PSTools, but I want to take into account that the computer I am trying to connect to may be turned off or on a different network, which is not reachable from the internal network. Basically if the machine is not able to be accessed through the admin share, this is the message that I am getting: Couldn't access MachineName: The network path was not found. Make sure that the default admin$ share is enabled on MachineName. This is the syntax I am using to try to capture the "Error Message" and then report back that if installation was successful or not (depending on if it can contact the machine) @echo off set /p name=What is the machine name?: psexec.exe \\\%name% -u *useraccount* -p *password* \\\ServerName\installation.bat FOR /F "tokens=*" %%A IN ('COMMAND ^| FIND "Couldn't access"') DO SET Error=%%A If "%Error%"=="Couldn't access" ECHO Installation Failed. Else ECHO Installtion complete. Pause exit Currently it hangs right at the point it's defining the Error Variable. Can't quite figure out what I am going wrong here.

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  • check for several conditions when a user logs in

    - by paul
    I would like to accomplish the following: If a username or password field is null, notify the user. If user name already exists, do not insert into the database and notify user to create a different name. if the username is unique and password is not null, return the username to the user. As of now it always returns "Please enter a different user name." I believe the issue has to do with the database query but I am not sure. If anyone can have a look and see if I am making an error, I greatly appreciate it, thanks. if ($userName or $userPassword = null) { echo "Please enter a user name and password or return to the homepage."; } elseif (mysql_num_rows(mysql_query("SELECT count(userName) FROM logininfo WHERE userName = '$userName'")) ==1) { echo "Please enter a different user name."; } elseif ($userName and $userPassword != null) { echo "Your login name is: $userName"; }

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  • retrieved upload images in php

    - by hunter
    i want to retrieve image from client (ipod) programmed in objective c i use the following code $TARGET_PATH = "pics/"; $image = $_FILES['photo']; $TARGET_PATH =$TARGET_PATH . basename( $_FILES['photo']['name']); $TARGET_PATH =$TARGET_PATH.".jpg"; if(file_exists($TARGET_PATH)) { $TARGET_PATH =$TARGET_PATH .uniqid() . ".jpg"; } if (move_uploaded_file($image['tmp_name'], $TARGET_PATH)) { $TARGET_PATH="http://www.".$_SERVER["SERVER_NAME"]."/abc/".$TARGET_PATH; echo $TARGET_PATH; echo "image upload successfully";} else{ echo "could not upload image"; } this code upload five to six images successfully and after that it gives me error i.e Notice: Undefined index: photo in /home/abc/public_html/abc.com/fish/mycatch_post.php on line 42 Notice: Undefined index: photo in /home/abc/public_html/abc.com/fish/mycatch_post.php on line 53 could not upload image

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  • convert Bash command to php

    - by Keverw
    This is the bash command echo -n x && (echo 618cf954-6576-491d-8ac6-a1b888c4705d |xxd -r -p |openssl base64|tr '/+' '_-') This is my php <? $uuid = "618cf954-6576-491d-8ac6-a1b888c4705d"; $voiceid = "x" . $uuid; $voiceid = base64_encode($voiceid); $voiceid = str_replace("+", "-", $voiceid); $voiceid = str_replace("/", "_", $voiceid); echo $voiceid; ?> The bash gives the right output, the php one isn't. I'm not sure what i need to do deferent in php.

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  • How to display the found matches of preg_match function in PHP?

    - by Eric
    I am using the following to check if links exist on file.php: $fopen = fopen('file.php', 'r'); $fread = fread($fopen, filesize('file.php')); $pattern = "/^<a href=/i"; if (preg_match($pattern, $fread)) { echo 'Match Found'; } else { echo 'Match Not Found'; } if I echo preg_match($pattern, $fread) I get a boolean value, not the found matches. I tried what was on the php.net manual and did this: preg_match($pattern, $fread, $matches); then when I echoed $matches I got "Array" message. So I tried a foreach loop and when that didn't display anything I tried $matches[0] and that too outputted nothing. So how does one go about displaying the matches found?

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  • generate php classes in bash

    - by Derek
    i have this script: #!/bin/bash if [[ -z "$1" ]] ; then echo "Class is required" exit 1; fi if [[ -z "$2" ]] ; then package="Default" else package=$2; fi echo "<?php /** * $1.class.php * * Vcard class file. * @name Project * @author Author * @link http://www.domain.com * @copyright Copyright © 2011 * @package $package * @version 1.0 */ /** * The main $1 class * @package $package */ class $1 { /** * Constructor setup. */ public function __construct() { } /** * Destructor setup. */ public function __destruct() { } } " > $1.class.php php -l $1.class.php echo "Done"; if i do: ./generate.sh my_class it creates everything with my_class. how can i modify this to: MyClass? i need to use MyClass for the filename, and the class name etc... later in the code i use the argument (in this case my_class) for some other purposes. thanks

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  • I want to store a php $_SESSION variable and retrieve it on another page.

    - by CDeanMartin
    <?php session_start(); // store session data $_SESSION['count']=0; ?> <html><head></head> <body> <?php include ("getElement.php"); echo getLinkButton("myscript.php", "myscript.php"); echo $_SESSION['count']++; ?> </body> </html> The code above works, but when I click the link to navigate to myscript.php: <?php echo $_SESSION['count']; ?> I get this error: Undefined variable: _SESSION in /home/ubuntu/public_html/myscript.php on line 2

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  • bash check if value in list

    - by tkoomzaaskz
    I've got a script with a variable taken from command line parameters. I want to check if it's value is one of dev, beta or prod. I've got following code snippet: #!/usr/bin/env bash ENV_NAME=$1 echo "env name = $ENV_NAME" ENVIRONMENTS=('dev','beta','prod') if [[ $ENVIRONMENTS =~ $ENV_NAME ]]; then echo 'correct' exit else echo 'incorrect' exit fi When I run my script, it doesn't matter which parameters I pass: ./script.sh beta or ./script.sh or ./script.sh whatever, I always get correct echoed. What is wrong in my script?

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  • Need help... how to add md5 to password field in php?

    - by jones
    Hi mates, i looking some help and nice attention here.. i bought some php script many years ago and now no suport anymore... i just want to add md5 to password field.. here my form: <?php $SQL = "SELECT * from USERS WHERE USERNAME = '$_SESSION[username]'"; $result = @mysql_query( $SQL ); $row = @mysql_fetch_array( $result ); include 'menu.php'; ?> <FORM METHOD="post" ACTION="?page=query_client"> <INPUT TYPE="hidden" NAME="controller" VALUE="USERS~update~account_details&up=1~<?php echo $row[ID]; ?>"> <TABLE CLASS="basictable"> <TR> <TD CLASS="tdmenu" WIDTH="40%">Username</TD> <TD CLASS="tdmenu" WIDTH="60%"> <b><?php echo $row[USERNAME]; ?></b> </TD> </TR> <TR> <TD CLASS="tdmenu" WIDTH="40%">Password *</TD> <TD CLASS="tdmenu" WIDTH="60%"> <INPUT TYPE="PASSWORD" NAME="PASSWORD" SIZE="40" VALUE="<?php echo $row[PASSWORD]; ?>"> </TD> </TR> <TR> <TD CLASS="tdmenu" WIDTH="40%">Email Address *</TD> <TD CLASS="tdmenu" WIDTH="60%"> <INPUT TYPE="text" NAME="EMAIL" SIZE="40" VALUE="<?php echo $row[EMAIL]; ?>"> </TD> </TR> <TR> <TD CLASS="tdmenu" WIDTH="40%">Full Name *</TD> <TD CLASS="tdmenu" WIDTH="60%"> <INPUT TYPE="text" NAME="FULLNAME" SIZE="40" VALUE="<?php echo $row[FULLNAME]; ?>"> </TD> <TR> <TD CLASS="tdmenu" WIDTH="40%">Address *</TD> <TD CLASS="tdmenu" WIDTH="60%"> <INPUT TYPE="text" NAME="ADDRESS1" SIZE="40" VALUE="<?php echo $row[ADDRESS1]; ?>"> </TD> </TR> <BR> <TABLE CLASS="basictable"> <TR> <TD CLASS="tdhead2" > <DIV ALIGN="CENTER"><B> <INPUT TYPE="submit" NAME="Submit" VALUE="Submit"> </B></DIV> </TD> </TR> </TABLE> </FORM> and the it self as query_client.php inside look like: <?PHP @session_start(); $controller = $_POST['controller']; $pieces = explode("~", $controller); $table = $pieces[0]; $qt = $pieces[1]; $return = $pieces[2]; $id = $pieces[3]; $hack = $pieces[4]; if ($qt == insert) $qt = 'INSERT INTO'; if ($qt == update) { $qt = 'UPDATE'; $end = "WHERE ID = '$id'"; } $pre = array_keys( $_POST ); mysql_query ("CREATE TABLE IF NOT EXISTS `$table` (`ID` INT NOT NULL AUTO_INCREMENT , PRIMARY KEY ( `id` ) )"); $count = count($pre); $count = $count - 2; $sql = "$qt $table SET"; for ($i=0; $i < $count; $i++) { $x=$i+1; $y = $_POST[$pre[$x]]; $d = $y; mysql_query ("ALTER TABLE `$table` ADD `$pre[$x]` TEXT NOT NULL"); $sql .= " `$pre[$x]` = '$d',"; } $sql .= " ID = '$id' $end"; $query = mysql_query($sql) or die("$sql_error" . mysql_error()); if (empty($hack)) { } else { $pieces = explode("/", $hack); $h0 = $pieces[0]; $h1 = $pieces[1]; $h2 = $pieces[2]; $h3 = $pieces[3]; $h4 = $pieces[4]; $h5 = $pieces[5]; mysql_query ("ALTER TABLE `$table` $h0 $h1 $h2 $h3 $h4 $h5"); $query = mysql_query($sql) or die("$sql_error" . mysql_error()); } if (isset($_GET[inc])) include "$_GET[inc].php"; ?> so please help me how to add md5 in PASSWORD field? thanks in advance..

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  • Jump to next page with qt_writeobject (embeded quicktime)

    - by Trikks
    Hi Im trying to script a website that will play a video, and when the video is done it should redirect to another page. However the instructions on apples website wont match. Any tips? Here is some of the code. <script language="JavaScript" type="text/javascript"> QT_WriteOBJECT('<?php echo $play_me; ?>','<?php echo $moduleWidth ?>','<?php echo $moduleHeight ?>','', 'enablejavascript','true', 'controller','false', 'bgcolor','000000', 'autoplay','true', 'cache','true', 'scale','tofit', 'qtnext1','<www.example.com> T<myself>' ); //'qtnext1','GOTO0' </script> Ideal would be if the link could be created with javascript, ie parent.location='www.example.com'; Thanks

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  • Error during data INSERT in php

    - by nectar
    here my code- $sql = "INSERT INTO tblpin ('pinId', 'ownerId', 'usedby', 'status') VALUES "; for($i=0; $i0) { $sql .= ", "; } $sql .= "('$pin[$i]', '$ownerid', 'Free', '1')"; } $sql .= ";"; echo $sql; mysql_query($sql); if(mysql_affected_rows() 0) { echo "done"; } else { echo "Fail"; } output: ** INSERT INTO tblpin ('pinId', 'ownerId', 'usedby', 'status') VALUES ('13837927', 'admin', 'Free', '1'), ('59576082', 'admin', 'Free', '1'); Fail why it is not inserting values when $sql query is right?

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  • Select unseen random row from MySQL table

    - by user1476925
    We have a list of questions in a MySQL database and want it to show a random approved question to the user. When you click the Random button, we want another random question to be shown, but not any of the ones the user has already seen. Right now the script looks like this: <?php mysql_connect("localhost", "username", "password") or die(mysql_error()); mysql_select_db("aldrig") or die(mysql_error()); $result = mysql_query("SELECT * FROM spg WHERE approved='1' ORDER BY RAND() LIMIT 1;") or die(mysql_error()); while($row = mysql_fetch_array( $result )) { echo "<div class='contentTitle'><h1>"; echo $row['text']; echo "</h1></div>"; } ?>

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  • Simple php if query to double check

    - by skarama
    Is anything wrong with this code? <?php $variable = ; if (isset($variable)) { echo $variable ; echo "also this" ; } else echo "The variable is not set" ; ?> also, the other potential value of the variable is : $variable = <a href="http://www.mysite.com/article">This Article</a>; To clarify, I have a variable that may hold one of two possible values : an a href tag with it's url, or notihng at all. I need to have two different printouts for each of these cases, maybe I'm not doing it the right way though!

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  • Bash file descriptor leak

    - by Charles Duffy
    I get a file descriptor leak when running the following code: function get_fd_count() { local fds cd /proc/$$/fd; fds=( * ) # avoid a StackOverflow source colorizer bug echo "${#fds[@]}" } function fd_leak_func() { echo ">> Current FDs: $(get_fd_count)" read retval new_state < <(set +e; new_state=$(echo foo); retval=$?; printf "%d %s\n" $retval $new_state) } function parent_func() { while fd_leak_func; do :; done } parent_func Tested on both 3.2.25 and 4.0.28. Taking the while loop out of parent_func and running it at top level makes the problem go away. What's going on here? More to the point, are workarounds available?

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  • Unable to replace & with &amp; in XML uses Preg_replace in PHP

    - by Raind
    Hi, Need help here. Am not able to replace the '&' with '&' uses Preg_replace in PhP. But, if i do it manually (edited) on the xml file, it works out fine. Here are the sample: $XMLCharactersPreg = "/[&<\"\'()^*@+]/"; $XMLPregReplace = "&"; $d_Description = "50% offer & 20% further reduction for member"; if (preg_match($XMLCharactersPreg, $d_Description)) { echo "A match was found."; $XMLDealDescription = preg_replace($XMLCharactersPreg , $XMLPregReplace, $d_Description); echo "$XMLDealDescription "; } else { echo "A match was not found."; } Thanks.

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