
anujphp
Members-
Posts
30 -
Joined
-
Last visited
Never
Profile Information
-
Gender
Not Telling
anujphp's Achievements

Newbie (1/5)
0
Reputation
-
hello in displaying the pages it is not showing any error.it works fine but when I try to log in it is not working but when I use mozilla it works fine the log in system is working fine,I dont get any error message neither in php nor in javascrpt,Everything is working fine except the log in system But the same code is working fine in my system when I am running from my localhost even in IE,i am using IE 7
-
Sorry again... the website is working fine when i am visiting it using Mozilla but it is not working when i uses Internet explorer
-
hello MadTechie I am sorry i dont no your name so calling you by your username. The issue is resolved,it was a problem with a php settion,i have changed the php settion in the server and the website is working fine. I would like to thank you for your time and effort,Thanks a lot.and yes I have learnt few thing from your about session .From next time i will not use the old way of registering session infact i will change the code in my present site also.Thank a lot once again
-
http://jaika.co.nr/post.php is there is the server it will show you the website page with a request to log in.as one cannot view the page without log in. i have made the changes but the issue is still there.. when i use Internet explorer i am nto been able to log in but when I use mozilla i am able to log in but the session variable is not showing i,e login_in which i register while someone log in
-
i am sorry for creating confusion.. $result=executeQuery($sql); its my created function which simple execute a query the code is function executeQuery($sql) { $result = mysql_query($sql) or die("query failed: $sql"); return $result; } I have changed the name of the file.. postrecipe.php is not the original name.its post.php same way its userlogin.php and checkuserlogin.php $sql= "select * query"; this means the query which i execute.as the website is running fine in my system so i thought it wont be necessary to give the code. in my system i am having php 5.2 and in the webserver they are also using php 5.2 still as you have suggested i have changed the code in my system and its not working but my old code is working
-
This is the code where i am checking the user and registering the variable in session <?php session_start(); require "config.php"; $login_id = $HTTP_POST_VARS['loginid']; $password = $HTTP_POST_VARS['password']; $sql= "select * query"; $result=executeQuery($sql); if($line=mysql_fetch_array($result)) { //$msg= "Login Successful"; session_register("login_id"); //session_register('msg'); header("Location: postrecipe.php "); exit; } else { $msg= "Login Failed : Check Login and Password"; session_register('msg'); header("Location: login.php "); exit; } ?> Thank you for your time and effort
-
i have replaced my code with yours and now nothing is showing up after i enter username and password. you can have a look to the website yourself the address is http://jaika.co.nr thanks
-
Thank you MadTechie The simple code is working fine i took few more test and found that this function is not working session_register("msg");
-
hello everyone Please help me solving this issue my website is working fine when i run the website in vertrigo environment but when i have uploaded it in a webser www.ueuo.com then the session is not working and i am not been able to log in,the session is not working at all. here is the code of my logincheck.php file <?session_start(); if(!session_is_registered('loginid')) { $loginmsg="Please Login to view Members Area"; session_register("loginmsg"); header("Location: login.php "); exit; } ?> Thanks in advance
-
hello Priti I am a beginer of php and javascript,i dont no how to track whether script is creating xmlHttp object or not but it is definitely entering the loop where the code for xmlHttp object is created i,e the code segmentgiven below //Internet Explorer try { xmlHttp=new ActiveXObject("Msxml2.XMLHTTP"); } I wrote a alert statement below this line and i got the alert. i think it is creating the object,if it does not create then it will give me an error message as browser does not support http request as i am checking xmlHttp by the below code if (xmlHttp==null) { alert ("Browser does not support HTTP Request") return } hope i am thinking and doing things correct. regarding the error message i have not received any error message in the status bar,status bar is blank thanks
-
display image in a particular width and height
anujphp replied to anujphp's topic in PHP Coding Help
thanks for the link its a good one but i ont ned all those feature,i have written a code to resize a picture..but in my project i need to display a picture in different sizes.so i cant keep a picture in sifferent size -
Hello i am displaying images from a paticular folder,the code is working fine but i want to display all the image in same width and height...the code for displaying images is givn below <?php // define absolute path to image folder $image_folder = 'IMAGE FOLDER NAME'; // get the image name from the query string if (isset($_GET['pic']) && basename($_GET['pic']) == $_GET['pic']) { $pic = $image_folder.$_GET['pic']; if (file_exists($pic) && is_readable($pic)) { // get the filename extension $ext = substr($pic, -3); // set the MIME type switch ($ext) { case 'jpg': $mime = 'image/jpeg'; break; case 'gif': $mime = 'image/gif'; break; case 'png': $mime = 'image/png'; break; default: $mime = false; } // if a valid MIME type exists, display the image // by sending appropriate headers and streaming the file if ($mime) { header('Content-type: '.$mime); header('Content-length: '.filesize($pic)); $file = @ fopen($pic, 'rb'); if ($file) { fpassthru($file); exit; } } } } ?> I am calling the php pages as below FILENAME.php?pic=$imgname THANK YOU IN ADVANCE....