Posts by Shanti C - Page 10

Never fear shadows, they simply mean that there is a light some where near by.
-Ruth E Renkei.

yes, you can do this by php,ajax,mysql combination.
check this link: http://www.w3schools.com/ajax/ajax_aspphp.asp

try this:

<?php

if(isset($_SESSION['userlevel']) || !empty($_SESSION['userlevel']))
{
if ($_SESSION['userlevel'] == 2)
{
echo "ADMIN";
}
else
{
header('Location:index.php?query=notadmin');
}
}
else
{
echo "session not exists..";
}
?>

yes,
by this

session_cache_expire(30);

but check that page is ideal or not.

what do you want to do after selecting your drop down?

water

Your work is to discover your work and then with all your heart to give yourself to it.
- Buddha

check this link:
http://stackoverflow.com/questions/520237/how-do-i-expire-a-php-session-after-30-minutes

like this:

<select name="con_st" onChange="getData('current_page.php',this.value);" >

in javascript

function getData(page,x)
{
	window.location=page+"?dropdownid="+x;
}

in current_page.php
write your code and with the query string $_GET
like this:

if(isset($_GET['dropdownid']))
{
   $q="select * from your_tbl where your_field="+$_GET['dropdownid'];
// your code here
}

you can use some editors like : http://www.openwebware.com/

robo songs

feeling so much hungry,
going to have some great indian road made stuff....

Team work divides the task and multiplies the success.

today i worked on writing a class in java...

every time you refresh the page then the session will keep storing the values you entered. thats why you are getting duplicate values again and again.

what your jobs/list-jobs-detail.php contains???

>> and when you use GET method in ajax send, then i think we cannot get url query strings in jobs/list-jobs-detail.php page.
so, you have to change GET to POST.
or put one hidden variable for album id and use GET.

Sorry for my confusing post..

for this you have written your correctly..

and follow prieateas tip...

by this:

foreach($_SESSION as $directory){
for($i=0;$i<count($directory);$i++)
{
	$r_values=$directory[$i];
	foreach($r_values as $value)
	{
		echo $value;echo "<br>";
	}
}
}

or create alias names for common fields like :

SELECT p.thumb as photo_thumb,....... FROM photo_album as p, video as v ORDER BY photo_album.p DESC LIMIT 10

>> check $row getting value or not. If it is same field id in both tables then you have to get it by alias name.
and try to echo $row;

>> try with a static query string :

<a class="pics" href="javascript<b></b>:ajaxpage('jobs/list-jobs-detail.php&album=2', 'mainwindow');">

>> and when you use GET method in ajax send, then i think we cannot get url query strings in jobs/list-jobs-detail.php page.
so, you have to change GET to POST.
or put one hidden variable for album id and use GET.

yes,
you can do this by using sessions...

try this:

session_start();
if(!isset($_SESSION['words']))
{
   session_register();
   $_SESSION['words']=array();
}
$iterator=count($_SESSION['words']);
$_SESSION['words'][$iterator][] = wordCheck($lname, "lastname");
$_SESSION['words'][$iterator][] = wordCheck($addr, "address");
.
.
.
.
print_r($_SESSION['words']);

http://www.daniweb.com/forums/thread252011.html

http://www.wampserver.com/en/

Decide how many levels of categories and sub categories you need for your products.
Search(Advanced search) a product is very important , end user to see the product easily.
Need payment gateway to purchase online.
Showing your products images , information user friendly .
Reduce number of navigation pages while shopping and do payment.
maintain all invoices.

$c =($a-$b)*($a+$b);
echo $c;

print your new password entered and database password in your browser , then check once...

we are developing a web app, in that i need to upload and download files from client machine to ftp.
For these i found some scripts for ftp open source supports for a browser. But these are in .jar format(no source files). I can not fully customize these scripts.

put this in process.php

<?php
if($_SERVER['REQUEST_METHOD']== "POST")
{
  $a = $_POST['a']; // Get the value submitted for "a"
  $b = $_POST['b']; // Get the value submitted for "b"
  if ($a > $b)
  {
    echo "a is greater than b";
  }
  else
  {
    echo "a is greater than b";
  }
}
?>

check this link.

yes,

check your inserted id once..

//
Make sure query inserted user successfully
echo mysql_insert_id();
 if ( mysql_insert_id()!=0)
  {
  die("Error: User not added to database.");
  }
 else
  {
  // Redirect to thank you page.
  Header("Location: register.php?op=thanks");

i does't work with return
it still print Y_constructor_finished

if you want to stop the execution at return statement,
place break statement there..

<?php 
class X { 
  function X(){ 
    if($_POST['break']=='yes')
    {
      //return;
      break;
    } 
    } 
  } 

class Y extends X{ 
  function Y(){ 
    parent::__construct(); 
    print("Y_constructor_finished"); 
    } 
  } 

$_POST['break']=='yes'; 
$new_Y=new Y();  // will print "Y_constructor_finished"

<frameset rows="50%,50%">
<frame noresize="noresize" src="limits.php" />
<frame noresize="noresize" src="limits.php" />
</frameset>

this code is perfectly fine...

check your limits.php whether there is any error...
execute this page individually in your browser.

thanks

pass table name, fields as parameters to function declaration,
and write db query in your function definition which is in class.
try coding ...

Yes, pritaeas is right,

change this line:

$querychange = mysql_query("UPDATE users SET password='$newpassword' WHERE username='$user'");

use mkdir to make a directory.
Give user login name for this directory.
and you can store all uploaded images by this user in this directory by giving this directory path like:
assume login name is stored in session
$dirname=$_SESSION;
after creating directory..
save the uploaded images in this folder..

$dirname=$_SESSION['login_name'];
 if (file_exists("upload/" . $_FILES["file"]["name"]))
      {
      echo $_FILES["file"]["name"] . " already exists. ";
      }
    else
      {
      move_uploaded_file($_FILES["file"]["tmp_name"],
      "$dirname/" . $_FILES["file"]["name"]);
      echo "Stored in: " . "$dirname/" . $_FILES["file"]["name"];
      }
    }

what will be the logout page code

logout page contains destroy all your sessions (created on login page) and redirect to index/thankyou page...

check this tutorial..
and also this.

no body replies what happened?

post your check login query here...

I don't think there would be any way a client would be able to directly access the FTP server without providing the client authentication details.

I will provide ftp details internally in coding , getting from database based on login details...
But i need source files for these to customize the feature according to my application..

thanks...

or give me some ideas how to integrate ftp features in web browser????

what is $result contains?

check this link .

This is what we are implementing now..
But the problem is how to place all these files in client machine drives(C:/test/ or D:/test/) from ftp through web server...
Browser doesn't allow to do this...

Thanks for your prompt replies..

make active column as integer data type in your data base table structure, then you will be sure getting correct result.

check this thread

Thanks again S.O.S...

But we don't like to give ftp login details to the client. Thats why we planned for web application.

Please give me any other options are there...

S.O.S thanks for your reply.

Then how would i download and save all these files in client machine hardware memory.
Means we don't have access to client machine hard ware from a web application.
then how would it possible???

Hello all,

I am working on a project where i save client machine path into my database.
And if end user clicks download files button, then the files from ftp server have to be download in client machine directly. My application resides in web server.
Please give me some suggestions to complete this task.

Thanks in Advance.

Thanks all....

mark it solved..

Try this:

echo "Book Names:<br>";
foreach($names as $key=>$value)
{
echo $value."<br>";
}

try this and tel me what printed on browser after submit button clicked..